RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5

RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5

Other Exercises

Question 1.
In the figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. (C.B.S.E. 1991)
Solution:
In the figure,
∆ACB ~ ∆APQ
BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 1

Question 2.
In the figure, AB || QR. Find the length of PB. (C.B.S.E. 1994)
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 2
Solution:
In the figure,
In ∆PQR, AB || QR
AB = 3 cm, QR =9 cm, PR = 6 cm
In ∆PAB and ∆PQR
∠P = ∠P (common)
∠PAB = ∠PQR (corresponding angles)
∠PBA = ∠PRQ (corresponding angles)
∠PAB = ∠PQR (AAA axiom)
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 3

Question 3.
In the figure, XY || BC. Find the length of XY. (C.B.S.E. 1994C)
Solution:
In the figure
In ∆ABC XY || BC
AX = 1 cm, BC = 6 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 4
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 5

Question 4.
In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on hypotenuse is x. Prove that ab = cx.
Solution:
Given : In right ∆ABC, ∠B is right angle BD ⊥ AC
Now AB = a, BC = b, AC = c and BD = x
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 6
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 7

Question 5.
In the figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, find CD.
Solution:
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 8

Question 6.
In the figure, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.
Solution:
In right ∆ABC, ∠B = 90°
BD ⊥ AC
AB = 5.7 cm, BD = 3.8 and CD = 5.4 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 9
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 10

Question 7.
In the figure, DE || BC such that AE = \(\frac { 1 }{ 4 }\) AC. If AB = 6 cm, find AD.
Solution:
In the figure, in ∆ABC, DE || BC
AE = \(\frac { 1 }{ 4 }\) AC, AB = 6 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 11

Question 8.
In the figure, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, prove that ∆CED ~ ∆ABC.
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 12
Solution:
Given : In the figure AB ⊥ BC, DC ⊥ BC and DE ⊥ AC
To prove : ∆CED ~ ∆ABC.
Proof: AB ⊥ BC
∠B = 90°
and ∠A + ∠ACB = 90° ….(i)
DC ⊥ BC
∠DCB = 90°
=> ∠ACB + ∠DCA = 90° ….(ii)
From (i) and (ii)
∠A = ∠DCA
Now in ∆CED and ∆ABC,
∠E = ∠B (each 90°)
∠DEA or ∠DCE = ∠A (proved)
∆CED ~ ∆ABC (AA axiom)
Hence proved.

Question 9.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion
for two triangles, show that \(\frac { OA }{ OC }\) = \(\frac { OB }{ OD }\)
Solution:
Given : ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at O
To Prove : \(\frac { OA }{ OC }\) = \(\frac { OB }{ OD }\)
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 13

Question 10.
In ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively such that ∠MAP = ∠BAC. Prove that
(i) ∆ABC ~ ∆AMP
(ii) \(\frac { CA }{ PA }\) = \(\frac { BC }{ MP }\).
Solution:
Given : In ∆ABC and ∆AMP,
∠B = ∠M = 90°
∠MAP = ∠BAC
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 14
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 15

Question 11.
A vertical stick 10 cm long casts a shadow of 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower. (CB.S.E. 1991)
Solution:
The shadows are casted by a vertical stick and a tower at the same time
Their angles will be equal
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 16

Question 12.
In the figure, A = CED, prove that ∠CAB ~ ∠CED. Also find the value of x.
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 17
Solution:
Given : In ∆ABC,
∠CED = ∠A
AB = 9 cm, BE = 2 cm, EC = 10 cm, AD = 7 cm and DC = 8 cm
To prove :
(i) ∆CAB ~ ∆CED
(ii) Find the value of x
Proof: BC = BE + EC = 2 + 10 = 12 cm
AC = AD + DC = 7 + 8 = 15 cm
(i) Now in ∆CAB and ∆CED,
∠A = ∠CED (given)
∠C = ∠C (common)
∆CAB ~ ∆CED (AA axiom)
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 18

Question 13.
The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, what is the corresponding side of the other triangle? (C.B.S.E. 2002C)
Solution:
Let perimeter of ∆ABC = 25 cm
and perimeter of ∆DEF = 15 cm
and side BC of ∆ABC = 9 cm
Now we have to find the side EF of ∆DEF
∆ABC ~ ∆DEF (given)
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 19

Question 14.
In ∆ABC and ∆DEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL ⊥ BC and DM ⊥ EF, find AL : DM.
Solution:
In ∆ABC and ∆DEF,
AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm and FD = 8.4 cm
AL ⊥ BC and DM ⊥ EF
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 20
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 21

Question 15.
D and E are the points on the sides AB and AC respectively of a ∆ABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = \(\frac { 5 }{ 2 }\) DE.
Solution:
Given : In ∆ABC, points D and E are on the sides AB and AC respectively
and AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 22
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 23
Hence proved

Question 16.
D is the mid-point of side BC of a ∆ABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE : EX = 3 : 1.
Solution:
Given : In ∆ABC, D is mid point of BC, and E is mid point of AD
BE is joined and produced to meet AC at X
To prove : BE : EX = 3 : 1
Construction : From D, draw DY || BX meeting AC at Y
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 24
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 25
Hence proved.

Question 17.
ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained AB and BC.
Solution:
Given : ABCD is a parallelogram.
APQ is a straight line which meets BC at P and DC on producing at Q
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 26
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 27

Question 18.
In ∆ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that :
(i) ∆OMA ~ ∆OLC
(ii) \(\frac { OA }{ OC }\) = \(\frac { OM }{ OL }\)
Solution:
Given : In ∆ABC, AL ⊥ BC, CM ⊥ AB which intersect each other at O
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 28
(ii) \(\frac { OA }{ OC }\) = \(\frac { OM }{ OL }\)
Hence proved.

Question 19.
ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.
Solution:
Given : In quadrilateral ABCD, AD = BC
P, Q, R and S are the mid points of AB, AC, CD and AD respectively
PQ, QR, RS, SP are joined
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 29

Question 20.
In an isosceles ∆ABC, the base AB is produced both the ways to P and Q such that AP x BQ = AC². Prove that ∆APC ~ ∆BCQ.
Solution:
Given : In ∆ABC, AC = BC
Base AB is produced to both sides and points P and Q are taken in such a way that
AP x BQ = AC²
CP and CQ are joined
To prove : ∆APC ~ ∆BCQ
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 30

Question 21.
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above ground, find the length of her shadow after 4 seconds.
Solution:
Let AB be the lamp post and CD be the girl and AB = 3.6 m, CD = \(\frac { 90 }{ 100 }\) = 9 m
Distance covered in 4 seconds = 1.2 m x 4 = 4.8 m
BD = 4.8 m
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 31
x = 1.6
Length of her shadow = 1.6 m

Question 22.
A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Let AB be stick and DE be tower.
A stick 6 m long casts a shadow of 4 m i.e., AB = 6 m and BC = 4 m
Let DE casts shadow at the same time which is EF = 28 m
Let height of tower DE = x
Now in ∆ABC and ∆DEF,
∠B = ∠E (each 90°)
∠C = ∠F (shadows at the same time)
∆ABC ~ ∆DEF (AA criterion)
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 32

Question 23.
In the figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 33
Solution:
Given: In the figure, ∆ABC is a right angled triangle right angle at C.
DE ⊥ AB
To prove:
(i) ∆ABC ~ ∆ADE
(ii) Find the length of AE and DE
Proof: In ∆ABC and ∆ADE,
∠ACB = ∠AED (each 90°)
∠BAC = ∠DAE (common)
∆ABC ~ ∆ADE (AA axiom)
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 34
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 35

Question 24.
In the figure, PA, QB and RC are each perpendicular to AC. Prove that
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 36
Solution:
Given : In the figure, PA, QB and RC are perpendicular on AC and PA = x, QB = z and RC = y
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 37
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 38
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 39
Hence proved.

Question 25.
In the figure, we have AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 40
Solution:
In the figure, AB || CD || EF
AB = 6 cm, EF = 10 cm, BD = 4 cm, CD = x cm and DE = y cm
In ∆ABE, CE || AB
∆CED ~ ∆AEB
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5 41

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RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4

RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4

Other Exercises

Question 1.
(i) In figure, if AB || CD, find the value of x.
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4 1
(ii) In figure, if AB || CD, find the value of x.
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4 2
(iii) In figure, AB || CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, find x. (C.B.S.E. 2000C)
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4 3
Solution:
(i) In the figure, AB || CD
The diagonals of a trapezium divides each other proportionally
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4 4
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4 5
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4 6
RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4 7
=> (x – 8)(x – 11) = 0
Either x – 8 = 0, then x = 8 or x – 11 =0, then x = 11
x = 8 or 11

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Other Exercises

Mark the correct alternative in each of the following :
Question 1.
The distance between the points (cosθ, sinθ) and (sinθ, -cosθ) is
(a) √3
(b) √2
(c) 2
(d) 1
Solution:
(b) Distance between (cosθ, sinθ) and (sinθ, -cosθ)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 1
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 2

Question 2.
The distance between the points (a cos 25°, 0) and (0, a cos 65°) is
(a) a
(b) 2a
(c) 3a
(d) None of these
Solution:
(a) Distance between (a cos 25°, 0) and (0, a cos 65°)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 3

Question 3.
If x is a positive integer such that the distance between points P (x, 2) and Q (3, -6) is 10 units, then x =
(a) 3
(b) -3
(c) 9
(d) -9
Solution:
(c) Distance between P (x, 2) and Q (3, -6) = 10 units
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 4
=> x (x – 9) + 3 (x – 9) = 0
(x – 9) (x + 3) = 0
Either x – 9 = 0, then x = 9 or x + 3 = 0, then x = -3
x is positive integer
Hence x = 9

Question 4.
The distance between the points (a cosθ + b sinθ, 0) and (0, a sinθ – b cosθ) is
(a) a² + b²
(b) a + b
(c) a² – b²
(d) √(a²+b²)
Solution:
(d) Distance between (a cosθ + b sinθ, 0) and (0, a sinθ – b cosθ)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 5

Question 5.
If the distance between the points (4, p) and (1, 0) is 5, then p =
(a) ±4
(b) 4
(c) -4
(d) 0
Solution:
(a) Distance between (4, p) and (1, 0) = 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 6
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 7

Question 6.
A line segment is of length 10 units. If the coordinates of its one end are (2, -3) and the abscissa of the other end is 10, then its ordinate is
(a) 9, 6
(b) 3, -9
(c) -3, 9
(d) 9, -6
Solution:
(b) Let the ordinate of other end = y
then distance between (2, -3) and (10, y) = 10 units
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 8

Question 7.
The perimeter of the triangle formed by the points (0, 0), (1, 0) and (0, 1) is
(a) 1 ± √2
(b) √2 + 1
(c) 3
(d) 2 + √2
Solution:
(d) Let the vertices of ∆ABC be A (0, 0), B(1, 0) and C (0, 1)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 9
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 10

Question 8.
If A (2, 2), B (-4, -4) and C (5, -8) are the vertices of a triangle, then the length of the median through vertices C is
(a) √65
(b) √117
(c) √85
(d) √113
Solution:
(c) Let mid point of A (2, 2), B (-4, -4) be D whose coordinates will be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 11

Question 9.
If three points (0, 0), (3, √3) and (3, λ) form an equilateral triangle, then λ =
(a) 2
(b) -3
(c) -4
(d) None of these
Solution:
(d) Let the points (0, 0), (3, √3) and (3, λ) from an equilateral triangle
AB = BC = CA
=> AB² = BC² = CA²
Now, AB² = (x2 – x1)² + (y2 – y1
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 12

Question 10.
If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then k
(a) \(\frac { 1 }{ 3 }\)
(b) \(\frac { -1 }{ 3 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) \(\frac { -2 }{ 3 }\)
Solution:
(b) Let the points A (k, 2k), B (3k, 3k) and C (3, 1) be the vertices of a ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 13
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 14

Question 11.
The coordinates of the point on x-axis which are equidistant from the points (-3, 4) and (2, 5) are
(a) (20, 0)
(b) (-23, 0)
(c) (\(\frac { 4 }{ 5 }\) , 0)
(d) None of these
Solution:
(d)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 15

Question 12.
If (-1, 2), (2, -1) and (3, 1) are any three vertices of a parallelogram, then
(a) a = 2, b = 0
(b) a = -2, b = 0
(c) a = -2, b = 6
(d) a = 0, b = 4
Solution:
(d) In ||gm ABCD, diagonals AC and AD bisect each other at O
O is mid-point of AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 16

Question 13.
If A (5, 3), B (11, -5) and P (12, y) are the vertices of a right triangle right angled at P, then y =
(a) -2, 4
(b) -2, 4
(c) 2, -4
(d) 2, 4
Solution:
(c) A (5, 3), B (11, -5) and P (12, y) are the vertices of a right triangle, right angle at P
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 17
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 18

Question 14.
The area of the triangle formed by (a, b + c), (b, c + a) and (c, a + b) is
(a) a + b + c
(b) abc
(c) (a + b + c)²
(d) 0
Solution:
(d) Vertices of a triangle are (a, b + c), (b, c + a) and (c, a + b)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 19

Question 15.
If (x, 2), (-3, -4) and (7, -5) are coliinear, then x =
(a) 60
(b) 63
(c) -63
(d) -60
Solution:
(c) Area of triangle whose vertices are (x, 2), (-3, -4) and (7, -5)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 20
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 21

Question 16.
If points (t, 2t), (-2, 6) and (3, 1) are collinear, then t =
(a) \(\frac { 3 }{ 4 }\)
(b) \(\frac { 4 }{ 3 }\)
(c) \(\frac { 5 }{ 3 }\)
(d) \(\frac { 3 }{ 5 }\)
Solution:
(b) The area of triangle whose vertices are (t, 2t), (-2, 6) and (3, 1)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 22

Question 17.
If the area of the triangle formed by the points (x, 2x), (-2, 6) and (3, 1) is 5 square units, then x =
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 3 }{ 5 }\)
(c) 2
(d) 5
Solution:
(c) Area of triangle whose vertices are (x, 2x), (-2, 6) and (3, 1)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 23

Question 18.
If points (a, 0), (0, b) and (1, 1) are collinear, then \(\frac { 1 }{ a }\) + \(\frac { 1 }{ b }\) =
(a) 1
(b) 2
(c) 0
(d) -1
Solution:
(a) The area of triangle whose vertices are (a, 0), (0, b) and (1, 1)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 24
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 25

Question 19.
If the centroid of a triangle is (1, 4) and two of its vertices are (4, -3) and (-9, 7), then the area of the triangle is
(a) 183 sq. units
(b) \(\frac { 183 }{ 2 }\) sq. units
(c) 366 sq. units
(d) \(\frac { 183 }{ 4 }\) sq. units
Solution:
(b) Centroid of a triangle = (1, 4)
and two vertices of the triangle are (4, -3) and (-9, 7)
Let the third vertex be (x, y), then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 26
= \(\frac { 183 }{ 2 }\) sq. units

Question 20.
The line segment joining points (-3, -4) and (1, -2) is divided by y-axis in the ratio
(a) 1 : 3
(b) 2 : 3
(c) 3 : 1
(d) 2 : 3
Solution:
(c) The point lies on y-axis
Its abscissa will be zero
Let the point divides the line segment joining the points (-3, -4) and (1, -2) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 27

Question 21.
The ratio in which (4, 5) divides the join of (2, 3) and (7, 8) is
(a) -2 : 3
(b) -3 : 2
(c) 3 : 2
(d) 2 : 3
Solution:
(d) Let the point (4, 5) divides the line segment joining the points (2, 3) and (7, 8) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 28

Question 22.
The ratio in which the X-axis divides the segment joining (3, 6) and (12, -3) is
(a) 2 : 1
(b) 1 : 2
(c) -2 : 1
(d) 1 : -2
Solution:
(a) The point lies on x-axis
Its ordinate is zero
Let this point divides the line segment joining the points (3, 6) and (12, -3) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 29

Question 23.
If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then a3 + b3 + c3 =
(a) abc
(b) 0
(c) a + b + c
(d) 3 abc
Solution:
(d) Centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is origin (0, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 30

Question 24.
If points (1, 2), (-5, 6) and (a, -2) are collinear, then a =
(a) -3
(b) 7
(c) 2
(d) -2
Solution:
(b) The area of a triangle whose vertices are (1, 2), (-5, 6) and (a, -2)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 31
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 32

Question 25.
If the centroid of the triangle formed by (7, x), (y, -6) and (9, 10) is at (6, 3), then (x, y) =
(a) (4, 5)
(b) (5, 4)
(c) (-5, -2)
(d) (5, 2)
Solution:
(d) Centroid of (7, x), (y, -6) and (9, 10) is (6, 3)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 33

Question 26.
The distance of the point (4, 7) from the x-axis is
(a) 4
(b) 7
(c) 11
(d) √65
Solution:
(b) The distance of the point (4, 7) from x-axis = 7

Question 27.
The distance of the point (4, 7) from the y-axis is
(a) 4
(b) 7
(c) 11
(d) √65
Solution:
(a) The distance of the point (4, 7) from y-axis = 4

Question 28.
If P is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point Q on OY such that OP = OQ, are
(a) (0, 3)
(b) (3, 0)
(c) (0, 0)
(d) (0, -3)
Solution:
(a) P is a point on x-axis and its distance from 0 is 3
Co-ordinates of P will be (3, 0)
Q is a point on OY such that OP = OQ
Co-ordinates of Q will be (0, 3)

Question 29.
If the point (x, 4) lies on a circle whose centre is at the origin and radius is 5, then x =
(a) ±5
(b) ±3
(c) 0
(d) ±4
Solution:
(b) Point A (x, 4) is on a circle with centre O (0, 0) and radius = 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 34

Question 30.
If the point P (x, y) is equidistant from A (5, 1) and B (-1, 5), then
(a) 5x = y
(b) x = 5y
(c) 3x = 2y
(d) 2x = 3y
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 35

Question 31.
If points A (5, p), B (1, 5), C (2, 1) and D (6, 2) form a square ABCD, then p =
(a) 7
(b) 3
(c) 6
(d) 8
Solution:
(c) Vertices of a square are A (5, p), B (1, 5), C (2, 1) and D (6, 2)
The diagonals bisect each other at O
O is the mid-point of AC and BD
O is mid-point of BD, then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 36

Question 32.
The coordinates of the circumcentre of the triangle formed by the points O (0, 0), A (a, 0) and B (0, b) are
(a) (a, b)
(b) (\(\frac { a }{ 2 }\) , \(\frac { b }{ 2 }\))
(c) (\(\frac { b }{ 2 }\) , \(\frac { a }{ 2 }\))
(d) (b, a)
Solution:
(b) Let co-ordinates of C be (x, y) which is the centre of the circumcircle of ∆OAB
Radii of a circle are equal
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 37
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 38

Question 33.
The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (-3, 4) are
(a) (0, 2)
(b) (3, 0)
(b) (0, 3)
(d) (2, 0)
Solution:
(d) The given point P lies on x-axis
Let the co-ordinates of P be (x, 0)
The point P lies on the perpendicular bisector of of the line segment joining the points A (7, 6), B (-3, 4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 39

Question 34.
If the centroid of the triangle formed by the points (3, -5), (-7, 4), (10, -k) is at the point (k, -1), then k =
(a) 3
(b) 1
(c) 2
(d) 4
Solution:
(c) O (k, -1) is the centroid of triangle whose vertices are
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 40

Question 35.
If (-2, 1) is the centroid of the triangle having its vertices at (x, 0), (5, -2), (-8, y), then x, y satisfy the relation
(a) 3x + 8y = 0
(b) 3x – 8y = 0
(c) 8x + 3y = 0
(d) 8x = 3y
Solution:
(-2, 1) is the centroid of triangle whose vertices are (x, 0), (5, -2), (-8, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 41

Question 36.
The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are
(a) (3, 0)
(b) (0, 2)
(c) (-2, 3)
(d) (3, 2)
Solution:
(c) Three vertices of a rectangle are A (0, 0), B (2, 0), C (0, 3)
Let fourth vertex be D (x, y)
The diagonals of a rectangle bisect eachother at O
O is the mid-point of AC, then
Coordinates of O will be (\(\frac { 0+0 }{ 2 }\) , \(\frac { 0+3 }{ 2 }\))
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 42

Question 37.
The length of a line segment joining A (2, -3) and B is 10 units. If the abscissa of B is 10 units, then its ordinates can be
(a) 3 or -9
(b) -3 or 9
(c) 6 or 27
(d) -6 or-27
Solution:
(a) Abscissa of B is 10 and co-ordinates of A are (2, -3)
Let ordinates of B be y, then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 43

Question 38.
The ratio in which the line segment joining P(x1, y1) and Q (x2, y2) is divided by x-axis is
(a) y1 : y2
(b) -y1 : y2
(c) x1 : x2
(d) -x1 : x2
Solution:
(b) Let a point A on x-axis divides the line segment joining the points P (x1, y1), Q (x2, y2) in the ratio m1 : m2 and
let co-ordinates of A be (x, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 44

Question 39.
The ratio in which the line segment joining points A (a1, b1) and B (a2, b2) is divided by y-axis is
(a) -a1 : a2
(b) a1 : a2
(c) b1 : b2
(d) -b1 : b2
Solution:
(a) Let the point P on y-axis, divides the line segment joining the point A (a1, b1) and B (a2, b2) is the ratio m1 : m2 and
let the co-ordinates of P be (0, y), then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 45

Question 40.
If the line segment joining the points (3, -4) and (1, 2) is trisected at points P
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 46
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 47
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 48

Question 41.
If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are [CBSE 2012]
(a) (-6, 7)
(b) (6, -7)
(c) (6, 7)
(d) (-6, -7)
Solution:
(a) Let AB be the diameter of a circle with centre O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 49

Question 42.
The coordinates of the point P dividing the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1 are
(a) (2, 4)
(b) (3, 5)
(c) (4, 2)
(d) (5, 3) [CBSE 2012]
Solution:
(b) Point P divides the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1
Let coordinates of P be (x, y), then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 50

Question 43.
In the figure, the area of ∆ABC (in square units) is [CBSE 2013]
(a) 15
(b) 10
(c) 7.5
(d) 2.5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 51
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 52

Question 44.
The point on the x-axis which is equidistant from points (-1, 0) and (5, 0) is
(a) (0, 2)
(b) (2, 0)
(c) (3, 0)
(d) (0, 3) [CBSE 2013]
Solution:
(c) Let the point P (x, 0) is equidistant from the points A (-1, 0), B (5, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 53

Question 45.
If A (4, 9), B (2, 3) and C (6, 5) are the vertices of ∆ABC, then the length of median through C is
(a) 5 untis
(b) √10 units
(c) 25 units
(d) 10 units [CBSE 2014]
Solution:
(b) A (4, 9), B (2, 3) and C (6, 5) are the vertices of ∆ABC
Let median CD has been drawn C (6, 5)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 54

Question 46.
If P (2, 4), Q (0, 3), R (3, 6) and S (5, y) are the vertices of a paralelogram PQRS, then the value of y is
(a) 7
(b) 5
(c) -7
(d) -8 [CBSE 2014]
Solution:
(a) P (2, 4), Q (0, 3), R (3, 6) and S (5, y) are the vertices of a ||gm PQRS
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 55
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 56

Question 47.
If A (x, 2), B (-3, -4) and C (7, -5) are collinear, then the value of x is
(a) -63
(b) 63
(c) 60
(d) -60 [CBSE 2014]
Solution:
(a) A (x, 2), B (-3, -4) and C (7, -5) are collinear, then area ∆ABC = 0
Now area of ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 57

Question 48.
The perimeter of a triangle with vertices (0, 4) and (0, 0) and (3, 0) is
(a) 7 + √5
(b) 5
(c) 10
(d) 12 [CBSE 2014]
Solution:
(d) A (0, 4) and B (0, 0) and C (3, 0) are the vertices of ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 58

Question 49.
If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
(a) AP = \(\frac { 1 }{ 3 }\) AB
(b) AP = BP
(c) BP = \(\frac { 1 }{ 3 }\) AB
(d) AP = \(\frac { 1 }{ 2 }\) AB
Solution:
(d) Given that, the point P (2, 1) lies on the line segment joining the points (4, 2) and B (8, 4), which shows in the figure below:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 59
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 60

Question 50.
A line intersects the y-axis and x-axis at P and Q, respectively. If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(a) (0, -5) and (2, 0)
(b) (0, 10) and (-4, 0)
(c) (0, 4) and (-10, 0)
(d) (0, -10) and (4, 0)
Solution:
(d) Let the coordinates of P and Q (0, y) and (x, 0), respectively.
So, the mid-point of P (0, y) and Q (x, 0) is
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 61
2 = \(\frac { x + 0 }{ 2 }\) and -5 = \(\frac { y + 0 }{ 2 }\)
=> 4 = x and -10 = y
=> x = 4 and y = -10
So, the coordinates of P and Q are (0, -10) and (4, 0).

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS

Other Exercises

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
Write the distance between the points A (10 cosθ, 0) and B (0, 10 sinθ).
Solution:
Distance between the points A (10 cosθ, 0) and B (0, 10 sinθ)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 1

Question 2.
Write the perimeter of the triangle formed by the points O (0, 0), A (a, 0), and B (0, b).
Solution:
The vertices of a ∆OAB, O (0, 0), A (a, 0), and B (0, b)
Now length of OA
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 2

Question 3.
Write the ratio in which the line segment joining points (2, 3) and (3, -2) is divided by x-axis.
Solution:
The required point is on x-axis
Its ordinate will be 0
Let the point be (x, 0) and let this point divides the join of the points (2, 3) and (3, -2) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 3

Question 4.
What is the distance between the points (5 sin 60°, 0) and (0, 5 sin 30°) ?
Solution:
Distance between the given points
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 4

Question 5.
If A (-1, 3), B (1, -1) and C (5, 1) are the vertices of a triangle ABC, what is the length of the median through vertex A ?
Solution:
The vertices of ∆ABC are A (-1, 3), B (1, -1) and C (5, 1)
Let AD be the median
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 6

Question 6.
If the distance between points (x, 0) and (0, 3) is 5, what are the value of x ?
Solution:
Distance between (x, 0) and (0, 3) = 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 7

Question 7.
What is the area of the triangle formed by the points O (0, 0), A (6, 0) and B (0, 4) ?
Solution:
The vertices of the triangle OAB are O (0, 0), A (6, 0) and B (0, 4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 8
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 9

Question 8.
Write the coordinates of the point dividing line segment joining points (2, 3) and (3, 4) internally in the ratio 1 : 5.
Solution:
Let the coordinates of the required point be (x, y), then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 10

Question 9.
If the centroid of the triangle formed by points P (a, b), Q (b, c) and R (c, a) is at the origin, what is the value of a + b + c ?
Solution:
Vertices of ∆PQR are P (a, b), Q (b, c) and R (c, a) and its centroid = O (0, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 11

Question 10.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 12
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 13
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 14

Question 11.
Write the coordinates of a point on x- axis which is equidistant from the points (-3, 4) and (2, 5).
Solution:
The point is on x-axis
Its ordinates of the point P is (x, 0)
P is equidistant from A (-3, 4) and B (2, 5)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 15

Question 12.
If the mid-point of the segment joining A (x, y + 1) and B (x + 1, y + 2) is C (\(\frac { 3 }{ 2 }\) , \(\frac { 5 }{ 2 }\)) find x, y.
Solution:
C (\(\frac { 3 }{ 2 }\) , \(\frac { 5 }{ 2 }\)) is mid point of the line segment
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 16

Question 13.
Two vertices of a triangle have co-ordinates (-8, 7) and (9, 4). If the centroid of the triangle is at the origin, what are the co-ordinates of the third vertex ?
Solution:
Two vertices of a triangle are (-8, 7) and (9, 4)
Let the third vertex be (x, y)
Centroid of the triangle is (0, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 17

Question 14.
Write the coordinates the reflections of points (3, 5) in x and y-axes.
Solution:
Reflection of P (3, 5) in x-axis is will be (3, -5)
and reflection of P in y-axis will be (-3, 5)

Question 15.
If points Q and R reflections of point P (-3, 4) in X and Y axes respectively, what is QR ?
Solution:
Reflection of point P (-3, 4) in X-axis will be Q with coordinates Q (-3, -4) and reflection in Y-axis will be R (3, 4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 18

Question 16.
Write the formula for the area of the triangle having its vertices at (x1, y1), (x2, y2) and (x3, y3).
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 19

Question 17.
Write the condition of collinearity of points (x1, y1), (x2, y2) and (x3, y3).
Solution:
Three points (x1, y1), (x2, y2) and (x3, y3). are said to be collinear if the area of the triangle formed by these point = 0 i.e.,
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 20

Question 18.
Find the values of x for which the distance between the point P (2, -3) and Q (x, 5) is 10.
Solution:
Distance between P (2, -3) and Q (x, 5) = 10
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 21
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 22

Question 19.
Write the ratio in which the line segment joining the points A (3, -6) and B (5, 3) is divided by X-axis.
Solution:
The point lies on x-axis
Its ordinate will be = 0
Let the point P (x, 0) divides the line segment joining the points A (3, -6) and B (5, 3) in the ratio m : n.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 23

Question 20.
Find the distance between the points (\(\frac { -8 }{ 5 }\) , 2) and (\(\frac { 2 }{ 5 }\) , 2). (C.B.S.E. 2009)
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 24
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 25

Question 21.
Find the value of a so that the point (3, a) lies on the line represented by 2x – 3y + 5 = 0. (C.B.S.E. 2009)
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 26

Question 22.
What is the distance between the points A (c, 0) and B (0, – c) ? [CBSE 2010]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 27

Question 23.
If P (2, 6) is the mid-point of the line segment joining A (6, 5) and B (4, y), find y. [CBSE 2010]
Solution:
P (2, 6) is the mid-point of the line segment A (6, 5) and b (4, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 28

Question 24.
If the distance between the points (3, 0) and (0, y) is 5 units and y is positive, then what is the value of y ? [CBSE 2010]
Solution:
Distance between (3, 0) and (0, y) is 5 units
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 29

Question 25.
If P (x, 6) is the mid-point of the line segment joining A (6, 5) and B (4, y), find y. [CBSE 2010]
Solution:
P (x, 6) is the mid-point of the line segment joining the points A (6, 5), B (4, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 30

Question 26.
If P (2, p) is the mid-point of the line segment joining the points A (6, -5) and B (-2, 11), find the value of p. [CBSE 2010]
Solution:
P (2, p) is the mid-point of the line segment joining the points A (6, -5) and B (-2, 11)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 31

Question 27.
If A (1, 2), B (4, 3) and C (6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of fourth vertex D. [CBSE 2010]
Solution:
vertices of a parallelogram Let co-ordinates of D be (x, y)
Diagonals AC and BD bisect each other at O
Co-ordinates of O will be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 32
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 33

Question 28.
What is the distance between the points A (sinθ – cosθ, 0) and B (0, sinθ + cosθ)? [CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 34

Question 29.
What are the coordinates of the point where the perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis?
Solution:
Firstly, we plot the points of the line segment on the paper and join them.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 35
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 36
Now, we draw a straight line on paper passes through the mid-point P.
We see that the perpendicular bisector cuts the y-axis at the point (0, 13).
Hence, the required point is (0, 13).

Question 30.
Find the area of the triangle with vertices (a, b + c), (b, c + a) and (c, a + b).
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 37

Question 31.
If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then find a : b.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 38
=> 2a = b
Hence, the required relation is 2a = b

Question 32.
Find the coordinates of the point which is equidistant from the three vertices A (2x, 0), O (0, 0) and B (0, 2y) of ∆AOB.
Solution:
Let the coordinate of the point which is equidistant from the three vertices O (0, 0), A (0, 2y) and B (2x, 0) is P (h, k).
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 39
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 40
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 41

Question 33.
If the distance between the points (4, k), and (1, 0) is 5, then what can be the possible value of k? [CBSE 2017]
Solution:
Let the points x (4, k) and y (1, 0)
It is given that the distance xy is 5 units.
By using the distance formula,
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 42

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS

Other Exercises

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
What is the ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal?
Solution:
Diameter of a circle and side of an equilateral triangle are same
Let the diameter of the circle = a
Then radius (r) = \(\frac { a }{ 2 }\)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 1

Question 2.
If the circumference of two circles are in the ratio 2 : 3, what is the ratio of their areas ?
Solution:
Let R and r be the radii of two circles, then the ratio between their circumferences = 2πR : 2πr
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 2
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 3

Question 3.
Write the area of the sector of a circle whose radius is r and length of the arc is l.
Solution:
Let arc l subtends angle 9 at the centre of the circle
Now radius of a circle = r
and length of arc =l
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 4

Question 4.
What is the length (in terms of π) of the arc that subtends an angle of 36° at the centre of a circle of radius 5 cm?
Solution:
Radius of the circle = 5 cm
Angle at the center = 36°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 5

Question 5.
What is the angle subtended at the centre of a circle of radius 6 cm by an arc of length 3π cm ?
Solution:
Let the arc subtends angle θ at the centre of a circle
Radius of circle (r) = 6 cm
Length of arc = 3π cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 6

Question 6.
What is the area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm ?
Solution:
Radius of the circle (r) = 5 cm
Length of arc (l) = 3.5 cm
Let angle 9 be subtended by the arc at the centre
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 7

Question 7.
In a circle of radius 10 cm, an arc subtends an angle of 108° at the centre. What is the area of the sector in terms of π ?
Solution:
Radius of the circle = 10 cm
Angle at the centre = 108°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 8

Question 8.
If a square is inscribed in a circle, what is the ratio of the areas of the circle and the square ?
Solution:
A square ABCD is inscribed in a circle with centre O
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 9
Let the radius of the circle = r
Then its area = πr²
Now diagonal of the square = diameter of the circle = 2r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 10

Question 9.
Write the formula for the area of a sector of angle θ (in degrees) of a circle of radius r.
Solution:
Area of a sector of a circle whose radius = r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 11

Question 10.
 Write the formula for the area of a segment in a circle of radius r given that the sector angle is 0 (in degrees).
Solution:
Radius of the circle = r
and angle subtended by the sector at the centre = θ
Area of the segment
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 12

Question 11.
If the adjoining figure is a sector of a circle of radius 10.5 cm, what is the perimeter of the sector ? (Take π= 22/7)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 13
Solution:
Radius of the circle = 10.5 cm
Angle at the centre of the circle = 60°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 14

Question 12.
If the diameter of a semi-circular protractor is 14 cm then find its perimeter. (C.B.S.E. 2009)
Solution:
Diameter of semicircular protractor = 14 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 15
∴ Radius (r) =  \(\frac { 14 }{ 2 }\) = 7 cm
Now perimeter of protractor
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 16

Question 13.
An arc subtends an angle of 90° at the centre of the circle of radius 14 cm. Write the area of minor sector thus formed in terms of π.
Solution:
AB is an arc of the circle with centre O and radius 14 cm and subtends an angle of 90° at the centre O.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 17

Question 14.
Find the area of the largest triangle that can be inscribed in a semi-circle of radius r units. [CBSE 2015]
Solution:
Radius of semicircle = r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 18
In semicircle ΔABC is the largest triangle whose base is AC = 2 x r = 2r units
and height OB = r units
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 19

Question 15.
Find the area of a sector of circle of radius 21 cm and central angle 120°.
Solution:
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 20

Question 16.
What is the area of a square inscribed in a circle of diameter p cm?
Solution:
Diameter AC of the circle is p.
Also AC is diagonal of square ABCD.
Each angle of square is of 90°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 21

Question 17.
Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?
Solution:
False.
It is true only in the case of minor segment. But in case of major segment, area is always greater than the area of sector.

Question 18.
If the numerical value of the area of a circle is equal to the numerical value of its circumference, find its radius.
Solution:
∵ Numerical value of area of circle = Numerical value of circumference
∴  πr² = 2πr
or r = 2 units

Question 19.
How many revolutions a circular wheel of radius r metres makes in covering a distance of s metres?
Solution:
Radius of circular of wheel (r) = r m
Circumference of a circular wheel = 2πr
Distance to be covered = Sm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 22

Question 20.
Find the ratio of the area of the circle circumscribing a square to the area of the circle inscribed in the square.
Solution:
Let each side of of square = x
∴ Diameter of inner circle = x
Radius r = \(\frac { x }{ 2 }\)
Diameter of outer circle = AD
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 23
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 24

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5

Other Exercises

Question 1.
Find the area of a triangle whose vertices are :
(i) (6, 3), (-3, 5) and (4, -2)
(ii) (\({ at }_{ 1 }^{ 2 }\), 2at1), (\({ at }_{ 2 }^{ 2 }\), 2at2) and (\({ at }_{ 3 }^{ 2 }\), 2at3)
(iii) (a, c + a), (a, c) and (-a, c – a)
Solution:
(i) Co-ordinates of ∆ABC are A (6, 3), B (-3, 5) and C (4, -2)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 1
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 2
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 3

Question 2.
Find the area of the quadrilaterals, the coordinates of whose vertices are
(i) (-3, 2), (5, 4), (7, -6) and (-5, -4)
(ii) (1, 2), (6, 2), (5, 3) and (3, 4)
(iii) (-4, -2), (-3, -5), (3, -2), (2, 3) (C.B.S.E. 2009)
Solution:
(i) Let vertices of quadrilateral ABCD be A (-3, 2), B (5, 4), C (7, -6) and D (-5, -4)
Join AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 4
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 6
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 7
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 8
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 9

Question 3.
The four vertices of a quadrilaterals are (1, 2), (-5, 6), (7, -4) and (k, -2) taken in order. If the area of the quadrilateral is zero, find the value of k ?
Solution:
Let the vertices of quadrilateral ABCD be
A (1, 2), B (-5, 6), C (7, -4) and D (k, -2)
Join AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 10
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 11
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 12

Question 4.
The vertices of ∆ABC are (-2, 1), (5, 4) and (2, -3) respectively. Find the area of the triangle and the length of the altitude through A.
Solution:
Vertices of ∆ABC are A (-2, 1), B (5, 4) and C (2, -3) and AD ⊥ BC, let AD = h
Now area of ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 13
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 14

Question 5.
Show that the following sets of points are collinear
(a) (2, 5), (4, 6) and (8, 8)
(b) (1, -1), (2, 1) and (4, 5)
Solution:
We know that points are collinear if the area of the triangle formed by them is zero
(a) Vertices of ∆ABC are (2, 5), (4, 6) and (8, 8)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 15
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 16

Question 6.
Find the area of a quadrilateral ABCD, the coordinates of whose varities are A (-3, 2), B (5, 4), C (7, -6) and D (-5, -4). [CBSE 2016]
Solution:
Area of quadrilateral ABCD
= area of ∆ABC + area of ∆ACD
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 17

Question 7.
In ∆ABC, the coordinates of vertex A are (0, -1) and D (1, 0) and E (0, 1) respectively the mid-points of the sides AB and AC. If F is the mid-point of side C, find the area of ∆DEF. [CBSE 2016]
Solution:
Let B (p, q), C (r, s) and F (x, y)
Mid-point of AB = Coordinates of D
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 18
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 19
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 20

Question 8.
Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2). [CBSE 2015]
Solution:
In ∆PQR, L and N are mid points of QR and QP respectively coordinates of Q are (3, 2) of L are (2, -1) and of N are (1, 2)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 21
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 22

Question 9.
If P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2) are the vertices of a quadrilateral PQRS, find its area. [CBSE 2015]
Solution:
P (-5, -3), Q (-4, -6), R (2, -3) and S (1,2) are the vertices of a quadrilateral PQRS
Join PR which forms two triangles PQR and PSR
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 23
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 24

Question 10.
If A (-3, 5), B (-2, -7), C (1, -8) and D (6, 3) are the vertices of a quadrilateral ABCD, find its area. [CBSE 2014]
Solution:
A (-3, 5), B (-2, -7), C (1,-8) and D (6, 3) are the vertices of a quadrilateral ABCD
Join AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 25
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 26

Question 11.
For what value of ‘a’ the points (a, 1), (1, -1) and (11, 4) are collinear?
Solution:
Let the vertices of ∆ABC are A (a, 1), B (1, -1) and C (11, 4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 27

Question 12.
Prove that the points (a, b), (a1, b1) and (a – a1, b – b1) are collinear if ab1 = a1b.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 28

Question 13.
If the vertices of a triangle are (1, -3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value(s) of p. [CBSE 2012]
Solution:
The vertices of a triangle are (1, -3), (4, p) and (-9, 7) and area of triangle = 15 sq. units
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 29
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 30

Question 14.
If (x, y) be on the line joining the two points (1, -3) and (-4, 2), prove that x + y + 2 = 0.
Solution:
Point (x, y) be on the line joining the two points (1, -3) and (-4, 2)
Points (x, y), (1, -3) and (-4, 2) are collinear
Let the points (x, y) (1, -3) and (-4, 2) are the vertices of a triangle, then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 31

Question 15.
Find the value of k if points (k, 3), (6, -2) and (-3, 4) are collinear. [CBSE 2008]
Solution:
Let the points (k, 3), (6, -2) and (-3, 4) be the vertices of a triangle, then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 32
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 33

Question 16.
Find the value of k, if the points A (7, -2), B (5, 1) and C (3, 2k) are collinear. [CBSE 2010]
Solution:
Points A (7, -2), B (5, 1) and C (3, 2k) are collinear
area of ∆ABC = 0
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 34

Question 17.
If the point P (m, 3) lies on the line segment joining the points A (\(\frac { -2 }{ 5 }\) , 6) and B (2, 8), find the value of m.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 35

Question 18.
If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that x + y = a + b. [CBSE 2010]
Solution:
Point R (x, y) lies on the line segment joining the points P (a, b) and Q (b, a)
Area of ∆PRQ = 0
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 36
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 37

Question 19.
Find the value of k, if the points A (8, 1), B (3, -4) and C (2, k) are collinear. [CBSE 2010]
Solution:
The points A (8, 1), B (3, -4) and C (2, k) are collinear
Area of ∆ABC = 0
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 38

Question 20.
Find the value of a for which the area of the triangle formed by the points A (a, 2a), B (-2, 6) and C (3, 1) is 10 square units.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 39

Question 21.
If a ≠ b ≠ 0, prove that the points (a, a²), (b, b²), (0, 0) are never collinear. [CBSE 2017]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 40
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 41

Question 22.
The area of a triangle is 5 sq. units. Two of its vertices are at (2, 1) and (3, -2). If the third vertex is (\(\frac { 7 }{ 2 }\) , y), find y. [CBSE 2017]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 42

Question 23.
Prove that the points (a, 0), (0, b) and (1, 1) are collinear if, \(\frac { 1 }{ a }\) + \(\frac { 1 }{ b }\) = 1.
Solution:
Let the points are A (a, 0), B (0, b) and C (1, 1) which form a triangle
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 43

Question 24.
The point A divides the join of P (-5, 1) and Q (3, 5) in the ratio k : 1. Find the two values of k for which the area of ∆ABC where B is (1, 5) and C (7, -2) is equal to 2 units.
Solution:
Let the coordinates of A be (x, y) which divides the join of P (-5, 1) and Q (3, 5) in the ratio. Then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 44
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 45
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 46

Question 25.
The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x + 3. Find the third vertex.
Solution:
Let the coordinates of third vertex of the triangle be (x, y) and other two vertices are (2, 1) and (3, 2)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 47
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 48

Question 26.
If a ≠ b ≠ c, prove that the points (a, a²), (b, b²), (c, c²) can never be collinear.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 49
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 50

Question 27.
Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that \(\frac { \triangle DBC }{ \triangle ABC } =\frac { 1 }{ 2 }\) , find x?
Solution:
Let A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are the vertices of quadrilateral ABCD
AC and BD are joined
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 51
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 52
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 53

Question 28.
If three points (x1, y1), (x2, y2), (x3, y3) lie on the same line, prove that
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 54
Solution:
Let the points (x1, y1), (x2, y2), (x3, y3) are the vertices of a triangle
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 55

Question 29.
Find the area of a parallelogram ABCD if three of its vertices are A (2, 4), B (2 + √3, 5) and C (2, 6). [CBSE 2013]
Solution:
Three vertices of a ||gm ABCD are A (2, 4), B (2 + √3 , 5) and C (2, 6).
Draw one diagonal AC of ||gm ABCD
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 56
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 57

Question 30.
Find the value (s) of k for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, -k – 2) are collinear. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 58

Question 31.
If the points A (-1, -4), B (b, c) and C (5, -1) are collinear and 2b + c = 4, find the values of b and c. [CBSE 2014]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 59
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 60

Question 32.
If the points A (-2, 1), B (a, b) and C (4, -1) are collinear and a – b = 1, find the values of a and 6. [CBSE 2014]
Solution:
Points A (-2, 1), B (a, b) and C (4, -1) are
collinear if area ∆ABC = 0
Now area of ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 61
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 62

Question 33.
If the points A (1, -2), B (2, 3), C (a, 2) and D (-4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base. [NCERT Exemplar]
Solution:
In parallelogram, we know that, diagonals bisects each other
i.e., mid-point of AC = mid-point of BD
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 63
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 64
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 65
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 66
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 67

Question 34.
A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, find the area of ∆ADE. [NCERT Exemplar]
Solution:
Given that, A (6,1), B (8,2) and C (9,4) are three vertices of a parallelogram ABCD.
Let the fourth vertex of parallelogram be (x, y).
We know that, the diagonal of a parallelogram bisect each other.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 68
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 69

Question 35.
If D (\(\frac { -1 }{ 2 }\), \(\frac { 5 }{ 2 }\)) E (7, 3) and F (\(\frac { 7 }{ 2 }\), \(\frac { 7 }{ 2 }\)) are the mid-points of sides of ∆ABC, find the area of ∆ABC. [NCERT Exemplar]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 70
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 71
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 72

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4

Other Exercises

Question 1.
Find the centroid of the triangle whose vertices are :
(i) (1, 4), (-1, -1), (3, -2)
(ii) (-2, 3), (2, -1), (4, 0)
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 1

Question 2.
Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the Co-ordinates of the third vertex.
Solution:
Centroid of a triangle is O(0, 0) ….(i)
Co-ordinates of two vertices of a ∆ABC are A (1, 2) and B (3, 5)
Let the third vertex be (x, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 2

Question 3.
Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.
Solution:
Let two vertices of a ∆ABC be A (-3, 1) and B (0, -2) and third vertex C be (x, y)
Centroid of the ∆ABC is O (0, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 3

Question 4.
A (3, 2) and B (-2, 1) are two vertices of a triangle ABC whose centroid G has the coordinates (\(\frac { 5 }{ 3 }\) , \(\frac { -1 }{ 3 }\)) . Find the coordinates of the third vertex C of the triangle. [CBSE 2004]
Solution:
A (3, 2) and B (-2, 1) are the two vertices of ∆ABC whose centroid is G (\(\frac { 5 }{ 3 }\) , \(\frac { -1 }{ 3 }\))
Let third vertex C be (x, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 4

Question 5.
If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, find the co-ordinates of its centroid.
Solution:
In ∆ABC, D, E and F are the mid-points of the sides BC, CA and AB respectively.
The co-ordinates of D are (-2, 3), of E are (4,-3) and of F are (4, 5)
Let the co-ordinates of A, B and C be (x1, y1), (x2, y2), (x3, y3) respectively
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 6
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 7
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 8

Question 6.
Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.
Solution:
In ∆ABC,
D and E are the mid points of the sides AB and AC respectively
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 9
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 10

Question 7.
Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.
Solution:
Let A (x1, y1), B (x2, y2), C (x3, y3) and D (x4, y4) be the vertices of quadrilateral ABCD
E and F are the mid points of side BC and AD respectively and EF is joined G and H are the mid points of diagonal AC and BD.
GH are joined
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 11
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 12
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 13

Question 8.
If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA² + PB² + PC² = GA² + GB² + GC² + 3GP².
Solution:
In AABC, G is the centroid of it Let P (h, x) is any point in the plane
Let co-ordinates of A are (x1, y1) of B are (x2, y2) and of C are (x3, y3)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 14
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 15
Hence proved.

Question 9.
If G be the centroid of a triangle ABC, prove that AB² + BC² + CA² = 3 (GA² + GB² + GC²)
Solution:
Let the co-ordinates of the vertices of ∆ABC be A (x1, y1), B (x2, y2), C (x3, y3) and let G be the centroid of the triangle
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 16
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 17
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 18
Hence proved.

Question 10.
In the figure, a right triangle BOA is given. C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A and B.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 19
In right ∆OAB, co-ordinates of O are (0, 0) of A are (2a, 0) and of B are (0, 2b)
C is the mid-point of AB
Co-ordinates of C will be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 20
We see that CO = CA = CB
Hence C is equidistant from the vertices O, A and B.
Hence proved.

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4

Other Exercises

Question 1.
A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in the figure. If AB = 60 m and BC = 28 m, find the area of the plot.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 1
Solution:
Plot is formed of a rectangle ABCD and one semicircle on BC as diameter
∴  AB (l) = 60 m,
BC (b) = 28 m
∴ Radius of semicircle (r) = \(\frac { 1 }{ 2 }\) BC
= \(\frac { 1 }{ 2 }\) x 28 = 14m
Area of plot = area of rectangle ABCD + area of semicircle
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 2
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 3

Question 2.
A play ground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = 22/7).
Solution:
Length of rectangle (l) = 36 m 49
and width (b) = 24.5 =  \(\frac { 49 }{ 2 }\) m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 4
Question 3.
Find the area of the circle in which a square of area 64 cm2 is inscribed. (Use π = 3.14)
Solution:
Area of square = 64 cm²
∴  Side of square = \(\sqrt { Area } \)  = \(\sqrt { 64 } \)  = 8 cm
∵ The square in inscribed in the circle
∴  Radius of the circle will be = \(\frac { 1 }{ 2 }\) diagonal of
the square (r)=\(\frac { 1 }{ 2 }\) x \(\sqrt { 2 } \)a = \(\frac { 1 }{ 2 }\) x \(\sqrt { 2 } \) x 8 cm = 4\(\sqrt { 2 } \)
∴ Area of the circle = πr²
= 3.14 x (4\(\frac { 1 }{ 2 }\) x \(\sqrt { 2 } \) x 8 cm = 4\(\sqrt { 2 } \))² cm²
= 3.14 x 32 = 100.48 cm²

Question 4.
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.
Solution:
Length of rectangular piece (l) = 20 m
and width (b) = 15m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 5
Radius of each quadrant (r) = 3.5 m = \(\frac { 7 }{ 2 }\) m
Now area of the rectangle = l x b = 20 x 15 = 300 m²
and area of 4 quadrants = 4 x \(\frac { 1 }{ 4 }\) πr²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 6

Question 5.
In the figure, PQRS is a square of side 4 cm. Find the area of the shaded square.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 7
Solution:
Side of the square PQRS (a) = 4 cm
∴ Area of total square = a² = 4×4=16 cm²
Radius of each of the four quadrants at the corners = 1 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 8

Question 6.
Four cows are tethered at four corners of a square plot of side 50 m, so that they just cannot reach one another. What area will be left ungrazed ? (See figure)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 9
Solution:
side of square (a) = 50 m
∴ Area of the square field = a² = (50)² m² = 2500 m²
Radius of each quadrant (r) = \(\frac { 50 }{ 2 }\) = 25 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 10

Question 7.
A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20 m x 16 m, find the area of the field in which the cow can graze. [NCERT Exemplar]
Solution:
Let ABCD be a rectangular field of dimensions 20 m x 16 m.
Suppose, a cow is tied at a point A.  Let length of rope AE = 14m =r (say).
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 11

Question 8.
A calf is teid with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5 m, find the increase in area of the grassy lawn in which the calf can graze. [NCERT Exemplar]
Solution:
Let the calf be tied at the corner A of the square lawn.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 12
Then, the increase in area = Difference of the two sectors of central angle 90° each and radii 11.5 m (6 m + 5.5 m) and 6 m, which is the shaded region in the figure.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 13

Question 9.
A square water tank has its side equal to 40 m. There are four semi-circular grassy plots all round it. Find the cost of turfing the plot at ?1.25 per square metre (Take π = 3.14).
Solution:
Side of square tank (a) = 40 m
Radius of each semicircular grassy plots = \(\frac { 40 }{ 2 }\) = 20
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 14

Question 10.
A rectangular park is 100 m by 50 m. It is surrounded by semi-circular flower beds all round. Find the cost of levelling the semi-circular flower beds at 60 paise per square metre. (Use π = 3.14).
Solution:
Length of rectangular park (l) = 100 m
and width (b) = 50 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 15
Radius of each semicircular beds along the
lengths side (R) = \(\frac { 100 }{ 2 }\)  = 50 m
and radius of each semicircular beds along
the width side (r) = \(\frac { 50 }{ 2 }\) = 25 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 16

Question 11.
The inside perimeter of a running track (shown in the figure) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide, find the area of the track. Also find the length of the outer running track.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 17
Solution:
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 18
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 19

Question 12.
 Find the area of the figure, in square cm, correct to one place of decimal. (Take π = 22/7).

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 20
Solution:
Join AD
ABCD is a square whose each side = 10 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 21
Area of square = a² = (10)² = 100 cm²
Area of half semicircle whose radius is \(\frac { 10 }{ 2 }\) = 5
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 22
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 23

Question 13.
In the figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. (Use π = 22/7). [CBSE 2014]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 24
Solution:
AB = 20 cm, AE = 9 cm, DE = 12 cm
∠AED = 90°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 25
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 26

Question 14.
From each of the two opposite corners of a square of side 8 cm, a quadrant of a cirlce of radius 1.4 cm is cut. Another circle of diameter 4.2 cm is also cut from the centre as shown in the figure. Find the area of the remaining (shaded) portion of the square. (Use π = 22/7). [CBSE 2010]

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 27
Solution:
Side of a square ABCD = 8 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 28
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 29

Question 15.
In the figure, ABCD is a rectangle with AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semi¬circles are drawn as shown in the figure. Find the area of the shaded region.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 30
Solution:
In the figure, ABCD is a rectangle AB = 14 cm abd BC = 7 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 31
Two semicircles are draw on AD and BC as diameter and thid semicircle is drawn on Cd as diameter
Now area of rectangle ABCD = l x b = 14 x 7 = 98 cm²
Area of two semicircles on AD and BC
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 32

Question 16.
In the figure, ABCD is a rectangle, having AB = 20 cm and BC = 14 cm. Two sectors of 180° have been cut off. Calculate :
(i) the area of the shaded region.
(ii) the length of the boundary of the shaded region.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 33
Solution:
ABCD is a rectangle whose Length AB = 20 cm
and width BC = 14 cm
∴ Area of the rectangle = l x b = 20 x 14 = 280 cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 34

Question 17.
In the figure, the square ABCD is divided into five equal parts, all having same area. The central part is circular and the lines AE, GC, BF and HD lie along the diagonals AC and BD of the square. If AB = 22 cm, find :
(i) the circumference of the central part.
(ii) the perimeter of the part ABEF.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 35
Solution:
Side of the square ABCD = 22 cm
∴  Area = (side)² = (22)² = 484 cm²
∵ The squre is divided into 5 parts equal in area
∴ Area of each part  = \(\frac { 484 }{ 5 }\) cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 36

Question 18.
In the given figure, Find the area of the shaded region. (Use π = 3.14).  [CBSE 2015]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 37
Solution:
Side of large square = 14 cm
Radius of each semicircle = \(\frac { 4 }{ 2 }\) = 2 cm
Side of square = 4 cm
Area of square =  4 x 4=16 cm²
∴ Area of semicircles = 4 x \(\frac { 1 }{ 2 }\) πr²
= 2 x 3.14 x 2 x 2
= 8 x 3.14
= 25.12 cm²
∴ Area of shaded region = Area of large square – Area of central portion
= (14)2-(16+ 25.12) cm²
= 196-41.12 cm²
= 154.88 .cm²

Question 19.
In the Figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB
(ii) shaded region.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 38
Solution:
Radius of outer quadrant (R) = 3.5 cm
and radius of inner quadrant = 2 cm
∴  Area of shaded portion
= Area of outer quadrant – area of inner quadrant
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 39
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 40

Question 20.
In the figure, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 21 cm, find the area of the shaded region.[CBSE 2013]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 41
Solution:
In the figure, OPQ is a quadrant in which
OABC is a square OA = 21 cm
Join OB,
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 42
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 43

Question 21.
In the figure, OABC is a square of side 7 cm. If OAPC is a quadrant of a cirice with centre O, then find the area of the shaded region. (Use π = 22/7) [CBSE 2012]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 44
Solution:
In square OABC, whose side is 7 cm, OAPC
is a quadrant
Area of square = (side)²
= (7)² = 49 cm²
and radius of quadrant = 7 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 45

Question 22.
In the figure, OE = 20 cm. In sector OSFT, square OEFG is inscribed. Find the area of the shaded region.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 46
Solution:
In the figure OSFT is a quadrant and OEFG
is a square inscribed in it
The side of the square is OE = 20 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 47
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 48

Question 23.
Find the area of the shaded region in the figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. (Use π = 3.14)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 49
Solution:
In right AABC
AB²=AC² + BC² (Pythagoras Theorem)
= (24)²+ (10)²
= 576 + 100 = 676
= (26)²
∴ AB = 26 cm
∴ Diameter of circle = 26 cm
and radius (r)= \(\frac { 26 }{ 2 }\) = 13 cm
Now area of shaded portion
= Area of semicircle – area of right triangle ABC
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 50

Question 24.
A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (see figure). Find the radius of the inscribed circle and the area of the shaded part.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 51
Solution:
Each side of the equilateral triangle ABC (a) = 12 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 52

Question 25.
In the figure, an equilateral triangle ABC of side 6 cm has been inscribed in a circle. Find the area of the shaded region. (Take π = 3.14)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 53
Solution:
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 54

Question 26.
A circular field has a perimeter of 650 m. A square plot having its vertices on the circumference of the field is marked in the field. Calculate the area of the square plot.
Solution:
Perimeter of the circular field = 650 m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 55

Question 27.
Find the area of a shaded region in the figure, where a circular arc of radius 7 cm has been drawn with vertex A of an equilateral triangle ABC of side 14 cm as centre, (use π = 22/7 and \(\sqrt { 3 } \)  = 1.73) [CBSE 2015]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 56
Solution:
Radius of circular arc (r) = 7 cm
and side of equilateral AABC (a) = 14 cm and each angle = 60°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 57
Area of shaded region
=Area of circle + Area of equilateral triangle – 2 area of sector EAF
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 58

Question 28.
A regular hexagon is inscribed in a circle. If the area of hexagon is 24\(\sqrt { 3 } \) cm², find the area of the circle. (Use π = 3.14) [CBSE 2015]
Solution:
A regular hexagon ABCDEF is inscribed in a circle
Area of hexagon = 24 \(\sqrt { 3 } \) cm²
Let r be the radius of circle
∴ Side of regular hexagon = r
Area of equilateral ΔOAB = \(\frac { \sqrt { 3 } }{ 3 }\) r² sq. units
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 59

Question 29.
ABCDEF is a regular hexagon with centre O (see figure). If the area of triangle OAB is 9 cm², find the area of :
(i) the hexagon and
(ii) the circle in which the hexagon is inscribed.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 60
Solution:
O is the centre of the regular hexagon ABCDEF and area of the AOAB = 9 cm²
∵ By joining the vertices of the hexagon with O,
(i) We get 6 equal equilateral triangles
∴ Area of hexagon = 9 cm² x 6 = 54 cm²
(ii) Radius of the circle when this hexagon is inscribed in it will be = OB =AB  =r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 61

Question 30.
Four equal circles, each of radius 5 cm, touch each other as shown in the figure. Find the area included between them. (Take π = 3.14).
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 62
Solution:
Radius of each circle = 5 cm
∵ The four circles touch eachother externally
∴ By joining their centres, we get a square whose side will be 5 + 5 = 10 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 63
Now area of square so formed = a² = (10)² = 100 cm²
and area of 4 quadrants = 4 x \(\frac { 1 }{ 4 }\) πr²= πr²
= 3.14 x (5)2 cm² = 3.14 x 25 cm² = 78.5 cm²
∴  Area of the part included between the circles
= 100 – 78.5
= 21.5 cm²

Question 31.
Four equal circles, each of radius ‘a’ touch each other. Show that the area between them is \(\frac { 6 }{ 7 }\)a² . (Take π = \(\frac { 22 }{ 7 }\)
Solution:
Four circles each of radius ‘a’ touch each other at A, B, C and D respectively.
Their centes are P, Q, R and S respectively
By joining PQ, QR, RS, SD
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 64
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 65

Question 32.
A child makes a poster on a chart paper drawing a square ABCD of side 14 cm. She draws four circles with centre A, B, C and D in which she suggests different ways to save energy. The circles are drawn in such a way that each circle touches externally two of the three remaining circles in the given figure. In the shaded region she write a message ‘Save Energy’. Find the perimeter and area of the shaded region. (Use π = 22/7) [CBSE 2015]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 66
Solution:
Radius of circular arc (r) = 7 cm
Side of square ABCD (a) = 14 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 67
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 68

Question 33.
The diameter of a coin is 1 cm (see figure). If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region (Take π = 3.1416).

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 69
Solution:
Diameter of each coin = 1 cm
∴ Radius (r) = \(\frac { 1 }{ 2 }\) cm = 0.5 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 70
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 71

Question 34.
Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm x 7 cm. Find the area of the remaining card board. (Use π = 22/7) [CBSE 2013]
Solution:
Length of rectangle = 14 cm
and breadth = 7 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 72

Question 35.
In the figure, AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 73
Solution:
Radius of larger circle (R) = 7 cm
and radius of smaller circle (r) =\(\frac { 7 }{ 2 }\) cm
Area of shaded portion = Area of larger circle – area of smaller circle
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 74

Question 36.
In the figure, PSR, RTQ and PAQ are three semi-circles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter of the shaded region. [CBSE 2014]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 75
Solution:
Radius of larger semicircle (r1) =\(\frac { 10 }{ 2 }\) = 5 cm
Radius of large semicircle = (r2) = \(\frac { 7 }{ 2 }\) cm
Radius of small semicircle (r3) = \(\frac { 3 }{ 2 }\) cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 76

Question 37.
In the figure, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 77
Solution:
In the figure, two circles with centres A and B touch each-other internally at C
AC = 8 cm and AB = 3 cm
∴  BC = 8 – 3 = 5 cm
∴  Radius of bigger circle (R) = 8 cm
and of smaller circle (r) = 5 cm
∴  Area of shaded portion =Area of bigger circle – area of smaller circle
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 78

Question 38.
In the figure, ABCD is a square of side 2a. Find the ratio between
(i) the circumferences
(ii) the areas of the incircle and the circum- circle of the square.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 79
Solution:
A square ABCD is inscribed a circle
The side of the square = 2a
and one circle is inscribed in the square ABCD
Now diameter of the outer circle is AC
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 80
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 81

Question 39.
 In the figure, there are three semicircles, A, B and C having diameter 3 cm each, and another semicircle E having a circle D with diameter 4.5 cm are shown. Calculate :
(i) the area of the shaded region
(ii) the cost of painting the shaded region at the rate of 25 paise per cm², to the nearest rupee.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 82
Solution:
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 83
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 84
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 85

Question 40.
 In the figure, ABC is a right-angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semicircle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 86
Solution:
In right ΔABC,
AB = 28 cm, BC = 21 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 87
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 88

Question 41.
In the figure, O is the centre of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 89
Solution:
A semicircle is drawn on the diameter AB
ΔACB is drawn in this semicircle in right ΔACB
AC = 12 cm and BC = 16 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 90
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 91

Question 42.
In the figure, the boundary of the shaded region consists of four semi-circular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, find
(i) the length of the boundary,
(ii) the area of the shaded region.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 92
Solution:
Diameter of the biggest semicircle = 14 cm 14
∴  Radius (R) = \(\frac { 14 }{ 2 }\) = 7 cm
Diameter of the small semicircle = 7 cm 7
∴  Radius (r1) = \(\frac { 7 }{ 2 }\) cm
and diameter of each smaller circles = 3.5 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 93

Question 43.
In the figure, AB = 36 cm and M is mid¬point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all the three circles. Find the area of the shaded region.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 94
Solution:
In the figure, there are three semicircles one bigger and two smaller There is one small circle also
Now diameter of bigger semicircle = 36 cm
∴   Radius (R) = \(\frac { 36 }{ 2 }\) = 18 cm
and diameter of each smaller semicircle = 18 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 95
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 96

Question 44.
In the figure, ABC is a right angled triangle in which ∠A = 90°, AB = 21 cm and AC = 28 cm. Semi-circles are described on AB, BC and AC as diameters. Find the area of the shaded region.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 97
Solution:
In the figure, ABC is a right triangle in a semicircle A = 90°, AB = 21 cm and AC  = 28 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 98
Semicircles are drawn on BC and AC as diameters
In right ΔABC
BC² = AB²+AC² (Pythagoras Theorem)
= 21²+ 28²
= 441 + 784= 1225 = (35)²
∴  BC = 35 cm
Now radius of bigger semicircle (R) = \(\frac { 35 }{ 2 }\) cm,
of semicircle at AB = \(\frac { 21 }{ 2 }\) cm and of
semicircle on AC = \(\frac { 28 }{ 2 }\) cm = 14 cm
Now area of shaded portion
= Area of semicircle on AB as diameter + area of semicircle on AC as diameter + area of ΔABC – area of semicircle on BC as diameter
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 99

Question 45.
In the figure, shows the cross-section of railway tunnel. The radius OA of the circular part is 2 m. If ∠AOB = 90°, calculate :
(i) the height of the tunnel
(ii) the perimeter of the cross-section
(iii) the area of the cross-section.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 100
Solution:
Radius of the circular part of the tunnel = 2 m
i.e. OA = OB = 2 m
and ∠AOB = 90°
OD ⊥ AB
∴ D in mid-point of AB
∴ In right ΔAOB,
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 101
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 102

Question 46.
In the figure, shows a kite in which BCD is the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and ΔCEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 103
Solution:
ABCD is a square with side = 42 cm
BCD is a quadrant in which ∠BCD = 90° and radius = 42 cm
ΔCEF is an isosceles right triangle in which CE = CF = 6 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 104

Question 47.
In the figure, ABCD is a trapezium of area 24.5 cm2. In it, AD || BC, ∠DAB = 90°, AD = 10 cm and BC = 4 cm. If ABE is a quadrant of a circle, find the area of the shaded region. (Take π – (22/7). [ cbse2014]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 105
Solution:
In the figure, ABCD is a trapezium
Area = 24.5 cm2, AD || BC
∠DAB = 90°, AD = 10 cm BC = 4 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 106
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 107

Question 48.
In the figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and the distance between AB and DC is 14 cm. Circles of equal radii 7 cm with centres A, B, C and D have been drawn. Then, find the area of the shaded region of the figure. (Use π  = 22/7). [CBSE 2014]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 108
Solution:
In trapezium ABCD
AB || DC
AB = 18 cm, DC = 32 cm Height = 14 cm
Radius of each at the corner of trapezium = 7 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 109
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 110

Question 49.
From a thin metallic piece, in the shape of a trapezium ABCD, in which AB || CD and ∠BCD = 90°, a quarter circle BEFC is removed (see figure). Given AB = BC = 3.5 cm and DE = 2 cm, calculate the area of the remaining piece of the metal sheet.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 111
Solution:
ABCD is a trapezium in shape in which
AB || CD and ∠BCD = 90°
AB = BC = 3.5 cm, DE = 2 cm
∴ DC = DE + EC = DE + BC = 2 + 3.5 = 5.5 cm
Now area of trapezium ABCD = \(\frac { 1 }{ 2 }\) (AB + CD) x BC (∵ BC is height)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 112
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 113

Question 50.
In the figure, ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of radius 4 cm. Find the area of the shaded region correct upto 2 decimal places. (Take π = 3.142
and \(\sqrt { 3 } \)  = 1.732).
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 114
Solution:
In the figure, ABC is an equilateral triangle with 8 cm as side with centres A, B and C,
circular arcs drawn of radius 4 cm
Each side of ΔABC = 8 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 115
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 116

Question 51.
Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by three animals. (NCERT Exemplar)
Solution:
Given that, a triangular field with the three comers of the field a cow, a buffalo and a horse are tied separately with ropes. So, each animal grazed the field in each corner of triangular field as a sectorial form.
Given, radius of each sector (r) = 7m
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 117
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 118
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 119
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 120
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 121

Question 52.
In the given figure, the side of a square is 28 cm, and radius of each circle is half of the length of the side of the square where O and O’ are centres of the circles. Find the area of shaded region. [CBSE 2017]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 122
Solution:
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 123

Question 53.
In a hospital used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park to hospital whose length is 25 m and breadth is 20 m. If tank is filled completely then what will be the height of standing water used for irrigating the park? [CBSE 2017]
Solution:
Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylindrical tank, Vc = = πr²H = π X (1) 2 X 5= 5πm
Length of the park (L) = 25 m
Now water from the tank is used to irrigated the park. So, volume of cylindrical tank = Volume of water in the park.
⇒ 5π = 25 x 20 x h
⇒ 5π/25 x 20 = h
⇒ h = π/100 m
⇒ h = 0.0314 m
Through recycling of water, better use of the natural resource occurs without wastage. It helps in  reducing and preventing pollution.
It thus helps in conserving water. This keeps the greenery alive in urban areas like in parks gardens etc.

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3

Other Exercises

 

Question 1.
Find the coordinates of the point which divides the line segment joining (-1, 3) and (4, -7) internally in the ratio 3 : 4.
Solution:
The line segment joining the points A (-1,3) and B (4, -7) is divided into the ratio 3 : 4
Let P (x, y) divides AB in the ratio 3 : 4
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 1
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 2

Question 2.
Find the points of trisection of the line segment joining the points :
(i) (5, -6) and (-7, 5)
(ii) (3, -2) and (-3, -4)
(iii) (2, -2) and (-7, 4) [NCERT]
Solution:
(i) The line segment whose end points are A (5, -6) and B (-7,5) which is trisected at C and D
C divides it in the ratio 1 : 2
i.e., AC : CB = 1 : 2
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 3
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 4
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 6

Question 3.
Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points (-2, -1), (1, 0) (4,3) and (1, 2) meet.
Solution:
Let the vertices of the parallelogram ABCD be A (-2, -1), B (1, 0), C (4, 3) and D (1, 2) in which AC and BD are its diagonals which bisect each other at O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 7

Question 4.
Prove that the points (3, -2), (4, 0), (6, -3) and (5, -5) are the vertices of a parallelogram.
Solution:
Let the vertices of the quadrilateral ABCD be A (3, -2), B (4, 0), C (6, -3) and D (5, -5)
Now co-ordinates of the mid-point of AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 8
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 9

Question 5.
If P (9a – 2, -b) divides the line segment joining A (3a + 1, -3) and B (8a, 5) in the ratio 3 : 1, find the values of a and b. [NCERT Exemplar]
Solution:
Let P (9a – 2, -b) divides AB internally in the ratio 3 : 1.
By section formula,
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 10
=> 9a – 9 = 0
a = 1

Question 6.
If (a, b) is the mid-point of the line segment joining the points A (10, -6), B (k, 4) and a – 2b = 18, find the value of k and the distance AB. [NCERT Exemplar]
Solution:
Since, (a, b) is the mid-point of line segment AB.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 11
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 12
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 13

Question 7.
Find the ratio in which the points (2, y) divides the line segment joining the points A (-2, 2) and B (3, 7). Also, find the value of y. (C.B.S.E. 2009)
Solution:
Let the point P (2, y) divides the line segment joining the points A (-2, 2) and B (3, 7) in the ratio m1 : m2
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 14

Question 8.
If A (-1, 3), B (1, -1) and C (5, 1) are the vertices of a triangle ABC, find the length of the median through A.
Solution:
In ∆ABC, the vertices are A (-1, 3), B (1, -1) and C (5, 1)
D is the mid-point of BC
Co-ordinates of D will be (\(\frac { 1 + 5 }{ 2 }\) , \(\frac { -1 + 1 }{ 2 }\))
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 15

Question 9.
If the points P, Q (x, 7), R, S (6, y) in this order divide the line segment joining A (2, p) and B (7,10) in 5 equal parts, find x, y and p. [CBSE 2015]
Solution:
Points P, Q (x, 7), R, S (6, y) in order divides a line segment joining A (2, p) and B (7, 10) in 5 equal parts
i.e., AP = PQ = QR = RS = SB
Q is the mid point of A and S
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 16

Question 10.
If a vertex of a triangle be (1, 1) and the middle points of the sides through it be (-2, 3) and (5, 2), find the other vertices.
Solution:
Let co-ordinates of one vertex A are (1, 1) and mid-points of AB and AC are D (-2, 3) and E (5, 2)
Let the co-ordinates of B be (x1, y1) and C be (x2, y2)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 17
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 18

Question 11.
(i) In what ratio is the line segment joining the points (-2, -3) and (3, 7) divided by the y-axis ? Also find the co-ordinates of the point of division. [CBSE 2006C]
(ii) In what ratio is the line segment joining (-3, -1) and (-8, -9) divided at the point (-5, \(\frac { -21 }{ 5 }\)) ?
Solution:
(i) The point lies on y-axis
Its abscissa is O
Let the point (0, y) intersects the line joining the points (-2, -3) and (3, 7) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 19
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 20
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 21

Question 12.
If the mid-point of the line joining (3, 4) and (k, 7) is (x, y) and 2x + 2y + 1 = 0, find the value of k.
Solution:
Mid-point of the line joining the points (3, 4) and (k, 7) is (x, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 22

Question 13.
Find the ratio in which the points P (\(\frac { 3 }{ 4 }\) , \(\frac { 5 }{ 12 }\)) divides the line segments joining the points A (\(\frac { 1 }{ 2 }\) , \(\frac { 3 }{ 2 }\)) and B (2, -5). [CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 23
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 24

Question 14.
Find the ratio in which the line segment joining (-2, -3) and (5, 6) is divided by (i) x-axis (ii) y-axis. Also, find the co-ordinates of the point of division in each case.
Solution:
(i) Ordinate of a point on x-axis is zero
Let the co-ordinate of the point on x-axis be (0, x)
But (x, 0) is a point which divides the line segment joining the points (-2, -3) and (5, 6) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 25
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 26
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 27

Question 15.
Prove that the points (4, 5), (7, 6), (6, 3), (3, 2) are the vertices of a parallelogram. Is it a rectangle ?
Solution:
The vertices of a parallelogram ABCD are A (4, 5), B (7, 6), C (6, 3), and D (3, 2)
The diagonals AC and BD bisect each other at O
O is the mid-point of AC as well as of BD
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 28
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 29
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 30

Question 16.
Prove that (4, 3), (6, 4), (5, 6) and (3, 5) are the angular points of a square.
Solution:
Let A (4, 3), B (6, 4), C (5, 6) and D (3, 5) are the vertices of a square ABCD.
AC and BD are its diagonals which bisects each other at O.
O is the mid-point of AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 31
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 32
The diagonals of the quadrilateral ABCD are equal and bisect eachother at O and sides are equal
ABCD is a square

Question 17.
Prove that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices of a rectangle.
Solution:
Let the vertices of a quadrilateral ABCD are A (-4, -1), B (-2, -4), C (4, 0) and D (2, 3)
Join AC and BD which intersect eachother at O
If O is the mid-point of AC then its co
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 33
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 34

Question 18.
Find the lengths of the medians of a triangle whose vertices are A (-1, 3), B (1, -1) and C (5, 1).
Solution:
The co-ordinates of the vertices of ∆ABC are A (-1, 3), B (1, -1) and C (5, 1)
D, E and F are the mid-points of sides BC, CA and AB respectively
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 35
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 36

Question 19.
Find the ratio in which the line segment joining the pionts A (3, -3) and B (-2, 7) is divided by x-axis. Also, find the coordinates of the point of division. [CBSE 2014]
Solution:
Let a point P (x, 0)
x-axis divides the line segment joining the points A (3, -3) and B (-2, 7) in the ratio m1 : m2
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 37
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 38

Question 20.
Find the ratio in which the point P (x, 2) divides the line segment joining the points A (12, 5) and B (4, -3). Also, find the value of x. [CBSE 2014]
Solution:
Let P (x, 2) divides the line segment joining the points A (12, 5) and B (4, -3) in the ratio m1 : m2
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 39

Question 21.
Find the ratio in which the point P (-1, y) lying on the line segment joining A (-3, 10) and B (6, -8) divides it. Also find the value of y.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 40

Question 22.
Find the coordinates of a point A, where AB is a diameter of the circle whose centre is (2, -3) and B is (1, 4).
Solution:
AB is the diameter of the circle and O is the centre of the circle
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 41
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 42

Question 23.
If the points (-2, -1), (1, 0), (x, 3) and (1, y) form a parallelogram, find the values of x and y.
Solution:
In ||gm ABCD, co-ordinates of A (-2, -1), B (1, 0),C(x, 3) and D(1, y)
AC and BD are its diagonals which bisect eachother at O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 43
x = 4, y = 2

Question 24.
The points A (2, 0), B (9, 1), C (11, 6) and D (4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.
Solution:
The co-ordinates of vertices of a quadrilateral ABCD are A (2,0), B (9,1), C (11,6) and D (4, 4)
AC and BD are its diagonals which intersect eachother at O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 44
The co-ordinates of O in both cases are not same.
It is not a parallelogram and also not a rhombus.

Question 25.
In what ratio does the point (-4, 6) divide the line segment joining the points A (-6, 10) and B (3, -8) ?
Solution:
Let the point P (-4, 6) divides the line segment joining the points A (-6, 10) and B (3, -8) in the ratio m1 : m2
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 45

Question 26.
Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also, find the coordinates of the point of division.
Solution:
The points lies on y-axis
Let its coordinates be (0, y)
and let it divides the line segment joining the points (5, -6) and (-1, -4) in the ratio m1 : m2
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 46

Question 27.
Show that A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4) are the vertices of a rhombus.
Solution:
Vertices of a quadrilateral ABCD are A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4)
Join the diagonals AC and BD which intersect each other at O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 47
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 48
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 49

Question 28.
Find the lengths of the medians of a ∆ABC having vertices A (0, -1), B (2, 1) and C (0, 3).
Solution:
A (0, -1), B (2, 1) and C (0, 3) are the vertices of ∆ABC
Let D, E and F are the mid points of BC, CA and AB respectively
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 50
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 51

Question 29.
Find the lengths of the medians of a ∆ABC, having the vertices at A (5, 1), B (1, 5) and C (3,-1).
Solution:
A (5, 1), B (1, 5) and C (3, -1) are the vertices of ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 52
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 53

Question 30.
Find the co-ordinates of the points which divide the line segment joining the points (-4, 0) and (0, 6) in four equal parts.
Solution:
AB is a line segment whose ends points are A (-4, 0) and B (0, 6)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 54
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 55

Question 31.
Show that the mid-point of the line segment joining the points (5, 7) and (3, 9) is also the mid-point of the line segment joining the points (8, 6) and (0, 10).
Solution:
Let M be the mid point of AB. Co-ordinates of the mid point of this line segment joining two points A (5, 7) and B (3, 9)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 56

Question 32.
Find the distance of the point (1, 2) from the mid-point of the line segment joining the points (6, 8) and (2, 4).
Solution:
Let M be the mid-point of the line segment joining the points (6, 8) and (2, 4)
Now co-ordinates of M will be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 57
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 58

Question 33.
If A and B are (1, 4) and (5, 2) respectively, find the co-ordinates of P When \(\frac { AP }{ BP }\) = \(\frac { 3 }{ 4 }\)
Solution:
Point P divides the line segment joining the points (1, 4) and (5, 2) in the ratio of AP : PB = 3 : 4
Co-ordinates of P will be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 59

Question 34.
Show that the points A (1, 0), B (5, 3), C (2, 7) and D (-2, 4) are the vertices of a parallelogram.
Solution:
If ABCD is a parallelogram, then its diagonal
AC and BD bisect eachother at O
Let O is the mid-point of AC, then co
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 60
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 61

Question 35.
Determine the ratio in which the point P (m, 6) divides the join of A (-4, 3) and B (2, 8). Also find the value of m. [CBSE 2004]
Solution:
Let the ratio be r : s in which P (m, 6) divides the line segment joining the points A (-4, 3) and B (2, 8)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 62
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 63

Question 36.
Determine the ratio in which the point (-6, a) divides the join of A (-3, -1) and B (-8, 9). Also find the value of a. [CBSE 2004]
Solution:
Let the point P (-6, a) divides the join of A (-3, -1) and B (-8, 9) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 64

Question 37.
ABCD is a rectangle formed by joining the points A (-1, -1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle ? or a rhombus ? Justify your answer.
Solution:
ABCD is a rectangle whose vertices are A (-1,-1), B (-1,4), C (5, 4) and D (5, -1) P, Q, R, and S are the mid-points of the sides AB, BC, CD and DA respectively and are joined PR and QS are also joined.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 65
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 66
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 67
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 68

Question 38.
Points P, Q, R and S divide the line segment joining the pionts A (1, 2) and B (6, 7) in 5 equal parts. Find the coordinates of the points P, Q and R. [CBSE 2014]
Solution:
Points P, Q, R and S divides AB in 5 equal parts and let coordinates of P, Q, R and S be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 69
Coordinates of R are (4, 5)

Question 39.
If A and B are two points having co-ordinates (-2, -2) and (2, -4) respectively, find the co-ordinates of P such that AP = \(\frac { 3 }{ 7 }\) AB
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 70
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 71

Question 40.
Find the co-ordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts.
Solution:
Let P, Q and R divides the line segment AB in four equal parts
Co-ordinates of A are (-2, 2) and of B are (2, 8)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 72
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 73

Question 41.
Three consecutive vertices of a parallelogram are (-2, -1), (1, 0) and (4, 3). Find the fourth vertex.
Solution:
Let the co-ordinates of three vertices are A (-2, -1), B (1, 0) and C (4, 3)
and let the diagonals AC and BD bisect each other at O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 74
and \(\frac { y }{ 2 }\) = 1 => y = 2
Co-ordinates of D will be (1, 2)

Question 42.
The points (3, -4) and (-6, 2) are the extremities of a diagonal of a parallelogram. If the third vertex is (-1, -3). Find the co-ordinates of the fourth vertex.
Solution:
Let the extremities of a diagonal AC of a parallelogram ABCD are A (3, -4) and C (-6, 2)
Let AC and BD bisect eachother at O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 75
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 76

Question 43.
If the co-ordinates of the mid-points of the sides of a triangle are (1, 1), (2, -3) and (3, 4), find the vertices of the triangle.
Solution:
Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of the ∆ABC
D, E and F are the mid-points of BC, CA and AB respectively such that their co-ordinates are D (1, 1), E (2, -3) and F (3, 4)
D is mid-point of BC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 77
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 78
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 79

Question 44.
Determine the ratio in which the straight line x – y – 2 = 0 divides the line segment joining (3, -1) and (8, 9).
Solution:
Let the straight line x – y – 2 = 0 divides the line segment joining the points (3, -1), (8, 9) in the ratio m : n
Co-ordinates of the point will be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 80

Question 45.
Three vertices of a parallelogram are (a + b, a – b), (2 a + b, 2a – b), (a – b, a + b). Find the fourth vertex.
Solution:
In parallelogram ABCD co-ordinates are of A (a + b, a – b), B (2a + b, 2a – b), C (a – b, a + b)
Let co-ordinates of D be (x, y)
Join diagonal AC and BD
Which bisect eachother at O
O is the mid-point of AC as well as BD
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 81
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 82

Question 46.
If two vertices of a parallelogram are (3, 2), (-1, 0) and the diagonals cut at (2, -5), find the other vertices of the parallelogram.
Solution:
Two vertices of a parallelogram ABCD are A (3,2), and B (-1, 0) and its diagonals bisect each other at O (2, -5)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 83
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 84

Question 47.
If the co-ordinates of the mid-points of the sides of a triangle ar6 (3, 4), (4, 6) and (5, 7), find its vertices. [CBSE 2008]
Solution:
The co-ordinates of the mid-points of the sides BC, CA and AB are D (3, 4), E (4, 6) and F (5, 7) of the ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 85
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 86
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 87

Question 48.
The line segment joining the points P (3, 3) and Q (6, -6) is trisected at the points A and B such that A is nearer to P. If A also lies on the line given by 2x + y + k = 0, find the value of k. [CBSE 2009]
Solution:
Two points A and B trisect the line segment joining the points P (3, 3) and Q (6, -6) and A is nearer to P
and A lies also on the line 2x + y + k = 0
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 88
=> k = -8
Hence k = -8

Question 49.
If three consecutive vertices of a parallelogram are (1, -2), (3, 6) and (5, 10), find its fourth vertex.
Solution:
A (1, -2), B (3, 6) and C (5, 10) are the three consecutive vertices of the parallelogram ABCD
Let (x, y) be its fourth vertex
AC and BD are its diagonals which bisect each other at O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 89

Question 50.
If the points A (a, -11), B (5, b), C (2, 15) and D (1, 1) are the vertices of a parallelogram ABCD, find the values of a and b.
Solution:
A (a, -11), B (5, b), C (2, 15) and D (1, 1) are the vertices of a parallelogram ABCD
Diagonals AC and BD bisect eachother at O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 90
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 91

Question 51.
If the co-ordinates of the mid-points of the sides of a triangle be (3, -2), (-3, 1) and (4, -3), then find the co-ordinates of its vertices.
Solution:
In a ∆ABC,
D, E and F are the mid-points of the sides BC, CA and AB respectively and co-ordinates of D, E and F are (3, -2), (-3, 1) and (4, -3) respectively
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 92
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 93
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 94

Question 52.
The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the co-ordinates of P and Q are (p, -2) and (\(\frac { 5 }{ 3 }\) , q) respectively, find the values of p and q. [CBSE 2005]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 95

Question 53.
The line joining the points'(2, 1), (5, -8) is trisected at the points P and Q. If point P lies on the line 2x – y + k = 0, find the value of k. [CBSE 2005]
Solution:
Points A (2, 1), and B (5, -8) are the ends points of the line segment AB
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 96

Question 54.
A (4, 2), B (6, 5) and C (1, 4) are the vertices of ∆ABC,
(i) The median from A meets BC in D. Find the coordinates of the point D.
(ii) Find the coordinates of point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of the points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe? [NCERT,CBSE, 2009, 10]
Solution:
In ∆ABC, co-ordinates of A (4, 2) of (6, 5) and of (1, 4) and AD is BE and CF are the medians such that D, E and F are the mid points of the sides BC, CA and AB respectively
P is a point on AD such that AP : PD = 2 : 1
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 97
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 98
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 99
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 100
(iv) We see that co-ordinates of P, Q and R are same i.e., P, Q and R coincides eachother. Medians of the sides of a triangle pass through the same point which is called the centroid of the triangle.

Question 55.
If the points A (6, 1), B (8, 2), C (9, 4) and D (k, p) are the vertices of a parallelogram taken in order, then find the values of k and p.
Solution:
The diagonals of a parallelogram bisect each other
O is the mid-point of AC and also of BD
O is the mid-point of AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 101

Question 56.
A point P divides the line segment joining the points A (3, -5) and B (-4, 8) such that \(\frac { AP }{ PB }\) = \(\frac { k }{ 1 }\). If P lies on the line x + y = 0, then find the value of k. [CBSE 2012]
Solution:
Point P divides the line segment by joining the points A (3, -5) and B (-4, 8)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 102
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 103

Question 57.
The mid-point P of the line segment joining the points A (-10, 4) and B (-2, 0) lies on the line segment joining the pionts C (-9, -4) and D (-4, y). Find the ratio in which P divides CD. Also, find the value of y. [CBSE 2014]
Solution:
P is the mid-point of line segment joining the points A (-10, 4) and B (-2, 0)
Coordinates of P will be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 104
=> y = \(\frac { 18 }{ 3 }\) = 6
y = 6

Question 58.
If the point C (-1, 2) divides internally the line segment joining the points A (2, 5) and B (x, y) in the ratio 3 : 4, find the value of x² + y². [CBSE 2016]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 105

Question 59.
ABCD is a parallelogram with vertices A (x1, y1), B (x2, y2) and C (x3, y3). Find the coordinates of the fourth vertex D in terms of x1, x2, x3, y1, y2 and y3. [NCERT Exemplar]
Solution:
Let the coordinates of D be (x, y). We know that diagonals of a parallelogram bisect each other.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 106
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 107

Question 60.
The points A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of ∆ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the points of coordinates Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What are the coordinates of the centroid of the triangle ABC? [NCERT Exemplar]
Solution:
Given that, the points A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of ∆ABC.
(i) We know that, the median bisect the line segment into two equal parts i.e., here D is the mid-point of BC.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 108
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 109
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 110
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 111

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1

Other Exercises

Question 1.
In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 1
Solution:
∴ Perpendicular BC – 2 units and
Hypotenuse AC = 3 units
By Phythagoras Theorem, in AABC,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
AC2 = AB2 + BC2
⇒ (3)2 = (AB)2 + (2)2
⇒ 9 = AB2 + 4 ⇒ AB2 = 9-4 = 5
AB = √5 units
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 2
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 3
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 4
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 5
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 6
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 7
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 8
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 9
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 10
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 11
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 12
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 13
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 14
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 15
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 16
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 17
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 18
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 19
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 20

Question 2.
In a ΔABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) sin A, cos A
(ii) sin C, cos C.

Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 21

Question 3.
In the figure, find tan P and cot R. Is tan P = cot R ?
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 22
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 23

Question 4.
If sin A = \(\frac { 9 }{ 41 }\), compute cos A and tan A.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 24

Question 5.
Given 15 cot A = 8, find sin A and sec A.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 25

Question 6.
In ΔPQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 26
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 27

Question 7.
If cot 0 = \(\frac { 7 }{ 8 }\), evaluate :
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 28
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 29
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 30
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 31

Question 8.
If 3 cot A = 4, check whether \(\frac { 1-{ tan }^{ 2 }A }{ 1+{ tan }^{ 2 }A }\) = cos2 A – sin2 A or not.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 32
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 33

Question 9.
If tan θ = a/b , Find the Value of \(\frac { cos\theta +sin\theta }{ cos\theta -sin\theta }\).
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 34
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 35

Question 10.
If 3 tan θ = 4, find the value of 4cos θ – sin θ \(\frac { 4cos\theta -sin\theta }{ 2cos\theta +sin\theta }\).
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 36
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 37

Question 11.
If 3 cot 0 = 2, find the value of \(\frac { 4sins\theta -3cos\theta }{ 2sin\theta +6cos\theta }\).
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 38

Question 12.
If tan θ = \(\frac { a }{ b }\), prove that
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 39
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 40

Question 13.
If sec θ = \(\frac { 13 }{ 5 }\), show that \(\frac { 2sins\theta -3cos\theta }{ 4sin\theta -9cos\theta }\) =3.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 41

Question 14.
If cos θ \(\frac { 12 }{ 13 }\), show that sin θ (1 – tan θ) \(\frac { 35 }{ 156 }\).
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 42
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 43

Question 15.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 44
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 45
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 46

Question 16.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 47
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 48
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 49

Question 17.
If sec θ = \(\frac { 5 }{ 4 }\), find the value of \(\frac { sins\theta -2cos\theta }{ tan\theta -cot\theta }\).
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 50
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 51

Question 18.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 52
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 53
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 54

Question 19.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 55
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 56
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 57

Question 20.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 58
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 59
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 60

Question 21.
If tan θ = \(\frac { 24 }{ 7 }\), find that sin θ + cos θ.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 61
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 62

Question 22.
If sin θ = \(\frac { a }{ b }\), find sec θ + tan θ in terms of a and b.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 63

Question 23.
If 8 tan A = 15, find sin A – cos A.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 64

Question 24.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 65
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 66
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 67

Question 25.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 68
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 69
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 70

Question 26.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
∠A and ∠B are acute angles and cos A = cos B
Draw a right angle AABC, in which ∠C – 90°
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 71

Question 27.
In a ∆ABC, right angled at A, if tan C =√3 , find the value of sin B cos C + cos B sin C. (C.B.S.E. 2008)
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 72
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 73

Question 28.
28. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = \(\frac { 12 }{ 5 }\) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = \(\frac { 4 }{ 3 }\) for some angle θ.
Solution:
(i) False, value of tan A 0 to infinity.
(ii) True.
(iii) False, cos A is the abbreviation of cosine A.
(iv) False, it is the cotengent of angle A.
(v) Flase, value of sin θ varies on 0 to 1.

Question 29.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 74
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 75
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 76
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 77

Question 30.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 78
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 79
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 80

Question 31.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 81
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 82
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 83

Question 32.
If sin θ =\(\frac { 3 }{ 4 }\), prove that
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 84
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 85
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 86

Question 33.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 87
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 88
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 89

Question 34.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 90
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 91
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 92
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 93

Question 35.
If 3 cos θ-4 sin θ = 2 cos θ + sin θ, find tan θ.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 94

Question 36.
If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P.
Solution:
∠A and ∠P are acute angles and tan A = tan P Draw a right angled AAPB in which ∠B = 90°
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 95

 

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2

Other Exercises

Question 1.
Find the distance between the following pair of points :
(i) (-6, 7) and (-1, -5)
(ii) (a + b, b + c) and (a – b, c – b)
(iii) (a sin α, -b cos α) and (-a cos α, -b sin α)
(iv) (a, 0) and (0, b)
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 1
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 2
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 3

Question 2.
Find the value of a when the distance between the points (3, a) and (4, 1) is √10
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 4
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 5

Question 3.
If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 6

Question 4.
Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.
Solution:
Distance between (x, y) and (-3, 0) is
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 7
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 8

Question 5.
The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.
Solution:
Let the ordinate of other end by y, then The distance between (2, -3) and (10, y) is
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 9
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 10

Question 6.
Show that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices points of a rectangle. (C.B.S.E. 2006C)
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 11
AB = CD and AD = BC
and diagonal AC = BD
ABCD is a rectangle

Question 7.
Show that the points A (1, -2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 12

Question 8.
Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square. [NCERT]
Solution:
Vertices A (1, 7), B (4, 2), C (-1,-1), D (-4, 4)
If these are the vertices of a square, then its diagonals and sides are equal
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 13
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 14

Question 9.
Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle. (C.B.S.E. 2006C)
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 15
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 16

Question 10.
Prove that (2, -2), (-2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 17
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 18

Question 11.
Prove that the points (2a, 4a), (2a, 6a) and (2a + √3 a , 5a) are the vertices of an equilateral triangle.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 19

Question 12.
Prove that the points (2, 3), (-4, -6) and (1, \(\frac { 3 }{ 2 }\) )do not form a triangle.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 20
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 21

Question 13.
The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC. [NCERT Exemplar]
Solution:
Given that, the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ∆ABC right angled at B.
By Pythagoras theorem, AC² = AB² + BC² ………(i)
Now, by distance formula,
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 22
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 23

Question 14.
Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2) is a rhombus.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 24
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 25

Question 15.
Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.
Solution:
Two vertices of an isosceles ∆ABC are A (2, 0) and B (2, 5). Let co-ordinates of third vertex C be (x, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 26
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 27
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 28
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 29

Question 16.
Which point on x-axis is equidistant from (5, 9) and (-4, 6) ?
Solution:
Let co-ordinates of two points are A (5, 9), B (-4, 6)
The required point is on x-axis
Its ordinates or y-co-ordinates will be 0
Let the co-ordinates of the point C be (x, 0)
AC = CB
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 30
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 31

Question 17.
Prove that the point (-2, 5), (0, 1) and (2, -3) are collinear.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 32
Now AB + BC = 2√5 +2√5
and CA = 4√5
AB + BC = CA
A, B and C are collinear

Question 18.
The co-ordinates of the point P are (-3,2). Find the co-ordinates of the point Q which lies on the line joining P and origin such that OP = OQ.
Solution:
Co-ordinates of P are (-3, 2) and origin O are (0, 0)
Let co-ordinates of Q be (x, y)
O is the mid point of PQ
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 33
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 34
= 9 + 4 = (±3)² + (±2)²
The point will be in fourth quadrant
Its y-coordinates will be negative
and x-coordinates will be positive
Now comparing the equation
x² = (±3)² => x = ±3
y² = (±2)² => y = ±2
x = 3, y = -2
Co-ordinates of the point Q are (3, -2)

Question 19.
Which point on y-axis is equidistant from (2, 3) and (-4, 1) ?
Solution:
The required point lies on y-axis
Its abscissa will be zero
Let the point be C (0, y) and A (2, 3), B (-4, 1)
Now,
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 35

Question 20.
The three vertices of a parallelogram are (3, 4), (3, 8) and (9, 8). Find the fourth vertex.
Solution:
Let ABCD be a parallelogram and vertices will be A (3, 4), B (3, 8), C (9, 8)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 36
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 37
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 38
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 39

Question 21.
Find a point which is equidistant from the point A (-5, 4) and B (-1, 6). How many such points are there? [NCERT Exemplar]
Solution:
Let P (h, k) be the point which is equidistant from the points A (-5, 4) and B (-1, 6).
PA = PB
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 40
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 41
So, the mid-point of AB satisfy the Eq. (i).
Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the point A and B.
Replacing h, k, by x, y in above equation, we have 2x + y + 1 = 0

Question 22.
The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10√2 units. [NCERT Exemplar]
Solution:
By given condition,
Distance between the centre C (2a, a-1) and the point P (11, -9), which lie on the circle = Radius of circle
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 42
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 43

Question 23.
Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching the office? (Assume that all distance covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in kilometers. [NCERT Exemplar]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 44
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 45
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 46

Question 24.
Find the value of k, if the point P (0, 2) is equidistant from (3, k) and (k, 5).
Solution:
Let P (0, 2) is equidistant from A (3, k) and B (k, 5)
PA = PB
=> PA² = PB²
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 47

Question 25.
If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the
(i) interior,
(ii) exterior of the triangle. [NCERT Exemplar]
Solution:
Let the third vertex of an equilateral triangle be (x, y).
Let A (-4, 3), B (4,3) and C (x, y).
We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 48
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 49
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 50
But given that, the origin lies in the interior of the ∆ABC and the x-coordinate of third vertex is zero.
Then, y-coordinate of third vertex should be negative.
Hence, the require coordinate of third vertex,
C = (0, 3 – 4√3). [C ≠ (0, 3 + 4√3)]

Question 26.
Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.
Solution:
Let the co-ordinates of the vertices A, B, C and D of a rhombus are A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 51
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 52

Question 27.
Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.
Solution:
Let ABC is a triangle whose vertices are A (3, 0), B (-1, -6) and C (4, -1)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 53
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 54

Question 28.
Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4). [CBSE 2005]
Solution:
The required point is on x-axis
Its ordinate will be O
Let the co-ordinates of the required point P (x, 0)
Let the point P is equidistant from the points A (7, 6) and B (-3, 4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 55
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 56

Question 29.
(i) Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square. [CBSE 2004]
(ii) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD. [CBSE 2013]
(iii) Name the type of triangle PQR formed by the point P(√2 , √2), Q(- √2, – √2) and R (-√6 , √6 ). [NCERT Exemplar]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 57
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 58
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 59
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 60

Question 30.
Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3). [CBSE 2004]
Solution:
The point P lies on x-axis
The ordinates of P will be 0 Let the point P be (x, 0)
Let P is equidistant from A (-2, 5) and B (2, -3)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 61
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 62

Question 31.
Find the value of x such that PQ = QR where the co-ordinates of P, Q and R are (6, -1) (1, 3) and (x, 8) respectively. [CBSE 2005]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 63

Question 32.
Prove that the points (0, 0), (5, 5) and (-5, 5) are the vertices of a right isosceles triangle. [CBSE 2005]
Solution:
Let the vertices of a triangle be A (0, 0), B (5, 5) and C (-5, 5)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 64
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 65

Question 33.
If the points P (x, y) is equidistant from the points A (5, 1) and B (1,5), prove that x = y. [CBSE 2005]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 66

Question 34.
If Q (0, 1) is equidistant from P (5, -3) and R (x, 6) find the values of x. Also find the distances QR and PR. [NCERT]
Solution:
Q (0, 1) is equidistant from P (5, -3) and R (x, 6)
PQ = RQ
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 67
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 68

Question 35.
Find the values ofy for which the distance between the points P (2, -3) and Q (10, y) is 10 units. [NCERT]
Solution:
Distance between P (2, -3) and Q (10, y) = 10
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 69

Question 36.
If the point P (k – 1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k. [CBSE 2014]
Solution:
Point P (k – 1, 2) is equidistant from A (3, k) and B (k, 5)
PA= PB
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 70

Question 37.
If the point A (0, 2) is equidistant from the point B (3, p) and C (p, 5), find p. Also, find the length of AB. [CBSE 2014]
Solution:
Point A (0, 2) is equidistant from B (3, p) and C (p, 5)
AB = AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 71

Question 38.
Name the quadrilateral formed, if any, by the following points, and give reasons for your answers :
(i) A (-1, -2), B (1, 0), C (-1, 2), D (-3, 0)
(ii) A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)
(iii) A (4, 5), B (7, 6), C (4, 3), D (1, 2)
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 72
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 73
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 74
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 75

Question 39.
Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3, 5).
Solution:
Let the given points are A (7, 1) and B (3, 5) and mid point be M
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 76
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 77

Question 40.
Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order, form a rhombus. Also find its area.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 78
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 79

Question 41.
In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at A (3, 1), B (6, 4) and C (8, 6). Do you think they are seated in a line ?
Solution:
A (3, 1), B (6, 4) and C (8, 6)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 80

Question 42.
Find a point ony-axis which is equidistant from the points (5, -2) and (-3, 2).
Solution:
The point lies on y-axis
Its x = 0
Let the required point be (0, y) and let A (5, -2) and B (-3, 2)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 81

Question 43.
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4). [NCERT]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 82

Question 44.
If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p. [CBSE 2012]
Solution:
Point A (0, 2) is equidistant from the points B (3, p) and C (p, 5)
AB = AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 83
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 84

Question 45.
Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle. [CBSE 2013]
Solution:
Let points are A (7, 10), B (-2, 5) and C (3, -4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 85

Question 46.
If the point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8), find the value of x and And the distance AP. [CBSE 2014]
Solution:
Point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8)
PA = PB
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 86
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 87

Question 47.
If A (3, y) is equidistant from points P (8, -3) and Q (7, 6), find the value of y and find the distance AQ. [CBSE 2014]
Solution:
Point A (3, y) is equidistant from P (8, -3) and Q (7, 6)
i.e., AP = AQ
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 88

Question 48.
If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex. [CBSE 2014]
Solution:
Let A (0, -3) and B (0, 3) are vertices of an equilateral triangle
Let the coordinates of the third vertex be C (x, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 89
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 90

Question 49.
If the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3), find k. Also, find the length of AP.
Solution:
Point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3)
AP = BP
Now,
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 91
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 92

Question 50.
Show that ∆ABC, where A (-2, 0), B (2, 0), C (0, 2) and ∆PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 93
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 94
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 95

Question 51.
An equilateral triangle has two vertices at the points (3, 4), and (-2, 3). Find the co-ordinates of the third vertex.
Solution:
Let two vertices of an equilateral triangle are A (3,4), and B (-2,3) and let the third vertex be C (x, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 96
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 97
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 98
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 99

Question 52.
Find the circumcentre of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 100
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 101
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 102

Question 53.
Find the angle subtended at the origin by the line segment whose end points are (0, 100) and (10, 0).
Solution:
Let co-ordinates of the end points of a line segment are A (0, 100), B (10, 0) and origin is O (0, 0)
Abscissa of A is 0
It lies on y-axis
Similarly, ordinates of B is 0
It lies on x-axis
But axes intersect each other at right angle
AB will subtended 90° at the origin
Angle is 90° or \(\frac { \pi }{ 2 }\)

Question 54.
Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 103
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 104
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 105

Question 55.
If two opposite vertices of a square are (5, 4) and (1, -6), find the coordinates of its remaining two vertices.
Solution:
Two opposite points of a square are (5, 4) and (1, -6)
Let ABCD be a square and A (5, 4) and C (1, -6) are the opposite points
Let the co-ordinates of B be (x, y). Join AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 106
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 107
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 108

Question 56.
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3).
Solution:
Let O is the centre of the circle is (x, 7) Join OA, OB and OC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 109
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 110

Question 57.
Two opposite vertices of a square are (-1, 2) and (3, 2). Find the co-ordinates of other two vertices.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 111
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 112
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 113

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