Category: CBSE Class 10

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

• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

 

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

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

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

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

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,

=> 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.

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

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 (\(\frac { 1 + 5 }{ 2 }\) , \(\frac { -1 + 1 }{ 2 }\))

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

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)

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

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)

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:

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

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

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


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

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

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

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

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:

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

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

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

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

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

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

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

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

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)

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)

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

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

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

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)

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

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.

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

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:

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)

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

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

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

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

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

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)

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

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

=> 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

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

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

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:

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

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




(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

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)

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

=> 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:

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.

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.

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

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

• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Question 1.
In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.

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

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:

Question 3.
In the figure, find tan P and cot R. Is tan P = cot R ?

Solution:

Question 4.
If sin A = \(\frac { 9 }{ 41 }\), compute cos A and tan A.
Solution:

Question 5.
Given 15 cot A = 8, find sin A and sec A.
Solution:

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:

Question 7.
If cot 0 = \(\frac { 7 }{ 8 }\), evaluate :

Solution:

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:

Question 9.
If tan θ = a/b , Find the Value of \(\frac { cos\theta +sin\theta }{ cos\theta -sin\theta }\).
Solution:

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:

Question 11.
If 3 cot 0 = 2, find the value of \(\frac { 4sins\theta -3cos\theta }{ 2sin\theta +6cos\theta }\).
Solution:

Question 12.
If tan θ = \(\frac { a }{ b }\), prove that

Solution:

Question 13.
If sec θ = \(\frac { 13 }{ 5 }\), show that \(\frac { 2sins\theta -3cos\theta }{ 4sin\theta -9cos\theta }\) =3.
Solution:

Question 14.
If cos θ \(\frac { 12 }{ 13 }\), show that sin θ (1 – tan θ) \(\frac { 35 }{ 156 }\).
Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.
If sec θ = \(\frac { 5 }{ 4 }\), find the value of \(\frac { sins\theta -2cos\theta }{ tan\theta -cot\theta }\).
Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.
If tan θ = \(\frac { 24 }{ 7 }\), find that sin θ + cos θ.
Solution:

Question 22.
If sin θ = \(\frac { a }{ b }\), find sec θ + tan θ in terms of a and b.
Solution:

Question 23.
If 8 tan A = 15, find sin A – cos A.
Solution:

Question 24.

Solution:

Question 25.

Solution:

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°

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:

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.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.
If sin θ =\(\frac { 3 }{ 4 }\), prove that

Solution:

Question 33.

Solution:

Question 34.

Solution:

Question 35.
If 3 cos θ-4 sin θ = 2 cos θ + sin θ, find tan θ.
Solution:

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°

 

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

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

• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

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:

Question 2.
Find the value of a when the distance between the points (3, a) and (4, 1) is √10
Solution:

Question 3.
If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.
Solution:

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

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

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:

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:

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

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:

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:

Question 11.
Prove that the points (2a, 4a), (2a, 6a) and (2a + √3 a , 5a) are the vertices of an equilateral triangle.
Solution:

Question 12.
Prove that the points (2, 3), (-4, -6) and (1, \(\frac { 3 }{ 2 }\) )do not form a triangle.
Solution:

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,

Question 14.
Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2) is a rhombus.
Solution:

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)

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

Question 17.
Prove that the point (-2, 5), (0, 1) and (2, -3) are collinear.
Solution:

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


= 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,

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)

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


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

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:

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

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.



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)

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)

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)

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:

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)

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:

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)

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:

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

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

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

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

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:

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

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:

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)

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)

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:

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

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)

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

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

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)

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,

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:

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)

Question 52.
Find the circumcentre of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).
Solution:

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:

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

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

Question 57.
Two opposite vertices of a square are (-1, 2) and (3, 2). Find the co-ordinates of other two vertices.
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 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 13.2

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 13.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3
• 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 VSAQS
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

Question 1.
Find, in terms of π the length of the arc that subtends an angle of 30° at the centre of a circle of radius 4 cm.
Solution:
Radius of the circle (r) = 4 cm
Angle at the centre subtended an arc = 30°

Question 2.
Find the angle subtended at the centre of a circle of radius 5 cm by an arc of length \((\frac { 5\pi }{ 3 } \) cm.
Solution:
Radius of the circle (r) = 5 cm 571
Length of arc = \(\frac { 5\pi }{ 3 }\) cm
Let θ be the angle subtended by the arc, then

Question 3.
An arc of length 20tc cm subtends an angle of 144° at the centre of a circle. Find the radius of the circle.
Solution:
Length of an arc = 20π cm
Angle subtended by the arc = 144°
Let r be the radius of the circle, then

Question 4.
An arc of length 15 cm subtends an angle of 45° at the centre of a circle. Find in terms of π ; the radius of the circle.
Solution:
Length of arc = 15 cm
Angle subtended at the centre = 45°
Let r be the radius of the circle, then

Question 5.
Find the angle subtended at the centre of a circle of radius ‘a’  by an arc of length \((\frac { a\pi }{ 4 } )\)  cm.
Solution:
Radius of the circle (r) = a cm
Length of arc = \(\frac { a\pi }{ 4 }\)   cm
Let θ be the angle subtended by the arc at the centre, then

Question 6.
A sector of a circle of radius 4 cm contains an angle of 30°. Find the area of the sector.
Solution:
Radius of the sector of a circle (r) = 4 cm
Angle at the centre (θ) = 30°

Question 7.
A sector of a circle of radius 8 cm contains an angle of 135°. Find the area of the sector.
Solution:
Radius of the sector of the circle (r) = 8 cm
Angle at the centre (θ) = 135°

Question 8.
The area of a sector of a circle of radius 2 cm is 7 is cm². Find the angle contained by the sector.
Solution:
Area of the sector of a circle =π cm²
Radius of the circle (r) = 2 cm
Let 0 be the angle at the centre, then

Question 9.
The area of a sector of a circle of radius 5 cm is 5π cm². Find the angle contained by the sector.
Solution:
Area of the sector of a circle = 5π cm²
Radius of the circle (r) = 5 cm
Let 9 be the angle at the centre, then

Question 10.
Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm. [NCERT Exemplar]
Solution:
Let the central angle of the sector be θ.
Given that, radius of the sector of a circle (r) = 5 cm
and arc length (l) = 3.5 cm

Question 11.
In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the length of the arc and area of the sector.
Solution:
Radius of the circle (r) = 25 cm
Angle at the centre (θ) = 72°

Question 12.
The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.
Solution:
Radius of the circle (r) = 5.7 m
Perimeter of the sector = 27.2 m
Length of the arc = Perimeter – 2r
= (27.2 – 2 x 5.7) m
= 27.2 – 11.4 = 15.8 m
Let θ be the central angle, then

Question 13.
The perimeter of a certain sector of a circle of radius 5.6 cm is 27.2 m. Find the area of the sector.
Solution:
Radius of the sector (r) = 5.6 cm
and perimeter of the sector = 27.2 cm
∴ Length of arc = Perimeter – 2r
= 27.2 – 2 x 5.6
= 27.2- 11.2= 16.0 cm
∴ θ be the angle at the centre, then

Question 14.
A sector is cut-off from a circle of radius 21 cm. The angle of the sector is 120°. Find the length of its arc and the area.
Solution:
Radius of the sector of a circle (r) = 21 cm
Angle at the centre = 120°

Question 15.
The minute hand of a clock is \(\sqrt { 21 } \)  cm long, Find the area described by the minute hand on the face of the clock between 7.00 A.M. and 7.05 A.M.
Solution:
Length of minute hand of a clock (r) = \(\sqrt { 21 }\)  cm
Period = 7 a.m. to 7.05 a.m. 5 minutes

Question 16.
The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8.00 A.M. and 8.25 A.M.
Solution:
Length of minute hand of a clock (r) = 10 cm
Period = 8 A.M. to 8.25 A.M. = 25 minutes

Question 17.
The sector of 56° cut out from a circle contains area 4.4 cm², Find the radius of the circle.
Solution:
Area of a sector = 4.4 cm²
Central angle = 56°                                        ‘
Let r be the radius of the sector of the circle, then

Question 18.
Area of a sector of central angle 200° of a circle s 770²cm. Find the length of the corresponding are of this sector.
Solution:
Let the radius of the sector AOBA be r.
Given that, Central angle of sector AOBA = θ = 200°
and area of the sector AOBA = 770 cm²

Question 19.
The length of minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6:05 am and 6:40 am.           [NCERT Exemplar]
Solution:
We know that, in 60 min, minute hand revolving = 360°
In 1 min, minute hand revolving = \(\frac { 360\circ }{ 60\circ }\)

Question 20.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. [CBSE 2013]
Solution:
Length of minute hand (r)= 14 cm
Area swept by the minute hand in 5 minutes

Question 21.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find (0 the length of the arc (ii) area of the secter formed by the arc. (Use π = 22/7) [CBSE 2013]
Solution:
Radius of a circle (r) = 21 cm
Angle at the centre = 60°

Question 22.
From a circular piece of cardboard of radius 3 cm two sectors of 90° have been cut off. Find the perimeter of the remaining portion nearest hundredth centimeters (Take π = 22/7).
Solution:
Radius of the circular piece of cardboard (r) = 3 cm

∴ Two sectors of 90° each have been cut off
∴ We get a semicular cardboard piece
∴ Perimeter of arc ACB

Question 23.
The area of a sector is one-twelfth that of the complete circle. Find the angle of the sector.
Solution:
Let r be the radius of the circle and 0 be the central angle of the sector of the circle Then area of circle = πr²

Question 24.
AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm. Find the area of the sector of the circle formed by the chord AB.
Solution:
Radius of the circle with centre O (r) = 4 cm
Length of chord AB = 4 cm

Question 25.
In a circle of radius 6 cm, a chord of length 10 cm makes an angle of 110° at the centre of the circle. Find 
(i)  the circumference of the circle,
(ii) the area of the circle,
(iii) the length of the arc AB,
(iv) the area of the sector OAB.
Solution:
Radius of the circle (r) = 6 cm
Length of chord = 10 cm
and central angle (θ) =110°

Question 26.
Figure, shows a sector of a circle, centre O, containing an angle θ°. Prove that :

Solution:
Radius of the circle = r
Arc AC subtends ∠θ at the centre of the
circle. OAB is a right triangle
In the right ΔOAB,

Question 27.
Figure, shows a sector of a circle of radius r cm containing an angle θ°. The area of the sector is A cm² and perimeter of the sector is 50 cm. Prove that

Solution:
Radius of the sector of the circle = r cm
and angle at the centre = 0
Area of sector OAB = A cm²
and perimeter of sector OAB = 50 cm

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2 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 6.1

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 6.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Question 1.
On which axis do the following points lie?
(i) P (5, 0)
(ii) Q (0 – 2)
(iii) R (-4, 0)
(iv) S (0, 5)
Solution:
(i) P (5, 0)
Its ordinate or y-axis is 0. It lies on x-axis
(ii) Q (0 – 2)
Its abscissa or x-axis is 0. It lies on y-axis
(iii) R (-4, 0)
Its ordinate is 0 It lies on x-axis
(iv) S (0, 5)
Its abscissa is 0. It lies on y-axis

Question 2.
Let ABCD be a square of side 2a. Find the coordinates of the vertices of this square when
(i) A coincides with the origin and AB and AD are along OX and OY respectively.
(ii) The centre of the square is at the-origin and coordinate axes are parallel to the sides AB and AD respectively.
Solution:
ABCD is a square whose side is 2a
(i) A coincides with origin (0, 0)
AB and AD are along OX and OY respectively
Co-ordinates of A are (0, 0), of B are (2a, 0) of C are (2a, 2a) and of D are (0, 2a)
(ii) The centre of the square is at the origin (0, 0) and co-ordinates axes are parallel to the sides AB and AD respectively.
Then the co-ordinates of A are (a, a) of B are (-a, a), of C are (-a, -a) and of D are (a, -a) as shown in the figure given below :

Question 3.
The base PQ of two equilateral triangles PQR and PQR’ with side 2a lies along y- axis such.that the mid-point of PQ is at the origin. Find the coordinates of the vertices R and R’ of the triangles.
Solution:
∆PQR and PQR’ are equilateral triangles with side 2a each and base PQ and mid of point of PQ is 0 (0, 0) and PQ lies along y-axis

 

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1 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 13 Areas Related to Circles Ex 13.1

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 13.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3
• 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 VSAQS
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

Question 1.
Find the circumference and area of a circle of radius 4.2 cm.
Solution:
Radius of a circle = 4.2 cm

Question 2.
Find the circumference of a circle whose area is 301.84 cm².
Solution:
Area of a circle = 301.84 cm²
Let r be the radius, then πr² = 301.84

Question 3.
Find the area of a circle whose circum­ference is 44 cm.
Solution:
Circumference of a circle = 44 cm
Let r be the radius,
then 2πr = circumference

Question 4.
The circumference of a circle exceeds the diameter by 16.8 cm. Find the circum­ference of the circle. (C.B.S.E. 1996)
Solution:
Let r be the radius of the circle
∴  Circumference = 2r + 16.8 cm
⇒  2πr = 2r + 16.8
⇒  2πr – 2r = 16.8

Question 5.
A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22/7)
Solution:
Radius of the circle (r) = Length of the rope = 28 m .
Area of the place where the horse can graze

Question 6.
A steel wire when bent in the form of a square encloses an area of 121 cm². If the same wire is bent in the form of a circle, find the area of the circle.  (C.B.S.E. 1997)
Solution:
Area of square formed by a wire =121 cm²
∴ Side of square (a) = \(\sqrt { Area } \)  = \(\sqrt { 121 } \)  = 11 cm Perimeter of the square = 4 x side = 4 x 11 = 44 cm
∴Circumference of the circle formed by the wire = 44cm
Let r be the radius

Question 7.
The circumference of two circles are in the ratio 2 : 3. Find the ratio of their areas.
Solution:
Let R and r be the radii of two circles and their ratio between them circumference = 2 : 3

Question 8.
The sum of radii of two circles is 140 cm and the difference of their circum­ferences is 88 cm. Find the diameters of the circles.
Solution:
Let R and r be the radii of two circles Then R + r = 140 cm  …….(i)
and difference of their circumferences

Question 9.
Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm. [NCERT Exemplar]
Solution:
Let the radius of a circle be r.
Circumference of a circle = 2πr
Let the radii of two circles are r₁ and r₂ whose
values are 15 cm and 18 cm respectively.
i.e., r₁ = 15 cm and r₂ = 18 cm
Now, by given condition,
Circumference of circle = Circumference of first circle + Circumference of second circle
⇒   2πr = 2π r₁ + 2πr₂ =
⇒  r = r₁ + r₂
⇒   r = 15 + 18
∴ r = 33 cm
Hence, required radius of a circle is 33 cm.

Question 10.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.
Solution:
Radius of first circle (r₁) = 8 cm
and radius of second circle (r₂) = 6 cm

Question 11.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circle which has its circumference equal to the sum of the circumferences of the two circles.
Solution:
Radius of the first circle (r₁) = 19 cm
and radius of the second circle (r₂) = 9 cm S
um of their circumferences = 2πr₁ + 2πr₂
= 2π (r₁ + r₂) = 2π (19 + 9) cm
= 2π x 28 = 56π cm
Let R be the radius of the circle whose circumference is the sum of the circum­ferences of given two circles, then

Question 12.
The area of a circular playground is 22176 m². Find the cost of fencing this ground at the rate of ₹50 per metre.  [NCERT Exempiar]
Solution:
Given, area of a circular playground  = 22176 m²

Question 13.
The side of a square is 10 cm. Find the area of circumscribed and inscribed circles.
Solution:
ABCD is a square whose each side is 10 cm
∴  AB = BC = CD = DA = 10 cm
AC and BD are its diagonals

Question 14.
If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.
Solution:
Let r be the radius of the circle a be the side of the square

Question 15.
The area of a circle inscribed in an equilateral triangle is 154 cm². Find the perimeter of the triangle. (Use π = 22/7 and \(\sqrt { 3 } \)  = 1.73)
Solution:
Area of the inscribed circle of ΔABC = 154 cm²

Question 16.
A field is in the form of a circle. A fence is to be erected around the field. The cost of fencing would be ₹2640 at the rate of ₹12 per metre. Then, the field is to be thoroughly ploughed at the cost of ₹0.50 per m². What is the amount required to plough the field ? (Take π = 22/7)
Solution:
Cost of the fencing the circular field = ₹2640
Rate = ₹12 per metre 2640
∴ Circumference = \((\frac { 2640 }{ 12 } )\) = 220 m
Let r be the radius of the field, then = 2πr = 220

Question 17.
A park is in the form of a rectangle 120 m x 100 m. At the centre of the park there is a circular lawn. The area of park excluding lawn is 8700 m². Find the radius of the circular lawn. (Use π = 22/7).
Solution:
Area of the park excluding lawn = 8700 m²
Length of rectangular park = 120 m
and width = 100 m

∴ Area of lawn = l x b
= 120 x 100 m² = 12000 m²
Let r be the radius of the circular lawn, then area of lawn = πr²

Question 18.
A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. Find the radius of the its wheels.
Solution:
Distance covered by the car in 450 revolutions = 1 km = 1000 m
∴ Distance covered in 1 revolution = \((\frac { 1000 }{ 450 } )\)
= \((\frac { 20 }{ 9 } )\) m

Question 19.
The area of enclosed between the concentric circles is 770 cm². If the radius of the outer circle is 21 cm, find the radius of the inner circle.
Solution:
Area of enclosed between two concentric circles = 770 cm²
Radius of the outer circle (R) = 21 cm

Question 20.
An archery target has three regions formed by the concentric circles as shown in the figure. If the diameters of the concentric circles are in the ratio 1:2:3, then find the ratio of the areas of three regions.[NCERT Exemplar]

Solution:
Let the diameters of concentric circles be k, 2k , 3k

Question 21.
The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/hr? [NCERT Exemplar]
Solution:
Given, radius of wheel, r = 35 cm
Circumference of the wheel = 2πr
= 2 x \((\frac { 22 }{ 7 } )\) x 35 = 220 cm
But speed of the wheel = 66 kmh^-1
= \((\frac { 66 x 1000 }{ 60 } )\) m/ mm
= 1100 x 100 cm min^-1
= 110000 cm min^-1
∴ Number of revolutions in 1 min
= \((\frac { 110000 }{ 220 } )\)= 500 revolution
Hence, required number of revolutions per minute is 500.

Question 22.
A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of ₹25 per m². [NCERT Exemplar]
Solution:
Given that, a circular pond is surrounded by a wide path.
The diameter of circular pond = 17.5 m

Question 23.
A circular park is surrounded by a rod 21 m wide. If the radius of the park is 105 m, find the area of the road. [NCERT Exemplar]
Solution:
Given that, a circular park is surrounded by a road.
Width of the road = 21 m
Radius of the park (r₁) = 105 m

.’. Radius of whole circular portion (park + road),
r_e = 105 + 21 = 126 m
Now, area of road = Area of whole circular portion – Area of circular park
= πr² – πr²             [∵ area of circle = πr²]

Question 24.
A square of diagonal 8 cm is inscribed in a circle. Find the area of the region lying outside the circle and inside the square.  [NCERT Exemplar]
Solution:
Let the side of a square be a and the radius of circle be r.
Given that, length of diagonal of square = 8 cm

Question 25.
A path of 4 m width runs round a semi­circular grassy plot whose circumference is 81 \((\frac { 5 }{ 7 } )\)m. Find:
(i) the area of the path
(ii) the cost of gravelling the path at the rate of ₹1.50 per square metre
(iii) the cost of turfing the plot at the rate of 45 paise per m².
Solution:
Width of path around the semicircular grassy plot = 4 m
Circumference of the plot = 81 \((\frac { 5 }{ 7 } )\)m
= \((\frac { 572 }{ 7 } )\) m
Let r be the radius of the plot, then

Question 26.
Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle, such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles. Find the radius of the third circle correct to one decimal place.
Solution:
Radius of first circle (r₁) = 3.5 cm
Radius of second circle (r₂) = 7 cm

Question 27.
A path of width 3.5 m runs around a semi­circular grassy plot whose perimeter is 72 m. Find the area of the path. (Use π = 22/7)                   [CBSE 2015]
Solution:
Perimeter of semicircle grassy plot = 72 m

Let r be the radius of the plot

Question 28.
A circular pond is of diameter 17.5 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of ₹25 per square metre (Use π = 3.14)               [CBSE 2014]
Solution:
Diameter of circular pond (d) = 17.5 m
Radius (r) =\((\frac { 1725 }{ 2 } )\) = 8.75 m
Width of path = 2m
∴  Radius of outer cirlce (R) = 8.75 + 2 = 10.75 m
Area of path = (R² – r²)π
= [(10.75)² – (8.75)²](3.14)
= 3.14(10.75 + 8.75) (10.75 – 8.75)
= 3.14 x 19.5 x 2 = 122.46 m²
Cost of 1 m² for constructing the path ₹25 m²
∴  Total cost = ₹ 122.46 x 25 = ₹3061.50

Question 29.
The outer circumference of a circular race-track is 528 m. The track is every­where 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre (Use π= 22/7).
Solution:
Let R and r be the radii of the outer and inner of track.
Outer circumference of the race track = 528 m

Question 30.
A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.
Solution:
Width of the road = 7 m

Circumference of the park = 352 m
Let r be the radius, then 2πr = 352

Question 31.
Prove that the area of a circular path of uniform width surrounding a circular region of radius r is πh(2r + h).
Solution:
Radius of inner circle = r
Width of path = h
∴ Outer radius (R) = (r + h)
∴ Area of path = πR² – πr²
= π {(r + h)² – r²}
= π {r² + h² + 2rh – r²}
= π {2rh + h²}
= πh (2r + h) Hence proved.

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1 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 5 Arithmetic Progressions MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Mark the correct alternative in each of the following :
Question 1.
If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
(a) 87
(b) 88
(c) 89
(d) 90
Solution:
(c) 7th term (a₇) = a + 6d = 34
13th term (a₁₃) = a + 12d = 64
Subtracting, 6d = 30 => d = 5
and a + 12 x 5 = 64 => a + 60 = 64 => a = 64 – 60 = 4
18th term (a₁₈) = a + 17d = 4 + 17 x 5 = 4 + 85 = 89

Question 2.
If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of (p + q) terms will be
(a) 0
(b) p – q
(c) p + q
(d) – (p + q)
Solution:
(d)

Question 3.
If the sum of n terms of an A.P. be 3n² + n and its common difference is 6, then its first term is
(a) 2
(b) 3
(c) 1
(d) 4
Solution:
(d)

Question 4.
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
(a) 5
(b) 6
(c) 7
(d) 8
Solution:
(b) First term of an A.P. (a) = 1
Last term (l) = 11
and sum of its terms = 36
Let n be the number of terms and d be the common difference, then

Question 5.
If the sum of n terms of an A.P. is 3n² + 5n then which of its terms is 164 ?
(a) 26th
(b) 27th
(c) 28th
(d) none of these
Solution:
(b)

Question 6.
If the sum of it terms of an A.P. is 2n² + 5n, then its nth term is
(a) 4n – 3
(b) 3n – 4
(c) 4n + 3
(d) 3n + 4
Solution:
(c)

Question 7.
If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is :
(a) 13
(b) 9
(c) 21
(d) 17
Solution:
(c)

Question 8.
If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are
(a) 5, 10, 15, 20
(b) 4, 10, 16, 22
(c) 3, 7, 11, 15
(d) None of these
Solution:
(a)
4 numbers are in A.P.
Let the numbers be
a – 3d, a – d, a + d, a + 3d
Where a is the first term and 2d is the common difference
Now their sum = 50
a – 3d + a – d + a + d + a + 3d = 50
and greatest number is 4 times the least number
a + 3d = 4 (a – 3d)
a + 3d = 4a – 12d
4a – a = 3d + 12d
=> 3a = 15d

Question 9.
Let S denotes the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = S_n – k S_n-1 + S_n-2 then k =
(a) 1
(b) 2
(c) 3
(d) None of these
Solution:
(b)

Question 10.
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by

(a) S
(b) 2S
(c) 3S
(d) None of these
Solution:
(b)

Question 11.
If the sum of first n even natural number is equal to k times the sum of first n odd natural numbers, then k =

Solution:
(d)

Question 12.
If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is

Solution:
(c)

Question 13.
If S₁ is the sum of an arithmetic progression of ‘n’ odd number of terms and S₂ is the sum of the terms of the

Solution:
(a)

Question 14.
If in an A.P., S_n = n²p and S_m = m²p, where S denotes the sum of r terms of the A.P., then S_p is equal to
(a) \(\frac { 1 }{ 2 }\) p³
(b) mnp
(c) p³
(d) (m + n) p²
Solution:
(c)

Question 15.
If Sn denote the sum of the first n terms of an A.P. If S_2n = 3S_n , then S_3n : S_n is equal to
(a) 4
(b) 6
(c) 8
(d) 10
Solution:
(b)

Question 16.
In an AP, S_p = q, S_q = p and S denotes the sum of first r terms. Then, S_p+q is equal to
(a) 0
(b) – (p + q)
(c) p + q
(d) pq
Solution:
(c) In an A.P. S_p = q, S_q = p
S_p+q = Sum of (p + q) terms = Sum of p term + Sum of q terms = q + p

Question 17.
If S_n denotes the sum of the first r terms of an A.P. Then, S_3n : (S_2n – S_n) is
(a) n
(b) 3n
(b) 3
(d) None of these
Solution:
(c)

Question 18.
If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 term is
(a) 3200
(b) 1600
(c) 200
(d) 2800
Solution:
(a)

Question 19.
The number of terms of the A.P. 3, 7,11, 15, … to be taken so that the sum is 406 is
(a) 5
(b) 10
(c) 12
(d) 14
Solution:
(d)

Question 20.
Sum of n terms of the series

Solution:
(c)

Question 21.
The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is
(a) 50th
(b) 502th
(c) 508th
(d) None of these
Solution:
(d)

Question 22.

Solution:
(c)

Question 23.

Solution:
(b) S_n is the sum of first n terms
Last term nth term = S_n – S_n-1

Question 24.
The common difference of an A.P., the sum of whose n terms is S_n, is

Solution:
(a)

Question 25.

Solution:
(b)
In first A.P. let its first term be a₁ and common difference d₁
and in second A.P., first term be a₂ and common difference d₂, then

Question 26.

Solution:
(b)

Question 27.
If the first term of an A.P. is a and nth term is b, then its common difference is

Solution:
(b)

Question 28.
The sum of first n odd natural numbers is
(a) 2n – 1
(b) 2n + 1
(c) n²
(d) n² – 1
Solution:
(c)

Question 29.
Two A.P.’s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th terms is
(a) 11
(b) 3
(c) 8
(d) 5
Solution:
(d) In two A.P.’s common-difference is same
Let A and a are two A.P. ’s
First term of A is 8 and first term of a is 3
A₃₀ – a₃₀ = 8 + (30 – 1) d – 3 – (30 – 1) d
= 5 + 29d – 29d = 5

Question 30.
If 18, a, b – 3 are in A.P., the a + b =
(a) 19
(b) 7
(c) 11
(d) 15
Solution:
(d) 18, a, b – 3 are in A.P., then a – 18 = -3 – b
=> a + b = -3 + 18 = 15

Question 31.
The sum of n terms of two A.P.’s are in the ratio 5n + 4 : 9n + 6. Then, the ratio of their 18th term is

Solution:
(a)

Question 32.

Solution:
(b)

Question 33.
The sum of n terms of an A.P. is 3n² + 5n, then 164 is its
(a) 24th term
(b) 27th term
(c) 26th term
(d) 25th term
Solution:
(b)

Question 34.
If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is
(a) n (n – 2)
(b) n (n + 2)
(c) n (n + 1)
(d) n (n – 1)
Solution:
(b)

Question 35.
If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
(a) 3 : 2
(b) 3 : 1
(c) 1 : 3
(d) 2 : 3
Solution:
(b)

Question 36.
The sum of first 20 odd natural numbers is
(a) 100
(b) 210
(c) 400
(d) 420 [CBSE 2012]
Solution:
(c)

Question 37.

Solution:
(a)

Question 38.

Solution:
(c)

Question 39.
The common difference of the A.P. \(\frac { 1 }{ 2b }\) ,

Solution:
(d)

Question 40.
If k, 2k – 1 and 2k + 1 are three consecutive terms of an AP, the value of k is
(a) -2
(b) 3
(c) -3
(d) 6 [CBSE 2014]
Solution:
(b) (2k – 1) – k = (2k + 1) – (2k- 1)
2k – 1 – k = 2
=> k = 3

Question 41.
The next term of the A.P. , √7 , √28, √63, …………
(a) √70
(b) √84
(c) √97
(d) √112 [CBSE 2014]
Solution:
(d)

= √(l6 x 7)= √112

Question 42.
The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals
(a) -3
(b) 4
(c) 5
(d) 2 [CBSE 2014]
Solution:
(c) 2 (3y + 5) = 3y – 1 + 5y + 1
(If a, b, c are in A.P., b – a = c – b=> 2b = a + c)
=> 6y + 10 = 8y
=> 10 = 2y
=> y = 5

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions 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 5 Arithmetic Progressions VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
Define an arithmetic progression.
Solution:
A sequence a₁, a₂, a₃, …, an is called an arithmetic progression of then exists a constant d
Such that a₂ – a₁ = d, a₃ – a₂ = d, ………… a_n – a_n-1 = d
and so on and d is called common difference

Question 2.
Write the common difference of an A.P. whose nth term is a_n = 3n + 7.
Solution:
a_n = 3n + 7
a₁ = 3 x 1 + 7 = 3 + 7 = 10
a₂ = 3 x 2 + 7 = 6 + 7 = 13
a₃ = 3 x 3 + 7 = 9 + 7 = 16
d = a₃ – a₂ or a₂ – a₁ = 16 – 13 = 3 or 13 – 10 = 3

Question 3.
Which term of the sequence 114, 109, 104, … is the first negative term ?
Solution:
Sequence is 114, 109, 104, …..
Let a_n term be negative

Question 4.
Write the value of a₃₀ – a₁₀ for the A.P. 4, 9, 14, 19, …………
Solution:

Question 5.
Write 5th term from the end of the A.P. 3, 5, 7, 9,…, 201.
Solution:

= 3 + 190 = 193
5th term from the end = 193

Question 6.
Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.
Solution:

Question 7.
Write the nth term of an A.P. the sum of whose n terms is S_n.
Solution:
Sum of n terms = S_n
Let a be the first term and d be the common difference a_n =S_n – S_n-1

Question 8.
Write the sum of first n odd natural numbers.
Solution:

Question 9.
Write the sum of first n even natural numbers.
Solution:
First n even natural numbers are
2, 4, 6, 8, ……….
Here a = 2, d = 2

Question 10.
If the sum of n terms of an A.P. is S_n = 3n² + 5n. Write its common difference.
Solution:

Question 11.
Write the expression for the common difference of an A.P. Whose first term is a and nth term is b.
Solution:
First term of an A.P. = a
and a_n = a + (n – 1) d = b .
Subtracting, b – a = (n – 1) d
d = \(\frac { b – a }{ n – 1 }\)

Question 12.
The first term of an A.P. is p and its common difference is q. Find its 10th term. [CBSE 2008]
Solution:
First term of an A.P. (a) = p
and common difference (d) = q
a₁₀ = a + (n – 1) d
= p + (10 – 1) q = p + 9q

Question 13.
For what value of p are 2p + 1, 13, 5p – 3 are three consecutive terms of an A.P.? [CBSE 2009]
Solution:

Question 14.
If \(\frac { 4 }{ 5 }\), a, 2 are three consecutive terms of an A.P., then find the value of a.
Solution:

Question 15.
If the sum of first p term of an A.P. is ap² + bp, find its common difference.
Solution:

Question 16.
Find the 9th term from the end of the A.P. 5, 9, 13, …, 185. [CBSE 2016]
Solution:
Here first term, a = 5
Common difference, d = 9 – 5 = 4
Last term, l = 185
nth term from the end = l – (n – 1) d
9th term from the end = 185 – (9 – 1) 4 = 185 – 8 x 4 = 185 – 32 = 153

Question 17.
For what value of k will the consecutive terms 2k + 1, 3k + 3 and 5k – 1 form on A.P.? [CBSE 2016]
Solution:
(3k + 3) – (2k + 1) = (5k – 1) – (3k + 3)
3k + 3 – 2k – 1 = 5k – 1 – 3k – 3
k + 2 = 2k – 4
2k – k = 2 + 4
k = 6

Question 18.
Write the nth term of the A.P.
\(\frac { 1 }{ m }\) , \(\frac { 1 + m }{ m }\) , \(\frac { 1 + 2m }{ m }\) , ……… [CBSE 2017]
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions 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 14 Surface Areas and Volumes VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.1
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS

Question 1.
The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. What is the ratio of their volumes ?
Solution:
Radii of the bases of a cylinder and a cone = 3:4
and ratio in their heights = 2:3
Let r_1, r₂ be the radii and h₁ and h₂ be their heights
heights of the cylinder and cone respectively,

Question 2.
If the heights of two right circular cones are in the ratio 1 : 2 and the perimeters of their bases are in the ratio 3 : 4. What is the ratio of their volumes ?
Solution:
Ratio in the heights of two cones =1:2 and ratio in the perimeter of their bases = 3:4
Let r₁, r₂ be the radii of two cones and ht and h2 be their heights

Question 3.
If a cone and sphere have equal radii and equal volumes what is the ratio of the diameter of the sphere to the height of the cone ?
Solution:
Let r be the radius of a cone, then
radius of sphere = r
Let h be the height of cone
Now volume of cone = volume of sphere

Question 4.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. What is the ratio of their volumes?
Solution:
Let r and h be the radius and heights of a cone, a hemisphere and a cylinder
∴ Volume of cone =  \((\frac { 1 }{ 3 } )\) πr²h
Volume of hemisphere = \((\frac { 2 }{ 3 } )\) πr³

Question 5.
The radii of two cylinders are in the ratio 3 : 5 and their heights are in the ratio 2 : 3. What is the ratio of their curved surface areas ?
Solution:
Radii of two cylinders are in the ratio = 3:5
and ratio in their heights = 2:3
Let r₁, r₂ be the radii and h₁, h₂ be the heights of the two cylinders respectively, then

Question 6.
Two cubes have their volumes in the ratio 1 : 27. What is the ratio of their surface areas ?
Solution:
Ratio in the volumes of two cubes = 1 : 27
Let a₁ and a₂ be the sides of the two cubes respectively then volume of the first area = a₁³
and volume of second cube = a_2³

Question 7.
Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii ?
Solution:
Ratio the heights of two right circular cylinders = 1:2
Let r₁,r₂ be their radii and h₁, h₂ be their

Question 8.
If the volumes of two cones are in the ratio 1 : 4, and their diameters are in the ratio 4 : 5, then write the ratio of their weights.
Solution:
Volumes of two cones are in the ratio =1:4 and their diameter are in the ratio = 4:5
Let r₁ and r₂ be the radii and h₁ ,h₂ be their

Question 9.
A sphere and a cube have equal surface areas. What is the ratio of the volume of the sphere to that of the cube ?
Solution:
Surface areas of a sphere and a cube are equal
Let r be the radius of sphere and a be the edge of cube,

Question 10.
What is the ratio of the volume of a cube to that of a sphere which will fit inside it?
Solution:
A sphere is fit inside the cube
Side of a cube = diameter of sphere
Let a be the side of cube and r be the radius of the sphere, then

Question 11.
What is the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height ?
Solution:
Diameters (or radii), and heights of a cylinder a cone and a sphere are equal,
Let r and h be the radius and height be the cone cylinder, cone and sphere respectively, thus their volumes will be

Question 12.
A sphere of maximum volume is cut-out from a solid hemisphere of radius r. What is the ratio of the volume of the hemisphere to that of the cut-out sphere?
Solution:
r is the radius of a hemisphere, then
the diameter of the sphere which is cut out of the hemisphere will be r

Question 13.
A metallic hemisphere is melted and recast in the shape of a cone with the same base radius R as that of the hemisphere. If H is the height of the cone, then write the value of \((\frac { H }{ R } )\).
Solution:
R is the radius of a hemisphere 2

Question 14.
A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and height are in the ratio 5 : 12, write the ratio of the total surface area of the cylinder to that of the cone.
Solution:
Radius and height of a cone and a cylinder be r and h respectively

Question 15.
A cylinder, a cone and a hemisphere are of equal base and have the same height. What is the ratio of their volumes ?
Solution:
Let r and h be the radii and heights of the cylinder cone and hemisphere respectively, then
Volume of cylinder = πr²h
Volume of cone = \((\frac { 1 }{ 3 } )\) πr²h
Volume of hemisphere = \((\frac { 2 }{ 3 } )\) πr³

Question 16.
The radii of two cones are in the ratio 2 : 1 and their volumes are equal. What is the ratio of their heights ?
Solution:
Radii of two cones are in the ratio = 2:1
Let r₁, r₂ be the radii of two cones and h₁, h₂ be their heights respectively,

Question 17.
Two cones have their heights in the ratio 1 : 3 and radii 3:1. What is the ratio of their volumes ?
Solution:
Ratio in heights of two cones = 1:3
and ratio in their ratio = 3:1
Let r₁, r₂ be their radii and h₁, h₂ be their
heights, then

Question 18.
A hemisphere and a cone have equal bases. If their heights are also equal, then what is the ratio of their curved surfaces ?
Solution:
Bases of a hemisphere and a cone are equal
and their heights are also equal
Let r and h be their radii and heights
respectively
∴ r = h₁

Question 19.
If r₁ and r₂ denote the radii of the circular bases of the frustum of a cone such that r₁ > r₂ then write the ratio of the height of the cone of which the frustum is a part to the height of the frustum.
Solution:
r₁ , r₂ are the radii of the bases of a frustum and r₁ > r₂
Let h₁ be the height of cone and h₂ be the height of smaller cone
∴ Height of frustum = h₁ – h₂

Question 20.
If the slant height of the frustum of a cone is 6 cm and the perimeters of its circular bases are 24 cm and 12 cm respectively. What is the curved surface area of the frustum ?
Solution:
Slant height of a frustum (l) = 6 cm
Perimeter of upper base (P₁) = 24 cm
and perimeter of lower base (P₂) = 12 cm

Question 21.
If the areas of circular bases of a frustum of a cone are 4 cm² and 9 cm² respectively and the height of the frustum is 12 cm. What is the volume of the frustum ?
Solution:
In a frustum,
Area of upper base (A₁) = 4 cm²
and area of lower base (A₂) = 9 cm²

Question 22.
The surface area of a sphere is 616 cm². Find its radius.
Solution:
Surface area of a sphere = 616 cm²
Let r be the radius, then

Question 23.
A cylinder and a cone are of the same base radius and of same height. Find the ratio of the value of the cylinder to that of the cone. [CBSE 2009]
Solution:
Let r be the radius of the base of the cylinder
small as of cone
and let height of the cylinder = h
Then height of cone = h
∴ Volume of cylinder =  πr²h
and volume of cone = \((\frac { 1 }{ 3 } )\)  πr²h

Question 24.
The slant height of the frustum of a cone is 5 cm. If the difference between the radii of its two circular ends is 4 cm, write the height of the frustum. [CBSE 2010]
Solution:
Slant height of frustum (l) = 5 cm
Difference between the upper and lower radii = 4 cm
Let h be height and upper radius r₁ and lower radius = r₂

Question 25.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?
Solution:
Volume of hemisphere = Surface area of hemisphere (given)

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes 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 14 Surface Areas and Volumes Revision Exercise

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise

Other Exercises

• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.1
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.2
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Ex 14.3
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes VSAQS
• RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes MCQS

Question 1.
A metallic sphere 1 dm in diameter is beaten into a circular sheet of uniform thickness equal to 1 mm. Find the radius of the sheet.
Solution:
Diameter of a sphere = 1 dm = 10 cm
∴  Radius (r) = \((\frac { 10 }{ 2 } )\) = 5 cm
Volume of metal used in the sphere = \((\frac { 4 }{ 3 } )\) πr³
RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise

Question 2.
Three solid spheres of radii 3,4 and 5 cm respectively are melted and converted into a single solid sphere. Find the radius of this sphere.
Solution:
Radius of first sphere (r₁) = 3 cm

Question 3.
A spherical shell of lead, whose external diameter is 18 cm, is melted and recast into a right circular cylinder, whose height is 8 cm and diameter 12 cm. Determine the internal diameter of the shell.
Solution:
Diameter of the cylinder = 12 cm
∴  Radius (r₁) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height (h) = 8 cm
∴ Volume = πr₁²h = π(6)² x 8 cm³
= π x 36 x 8 = 288π cm³
Now volume of metal used in spherical shell = 288π cm
External diameter = 18 cm 18
∴  External radius (R) = \((\frac { 18 }{ 2 } )\) = 9 cm
Let r be the internal radius, then
Volume of the metal = \((\frac { 4 }{ 3 } )\) π (R³ – r³ )

Question 4.
A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of two well = 10 m
∴ Radius (r) = \((\frac { 10 }{ 2 } )\) = 5 m
Depth (h) = 8.4 m
∴ Volume of earth dug out = πr²h

Question 5.
 In the middle of a rectangular field measuring 30 m x 20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.
Solution:
Diameter of well = 7 m
∴ Radius (r) = \((\frac { 7 }{ 2 } )\) m
Depth (h) = 10 m
∴ Volume of earth dug out = πr²h

Question 6.
The inner and outer radii of a hollow cylinder are 15 cm and 20 cm, respectively. The cylinder is melted and recast into a solid cylinder of the same height. Find the radius of the base of new cylinder.
Solution:
Inner radius of hollow cylinder (r) = 15 cm
Outer radius (R) = 20 cm
Let h be the height of the hollow cylinder,
Then volume of metal used = πR (R² – r²)
= πh (20²- 15²) cm³
=  πh (400 – 225) cm³
= 175 πh cm³
Volume of the new cylinder = 175nh cm³
Height = h
Let R be the radius of new cylinder,
then πR²h = 175 πh
⇒ R²= 175
⇒ R = \(\sqrt { 175 } \)
= 13.2
∴ Radius = 13.2 cm

Question 7.
Two cylindrical vessels are filled with oil. Their radii are 15 cm, 12 cm and heights 20 cm, 16 cm respectively. Find the radius of a cylindrical vessel 21 cm in height, which will just contain the oil of the two given vessels.
Solution:
Radius of first cylinder (r₁) = 15 cm
and radius of second cylinder (r₂) = 12 cm
Height of the first cylinder (h₁) = 20 cm
and height of second cylinder (h₂) = 16 cm
∴ Volume of both of cylinders

Question 8.
A cylindrical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.
Solution:
Diameter of cylindrical bucket = 28 cm
∴ Radius (r) = \((\frac { 28 }{ 2 } )\) = 14 cm
Height (h) = 72 cm
∴ Volume of water filled in it = πr²h
= \((\frac { 22 }{ 7 } )\) x 14 x 14 x 72 cm³ = 44352 cm³
∴ Volume of water in rectangular tank = 44352 cm³
Length of tank (l) = 66 cm
and breadth (b) = 28 cm
Let h₁ be its height
∴  Ibh₁ = 44352
⇒  66 x 28 h₁= 44352
⇒ h₁ = \((\frac { 44352 }{ 66×28 } )\) = 24
∴Height of water in the tank = 24 cm

Question 9.
A cubic cm of gold is drawn into a wire 0.1 mm in diameter, find the length of the wire.
Solution:
Volume of solid gold = 1 cm³
Diameter of cylinderical wire = 0.1 mm

Question 10.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 3 m
Radius (r) = \((\frac { 3 }{ 2 } )\) m
Depth (h) = 14 m

Question 11.
A conical vessel whose internal radius is 10 cm and height 48 cm is full of water. Find the volume of water. If this water is poured into a cylindrical vessel with internal radius 20 cm, find the height to which the water level rises in it.
Solution:
Internal radius of the conical vessel (r₁) = 10 cm
Height (h₁) = 48 cm

Question 12.
The vertical height of a conical tent is 42 dm and the diameter of its base is 5.4 m. Find the number of persons it can accommodate if each person is to be allowed 29.16 cubic dm.
Solution:
Vertical height of conical tent (h) = 42 dm
and diameter base (b) = 5.4 dm
∴ Radius (r) = \((\frac { 5.4 }{ 2 } )\) = 2.7 dm

Question 13.
A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them.
Solution:
Let r and h be the radius and height of a circular cylinder and also of a cone, then curved surface area of the cylinder = 2πrh
and curved surface area of cone

Question 14.
A sphere of diameter 5 cm is dropped into a cylindrical vessel partly filled with water. The diameter of the base of the vessel is 10 cm. If the sphere is completely submerged, by how much will the level of water rise ?
Solution:
Diameter of sphere = 5 cm

Question 15.
A spherical ball of iron has been melted – and made into smaller balls. If the radius of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made ?
Solution:
Let the radius of larger ball = r

Question 16.
Find the depth of a cylindrical tank of radius 28 m, if its capacity is equal to that of a rectangular tank of size 28 m x 16 m x 11 m.
Solution:
Dimensions of a rectangular tank = 28m x 16m x 11m
∴ Volume = 28 x 16 x 11 m³ = 4928 m³
∴  Volume of cylindrical tank = 4928 m³
Radius of the cylindrical tank = 28 m
Let h be depth of the tank, then

Question 17.
A hemispherical bowl of internal radius IS cm contains a liquid. The liquid is to be filled into cylindrical-shaped bottles of diameter S cm and height 6 cm. How many bottles are necessary to empty the bowl? (C.B.S.E. 2001C)
Solution:
Internal radius of hemispherical bowl (r) = 15 cm

Question 18.
In a cylindrical vessel of diameter 24 cm, filled up with sufficient quantity of water, a solid spherical ball of radius 6 cm is completely immersed. Find the increase in height of water level.
Solution:
Diameter of the cylindrical vessel = 24 cm

Question 19.
 A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Solution:
Radius of hemisphere of lead (r₁) = 7 cm

Question 20.
A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4\((\frac { 2 }{ 3 } )\)cm and height 3 cm. Find the number of cones so formed. (C.B.S.E. 2004)
Solution:

Question 21.
The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross section. If the length of the wire is 108 m, find its diameter. (C.B.S.E. 1994)
Solution:
Diameter of copper sphere – 18 cm 18
Radius (R) = \((\frac { 18 }{ 2 } )\) = 9 cm 4
Volume = \((\frac { 4 }{ 3 } )\) π (R³)

Question 22.
A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Solution:
Radius of hemisphere (R) = 7 cm

Question 23.
A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained. (C.B.S.E. 2004)
Solution:
Radius of sphere (R) = 10.5 cm
∴ Volume of sphere =\((\frac { 4 }{ 3 } )\) πR³

Question 24.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1:2:3.
Solution:
Let radius of a cone, a hemisphere and a cylinder be r
and height in each case = h
∴ h = r

Question 25.
A hollow sphere of internal and external diameters 4 and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Solution:
Outer diameter of a hollow sphere = 8 cm
∴ Outer radius (R) = \((\frac { 8 }{ 2 } )\) = 4 cm
and inner diameter = 4 cm
∴ Inner radius (r)=\((\frac { 4 }{ 2 } )\) =2 cm

Question 26.
The largest sphere is carved out of a cube of the side 10.5 cm. Find the volume of the sphere.
Solution:
Side of a cube = 10.5 cm
∵ The largest sphere is carved out of the cube,
∴ Diameter of the cube = side of the cube = 10.5 cm

Question 27.
Find the weight of a hollow sphere of metal having internal and external diameters as 20 cm and 22 cm, respectively if 1 cm³ of metal weighs 21 g.
Solution:
Internal diameter of a hollow sphere = 20 cm
and external diameter = 22 cm
∴ Outer radius (R) = \((\frac { 22 }{ 2 } )\) = 11 cm 20
and inner radius = \((\frac { 20 }{ 2 } )\) = 10 cm

Question 28.
A solid sphere of radius ‘r’ is melted and recast into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm, its height 24 cm and thinkness 2 cm, find the value of ‘r’.
Solution:
Radius of solid sphere = r

Question 29.
Lead spheres of diameter 6 cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18 cm and water rises by 40 cm find the number of lead spheres dropped in the water.
Solution:
Diameter of cylindrical diameter = 18 cm

Question 30.
The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm and height 4 cm arc cut off. Find the volume of the remaining solid.
Solution:
Diameter of right solid cylinder = 7 cm

Question 31.
A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. (Use TC = 3.1416)
Solution:
Total height of the solid =108 cm
Each diameter of base of hemispherical part = 36 cm

Question 32.
The surface area of a sphere is the same as the curved surface area of a cone having the radius of the bases as 120 cm and height 160 cm. Find the radius of the sphere.
Solution:

Question 33.
 A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them.
Solution:

Question 34.
A rectangular vessel of dimensions 20 cm x 16 cm x 11 cm is full of water. This water is poured into a conical vessel. The top of the conical vessel has its radius 10 cm. If the conical vessel is filled completely, determine its height. (Use π = 22/7)
Solution:
Dimension of rectangular vessel are 20 cm x 16 cm x 11 cm
Volume of vessel = 20 x 16 x 11 cm³= 3520 cm³
∴ Volume of water in conical vessel = 3520 cm³
Radius of the top of vessel = 10 cm
Let h be its height, then

Question 35.
If r₁ and r₂ be the radii of two solid metallic spheres and if they are melted into one solid sphere, prove that the radius of the new sphere is (r₁³ + r₁³ )1/3.
Solution:

Question 36.
A solid metal sphere of 6 cm diameter is melted and a circular sheet of thickness 1 cm is prepared. Determine the diameter of the sheet.
Solution:
Diameter of solid sphere = 6 cm 6

Question 37.
A hemispherical tank full of water is \((\frac { 25 }{ 7 } )\) emptied by a pipe at the rate of  litres per second. How much time will it take to half-empty the tank, if the tank is 3 metres in diameter ?
Solution:

Question 38.
Find the number of coins, 1.5 cm is diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:

Question 39.
The radius of the base of a right circular cone of semi-vertical angle a is r. Show that its volume is \((\frac { 1 }{ 3 } )\) πr³ cot a and curved surface area is πr² cosec α.
Solution:
Radius of circular cone = r
and semi vertical angle = α
Let AO = h and slant height AC = l
In ΔAOC, AO ⊥ BC

Question 40.
An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cubic cm of iron weighs 7.5 gm.
Solution:
Diameter of cylindrical portion = 20 cm
∴ Radius (r) = \((\frac { 20 }{ 2 } )\)  = 10 cm
Height of (h₁) = 2.8 m = 280 cm
and height of cone (h₂) = 42 cm

Question 41.
A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.
Solution:
Diameter of the tent = 105 m
∴ Radius (r) = \((\frac { 105 }{ 2 } )\)  m
Height of cylindrical part (h₁) = 3m
Slant height of conical part (h₂) – 53 m

∴ Total surface area of the tent = curved surface area of the conical part + curved surface area of the cylindrical area

Question 42.
Height of a solid cylinder is 10 cm and diameter 8 cm. Two equal conical hole have been made from its both ends. If the diameter of the holes is 6 cm and height 4 cm, find (i) volume of the cylinder, (ii) volume of one conical hole, (iii) volume of the remaining solid.
Solution:
Height of the solid cylinder (h₁) = 10 cm
Diameter = 8 cm
∴Radius (r₁) = \((\frac { 8 }{ 2 } )\) = 4 cm

Question 43.
The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm, and height 4 cm are cut off. Find the volume of the remaining solid.
Solution:
Diameter of the base of a cylinder = 7 cm
∴ Radius (r₁) = \((\frac { 7 }{ 2 } )\) cm
Height of cylinder (h₁) = 15 cm

Question 44.
A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. (Useπ = 3.1416)
Solution:
Total height of the solid =108 cm
Diameter of base of each hemisphere = 36 cm

Question 45.
The largest sphere is to be curved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.
Solution:
Radius of cylinder (r) = 7 cm
and height (h) = 14 cm

The diameter of the largest sphere curved out of the given cylinder = diameter of the cylinder
= 2 x 7 = 14 cm

Question 46.
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the base of the cylinder or the cone is 24 m. The height of the cylinder is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas required for making the tent. (Use π = 22/7)
Solution:
Diameter of the base of the cone = 24 m
∴  Radius (r) = \((\frac { 24 }{ 2 } )\) = 12 m
Height of the cylindrical part (h₁) = 11 m
Total height of the tent = 16 m
Height of the conical part (h₂)
= 16- 11 = 5 m

Question 47.
Area of the canvas required for the tent = 1320 m2 47. A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm find the total surface area and volume of the toy. (C.B.S.E. 2000, 2002)
Solution:
Radius of the toy (r) = 3.5 cm
Total height of the toy = 15.5 cm
∴ Height of the conical part = 15.5 – 3.5 = 12 cm
Slant height of the conical part (l)

Question 48.
A cylindrical container is filled with ice-cream, whose diameter is 12 cm and height is 15 cm. The whole ice-cream is distributed to 10 children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of its base, find the diameter of the ice-cream.
Solution:
Diameter of the cylindrical container = 12 cm
Radius (r₁) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height (h₁) = 15 cm

Question 49.
Find the volume of a solid in the form of a right circular cylinder with hemi-spherical ends whose total length is 2.7 m and the diameter of each hemispherical end is 0.7 m.
Solution:
Total length of solid = 2.7 m
Diameter of each hemisphere at the ends = 0.7 cm

Question 50.
A tent of height 8.25 m is in the form of a right circular cylinder with diameter of base 30 m and height 5.5 m, surmounted by a right circular cone of the same base. Find the cost of the canvas of the tent at the rate of Rs. 45 per m².
Solution:
Total height of the tent = 8.25 m
Height of cylindrical part (h₁) = 5.5 m
∴ Height of conical part (h₂) = 8.25 – 5.5 = 2.75m

Question 51.
An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm³ of iron has 8 gram mass approximately. (Use π = 355/115)
Solution:
Diameter of the base of the cylindrical pole = 12 cm
∴ Radius (r) = \((\frac { 12 }{ 2 } )\) = 6 cm
Height of cylindrical portion (h₁) = 110 cm
and height of conical portion (h₂) = 9 cm

Question 52.
The interior of a building is in the form of a cylinder of base radius 12 m and height 3.5 m, surmounted by a cone of equal base and slant height 12.5 m. Find the internal curved surface area and the capacity of the building.
Solution:
Radius of the building (r) = 12m
Height of the cylindrical portion (h₁) = 3.5 m and
slant height of conical portion (l) = 12.5 m

Question 53.
A right angled triangle with sides 3 cm and 4 cm is revolved around its hypotenuse. Find the volume of the double cone thus generated.
Solution:
In right angled ΔABC, ∠B = 90°
AB = 3 cm and BC = 4 cm

Now revolving the triangle along CA,

Question 54.
A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical portion is 6 cm and its height is 4 cm. Determine the surface area of the toy (Use π = 3.14).
Solution:
Diameter of the base of the toy = 6 cm
∴ Radius (r) = \((\frac { 6 }{ 2 } )\) = 3 cm
Height of conical portion (h) = 4 cm

Question 55.
Find the mass of a 3.5 m long lead pipe, if the external diameter of the pipe is 2.4 cm, thickness of the metal is 2 mm and the mass of 1 cm3 of lead is 11.4 grams.
Solution:
External diameter of a cylindrical pipe = 2.4 cm
Radius (R) = \((\frac { 2.4 }{ 2 } )\) = 1.2 cm
Thickness of the pipe = 2 mm =\((\frac { 2 }{ 10 } )\) = 0.2 cm
∴ Inner radius (r) = 1.2 – 0.2 = 1.0 cm
Height (length) of the pipe (h) = 3.5 m
= 350 cm
Volume of the mass of the pipe = πh (R² – r²)

Question 56.
A solid is in the form of a cylinder with hemispherical ends. Total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area of the solid.
Solution:
Total height of the solid = 19 cm
Diameter of the cylinder = 7 cm
Radius (r) = \((\frac { 7 }{ 2 } )\) cm
Height of the cylinder = 19 – 2 x \((\frac { 7 }{ 2 } )\) cm
= 19-7 =12cm

Question 57.
A golf ball has diameter equal to 4.2 cm. Its surface has 200 dimples each of radius 2 mm. Calculate the total surface area which is exposed to the surroundings assuming that the dimples are hemi-spherical.
Solution:
Diameter of the golf ball = 4.2 cm
∴  Radius (R) = \((\frac { 4.2 }{ 2 } )\) =2.1 cm
Radius of each hemispherical dimples (r)  = 2 mm = \((\frac { 2 }{ 10 } )\) = \((\frac { 2 }{ 5 } )\) cm
Curved surface area of one dimple = 2πr²

Question 58.
The radii of the ends of a bucket of height 24 cm are 15 cm and 5 cm. Find its capacity. (Take π = 22/7).
Solution:
Height of the bucket (frustum) (h) = 24 cm
Upper radius (r₁) = 15 cm
and lower radius (r₂) = 5 cm

Question 59.
The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm. Find its capacity in litres and the amount of sheet required to make this bucket.
Solution:
Height of the bucket (frustum) (h) = 30 cm
Upper radius (r₁) = 21 cm
and lower radius (r₂) = 7 cm

Question 60.
The radii of the ends of a frustum of a right circular cone are 5 metres and 8 metres and its lateral height is 5 metres. Find the lateral surface and volume of the frustum.
Solution:
Upper radius of a frustum (r₁) = 8 m
and lower radius (r₂) = 5 m
Lateral height (l) = 5m

Question 61.
A frustum of a cone is 9 cm thick and the diameters of its circular ends are 28 cm and 4 cm. Find the volume and lateral surface area of the frustum. (Take π = 22/7)
Solution:
Upper diameter = 28 cm
and lower diameter = 4 cm
Height (h) = 9 cm

Question 62.
A bucket is in the form of a frustum of a cone and holds 15.25 litres of water. The diameters of the top and bottom are 25 cm and 20 cm respectively. Find its height and area of tin used in its construction.
Solution:
Water in a bucket (frustum) = 15.25l
Upper diameter = 25 cm
and lower diameter = 20 cm
∴ Upper radius (r₁) = \((\frac { 25 }{ 2 } )\) cm
and lower radius (r₂) = \((\frac { 20 }{ 2 } )\) cm =10 cm
Volume = 15.25 /= 1525 x 10 cm³ = 15250 cm³
Let h be its height

Question 63.
If a cone of radius 10 cm is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volumes of the two parts. (C.B.S.E. 2000C)
Solution:
Radius of the cone (r₁) = 10 cm
Cone is divided into 2 parts Such that PQ || AB

Question 64.
A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 pnetres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per square metre, if the radius of the base is 14 metres
Solution:
Radius of the cylinder (r) = 14 m
and total height of the tent = 13.5 m
Height of the cylindrical part (h₁) = 3 m
Height of conical part (h₂) = 13.5-3.0 = 10.5m

Question 65.
An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height be 22 cm, the diameter of the cylindrical portion 8 cm and the diameter of the top of the funnel 18 cm, find the area of the tin required. (Use : π = 22/7).
Solution:
Upper diameter of the frustum = 18 cm
and  lower diameter = 8m

Question 66.
A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys with the shape of a right circular cone mounted on a hemisphere of radius 3 cm. If the height of the toy is 12 cm, find the number of toys so formed. (C.B.S.E. 2006C)
Solution:
Diameter of solid cylinder = 12 cm
and height (h₁) = 15 cm

Question 67.
A container open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of ₹21 per litre. (Use π = 22/7)
Solution:
Upper radius (R) = 20 cm
Lower radius (r) = 8 cm
Height (h) = 24 cm

Question 68.
 A cone of maximum size is carved out from a cube of edge 14 cm. Find the . surface area of the cone and of the remaining solid left out after the cone carved out. [NCERT Exemplar]
Solution:
The cone of maximum size that is carved out from a cube of edge 14 cm will be of base radius 7 cm and the height 14 cm.

Question 69.
A cone of radius 4 cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base. Compare the volumes of two parts. [NCERT Exemplar]
Solution:
Let h be the height of the given cone. One dividing the cone through the mid-point of its axis and parallel to its base into two parts, we obtain the following figure:

Question 70.
A wall 24 m, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm x 16 cm x 10 cm. If the mortar occupies \((\frac { 1 }{ 10 } )\) th of the volume of the wall, then find the number of bricks used in constructing the wall. [NCERT Exemplar]
Solution:
Given that, a wall is constructed with the help of bricks and mortar.
∴ Number of bricks

Question 71.
A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket. [NCERT Exemplar] 
Solution:

Question 72.
Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm. [NCERT Exemplar]
Solution:

Question 73.
Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape formed. [NCERT Exemplar]
Solution:
If two cones with same base and height are joined together along their bases, then the shape so formed is look like as figure shown.

Question 74.
From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid. [NCERT Exemplar]
Solution:
Given that, side of a solid cube (a) = 1 cm
Height of conical cavity i.e., cone, h = 7 cm

Since, the height of conical cavity and the side of cube is equal that means the conical cavity fit vertically in the cube.
Radius of conical cavity i. e., cone, r = 3 cm
⇒ Diameter = 2 x r = 2 x 3 = 6 cm
Since, the diameter is less than the side of a cube that means the base of a conical cavity is not fit inhorizontal face of cube.
Now, volume of cube = (side)³ = a³ = (7)³ = 34³ cm³
and volume of conical cavity i.e., cone

Question 75.
Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacitites are 2 : 1. Find the heights and capacities of the cones. Also, find the volume of the remaining portion of the cylinder. [NCERT Exemplar]

Solution:

Question 76.
An icecream cone full of icecream having radius 5 cm and height 10 cm as shown’in the figure. Calculate the volume of icecream, provided that its 1/6 parts is left unfilled with icecream.

Solution:

 

Hope given RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise 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 5 Arithmetic Progressions Ex 5.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.1
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.4
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.5
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.6
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions VSAQS
• RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions MCQS

Question 1.
Show that the sequence defined by a_n = 5n – 7 is an A.P., find its common difference.
Solution:

Question 2.
Show that the sequence defined by a_n = 3n² – 5 is not an A.P.
Solution:

Question 3.
The general term of a sequence is given by a_n = -4n + 15. Is the sequence an A.P.? If so, find its 15th term and the common difference.
Solution:
General term of a sequence
a_n = -4n + 15

Question 4.
Write the sequence with nth term :
(i) a_n = 3 + 4n
(ii) a_n = 5 + 2n
(iii) a_n = 6 – n
(iv) a_n = 9 – 5n
Show that all of the above sequences form A.P.
Solution:

Question 5.
The nth term of an A.P. is 6n + 2. Find the common difference. [CBSE 2008]
Solution:

Question 6.
Justify whether it is true to say that the sequence, having following nth term is an A.P.
(i) a_n = 2n – 1
(ii) a_n = 3n² + 5
(iii) a_n = 1 + n + n²
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.2 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.