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

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

Other Exercises

• 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.
A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in the figure. If AB = 60 m and BC = 28 m, find the area of the plot.

Solution:
Plot is formed of a rectangle ABCD and one semicircle on BC as diameter
∴  AB (l) = 60 m,
BC (b) = 28 m
∴ Radius of semicircle (r) = \(\frac { 1 }{ 2 }\) BC
= \(\frac { 1 }{ 2 }\) x 28 = 14m
Area of plot = area of rectangle ABCD + area of semicircle

Question 2.
A play ground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = 22/7).
Solution:
Length of rectangle (l) = 36 m 49
and width (b) = 24.5 =  \(\frac { 49 }{ 2 }\) m

Question 3.
Find the area of the circle in which a square of area 64 cm2 is inscribed. (Use π = 3.14)
Solution:
Area of square = 64 cm²
∴  Side of square = \(\sqrt { Area } \)  = \(\sqrt { 64 } \)  = 8 cm
∵ The square in inscribed in the circle
∴  Radius of the circle will be = \(\frac { 1 }{ 2 }\) diagonal of
the square (r)=\(\frac { 1 }{ 2 }\) x \(\sqrt { 2 } \)a = \(\frac { 1 }{ 2 }\) x \(\sqrt { 2 } \) x 8 cm = 4\(\sqrt { 2 } \)
∴ Area of the circle = πr²
= 3.14 x (4\(\frac { 1 }{ 2 }\) x \(\sqrt { 2 } \) x 8 cm = 4\(\sqrt { 2 } \))² cm²
= 3.14 x 32 = 100.48 cm²

Question 4.
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.
Solution:
Length of rectangular piece (l) = 20 m
and width (b) = 15m

Radius of each quadrant (r) = 3.5 m = \(\frac { 7 }{ 2 }\) m
Now area of the rectangle = l x b = 20 x 15 = 300 m²
and area of 4 quadrants = 4 x \(\frac { 1 }{ 4 }\) πr²

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


Solution:
Side of the square PQRS (a) = 4 cm
∴ Area of total square = a² = 4×4=16 cm²
Radius of each of the four quadrants at the corners = 1 cm

Question 6.
Four cows are tethered at four corners of a square plot of side 50 m, so that they just cannot reach one another. What area will be left ungrazed ? (See figure)

Solution:
side of square (a) = 50 m
∴ Area of the square field = a² = (50)² m² = 2500 m²
Radius of each quadrant (r) = \(\frac { 50 }{ 2 }\) = 25 m

Question 7.
A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20 m x 16 m, find the area of the field in which the cow can graze. [NCERT Exemplar]
Solution:
Let ABCD be a rectangular field of dimensions 20 m x 16 m.
Suppose, a cow is tied at a point A.  Let length of rope AE = 14m =r (say).

Question 8.
A calf is teid with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5 m, find the increase in area of the grassy lawn in which the calf can graze. [NCERT Exemplar]
Solution:
Let the calf be tied at the corner A of the square lawn.

Then, the increase in area = Difference of the two sectors of central angle 90° each and radii 11.5 m (6 m + 5.5 m) and 6 m, which is the shaded region in the figure.

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

Question 10.
A rectangular park is 100 m by 50 m. It is surrounded by semi-circular flower beds all round. Find the cost of levelling the semi-circular flower beds at 60 paise per square metre. (Use π = 3.14).
Solution:
Length of rectangular park (l) = 100 m
and width (b) = 50 m

Radius of each semicircular beds along the
lengths side (R) = \(\frac { 100 }{ 2 }\)  = 50 m
and radius of each semicircular beds along
the width side (r) = \(\frac { 50 }{ 2 }\) = 25 m

Question 11.
The inside perimeter of a running track (shown in the figure) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide, find the area of the track. Also find the length of the outer running track.

Solution:

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


Solution:
Join AD
ABCD is a square whose each side = 10 cm

Area of square = a² = (10)² = 100 cm²
Area of half semicircle whose radius is \(\frac { 10 }{ 2 }\) = 5

Question 13.
In the figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. (Use π = 22/7). [CBSE 2014]

Solution:
AB = 20 cm, AE = 9 cm, DE = 12 cm
∠AED = 90°

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


Solution:
Side of a square ABCD = 8 cm

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


Solution:
In the figure, ABCD is a rectangle AB = 14 cm abd BC = 7 cm

Two semicircles are draw on AD and BC as diameter and thid semicircle is drawn on Cd as diameter
Now area of rectangle ABCD = l x b = 14 x 7 = 98 cm²
Area of two semicircles on AD and BC

Question 16.
In the figure, ABCD is a rectangle, having AB = 20 cm and BC = 14 cm. Two sectors of 180° have been cut off. Calculate :
(i) the area of the shaded region.
(ii) the length of the boundary of the shaded region.

Solution:
ABCD is a rectangle whose Length AB = 20 cm
and width BC = 14 cm
∴ Area of the rectangle = l x b = 20 x 14 = 280 cm²

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

Solution:
Side of the square ABCD = 22 cm
∴  Area = (side)² = (22)² = 484 cm²
∵ The squre is divided into 5 parts equal in area
∴ Area of each part  = \(\frac { 484 }{ 5 }\) cm²

Question 18.
In the given figure, Find the area of the shaded region. (Use π = 3.14).  [CBSE 2015]

Solution:
Side of large square = 14 cm
Radius of each semicircle = \(\frac { 4 }{ 2 }\) = 2 cm
Side of square = 4 cm
Area of square =  4 x 4=16 cm²
∴ Area of semicircles = 4 x \(\frac { 1 }{ 2 }\) πr²
= 2 x 3.14 x 2 x 2
= 8 x 3.14
= 25.12 cm²
∴ Area of shaded region = Area of large square – Area of central portion
= (14)2-(16+ 25.12) cm²
= 196-41.12 cm²
= 154.88 .cm²

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

Solution:
Radius of outer quadrant (R) = 3.5 cm
and radius of inner quadrant = 2 cm
∴  Area of shaded portion
= Area of outer quadrant – area of inner quadrant

Question 20.
In the figure, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 21 cm, find the area of the shaded region.[CBSE 2013]

Solution:
In the figure, OPQ is a quadrant in which
OABC is a square OA = 21 cm
Join OB,

Question 21.
In the figure, OABC is a square of side 7 cm. If OAPC is a quadrant of a cirice with centre O, then find the area of the shaded region. (Use π = 22/7) [CBSE 2012]

Solution:
In square OABC, whose side is 7 cm, OAPC
is a quadrant
Area of square = (side)²
= (7)² = 49 cm²
and radius of quadrant = 7 cm

Question 22.
In the figure, OE = 20 cm. In sector OSFT, square OEFG is inscribed. Find the area of the shaded region.

Solution:
In the figure OSFT is a quadrant and OEFG
is a square inscribed in it
The side of the square is OE = 20 cm

Question 23.
Find the area of the shaded region in the figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. (Use π = 3.14)

Solution:
In right AABC
AB²=AC² + BC² (Pythagoras Theorem)
= (24)²+ (10)²
= 576 + 100 = 676
= (26)²
∴ AB = 26 cm
∴ Diameter of circle = 26 cm
and radius (r)= \(\frac { 26 }{ 2 }\) = 13 cm
Now area of shaded portion
= Area of semicircle – area of right triangle ABC

Question 24.
A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (see figure). Find the radius of the inscribed circle and the area of the shaded part.

Solution:
Each side of the equilateral triangle ABC (a) = 12 cm

Question 25.
In the figure, an equilateral triangle ABC of side 6 cm has been inscribed in a circle. Find the area of the shaded region. (Take π = 3.14)

Solution:

Question 26.
A circular field has a perimeter of 650 m. A square plot having its vertices on the circumference of the field is marked in the field. Calculate the area of the square plot.
Solution:
Perimeter of the circular field = 650 m

Question 27.
Find the area of a shaded region in the figure, where a circular arc of radius 7 cm has been drawn with vertex A of an equilateral triangle ABC of side 14 cm as centre, (use π = 22/7 and \(\sqrt { 3 } \)  = 1.73) [CBSE 2015]

Solution:
Radius of circular arc (r) = 7 cm
and side of equilateral AABC (a) = 14 cm and each angle = 60°

Area of shaded region
=Area of circle + Area of equilateral triangle – 2 area of sector EAF

Question 28.
A regular hexagon is inscribed in a circle. If the area of hexagon is 24\(\sqrt { 3 } \) cm², find the area of the circle. (Use π = 3.14) [CBSE 2015]
Solution:
A regular hexagon ABCDEF is inscribed in a circle
Area of hexagon = 24 \(\sqrt { 3 } \) cm²
Let r be the radius of circle
∴ Side of regular hexagon = r
Area of equilateral ΔOAB = \(\frac { \sqrt { 3 } }{ 3 }\) r² sq. units

Question 29.
ABCDEF is a regular hexagon with centre O (see figure). If the area of triangle OAB is 9 cm², find the area of :
(i) the hexagon and
(ii) the circle in which the hexagon is inscribed.

Solution:
O is the centre of the regular hexagon ABCDEF and area of the AOAB = 9 cm²
∵ By joining the vertices of the hexagon with O,
(i) We get 6 equal equilateral triangles
∴ Area of hexagon = 9 cm² x 6 = 54 cm²
(ii) Radius of the circle when this hexagon is inscribed in it will be = OB =AB  =r

Question 30.
Four equal circles, each of radius 5 cm, touch each other as shown in the figure. Find the area included between them. (Take π = 3.14).

Solution:
Radius of each circle = 5 cm
∵ The four circles touch eachother externally
∴ By joining their centres, we get a square whose side will be 5 + 5 = 10 cm

Now area of square so formed = a² = (10)² = 100 cm²
and area of 4 quadrants = 4 x \(\frac { 1 }{ 4 }\) πr²= πr²
= 3.14 x (5)2 cm² = 3.14 x 25 cm² = 78.5 cm²
∴  Area of the part included between the circles
= 100 – 78.5
= 21.5 cm²

Question 31.
Four equal circles, each of radius ‘a’ touch each other. Show that the area between them is \(\frac { 6 }{ 7 }\)a² . (Take π = \(\frac { 22 }{ 7 }\)
Solution:
Four circles each of radius ‘a’ touch each other at A, B, C and D respectively.
Their centes are P, Q, R and S respectively
By joining PQ, QR, RS, SD

Question 32.
A child makes a poster on a chart paper drawing a square ABCD of side 14 cm. She draws four circles with centre A, B, C and D in which she suggests different ways to save energy. The circles are drawn in such a way that each circle touches externally two of the three remaining circles in the given figure. In the shaded region she write a message ‘Save Energy’. Find the perimeter and area of the shaded region. (Use π = 22/7) [CBSE 2015]

Solution:
Radius of circular arc (r) = 7 cm
Side of square ABCD (a) = 14 cm

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


Solution:
Diameter of each coin = 1 cm
∴ Radius (r) = \(\frac { 1 }{ 2 }\) cm = 0.5 cm

Question 34.
Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm x 7 cm. Find the area of the remaining card board. (Use π = 22/7) [CBSE 2013]
Solution:
Length of rectangle = 14 cm
and breadth = 7 cm

Question 35.
In the figure, AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
Radius of larger circle (R) = 7 cm
and radius of smaller circle (r) =\(\frac { 7 }{ 2 }\) cm
Area of shaded portion = Area of larger circle – area of smaller circle

Question 36.
In the figure, PSR, RTQ and PAQ are three semi-circles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter of the shaded region. [CBSE 2014]

Solution:
Radius of larger semicircle (r₁) =\(\frac { 10 }{ 2 }\) = 5 cm
Radius of large semicircle = (r₂) = \(\frac { 7 }{ 2 }\) cm
Radius of small semicircle (r₃) = \(\frac { 3 }{ 2 }\) cm

Question 37.
In the figure, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.

Solution:
In the figure, two circles with centres A and B touch each-other internally at C
AC = 8 cm and AB = 3 cm
∴  BC = 8 – 3 = 5 cm
∴  Radius of bigger circle (R) = 8 cm
and of smaller circle (r) = 5 cm
∴  Area of shaded portion =Area of bigger circle – area of smaller circle

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

Solution:
A square ABCD is inscribed a circle
The side of the square = 2a
and one circle is inscribed in the square ABCD
Now diameter of the outer circle is AC

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


Solution:

Question 40.
 In the figure, ABC is a right-angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semicircle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

Solution:
In right ΔABC,
AB = 28 cm, BC = 21 cm

Question 41.
In the figure, O is the centre of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)

Solution:
A semicircle is drawn on the diameter AB
ΔACB is drawn in this semicircle in right ΔACB
AC = 12 cm and BC = 16 cm

Question 42.
In the figure, the boundary of the shaded region consists of four semi-circular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, find
(i) the length of the boundary,
(ii) the area of the shaded region.

Solution:
Diameter of the biggest semicircle = 14 cm 14
∴  Radius (R) = \(\frac { 14 }{ 2 }\) = 7 cm
Diameter of the small semicircle = 7 cm 7
∴  Radius (r₁) = \(\frac { 7 }{ 2 }\) cm
and diameter of each smaller circles = 3.5 cm

Question 43.
In the figure, AB = 36 cm and M is mid¬point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all the three circles. Find the area of the shaded region.

Solution:
In the figure, there are three semicircles one bigger and two smaller There is one small circle also
Now diameter of bigger semicircle = 36 cm
∴   Radius (R) = \(\frac { 36 }{ 2 }\) = 18 cm
and diameter of each smaller semicircle = 18 cm

Question 44.
In the figure, ABC is a right angled triangle in which ∠A = 90°, AB = 21 cm and AC = 28 cm. Semi-circles are described on AB, BC and AC as diameters. Find the area of the shaded region.

Solution:
In the figure, ABC is a right triangle in a semicircle ∠A = 90°, AB = 21 cm and AC  = 28 cm

Semicircles are drawn on BC and AC as diameters
In right ΔABC
BC² = AB²+AC² (Pythagoras Theorem)
= 21²+ 28²
= 441 + 784= 1225 = (35)²
∴  BC = 35 cm
Now radius of bigger semicircle (R) = \(\frac { 35 }{ 2 }\) cm,
of semicircle at AB = \(\frac { 21 }{ 2 }\) cm and of
semicircle on AC = \(\frac { 28 }{ 2 }\) cm = 14 cm
Now area of shaded portion
= Area of semicircle on AB as diameter + area of semicircle on AC as diameter + area of ΔABC – area of semicircle on BC as diameter

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

Solution:
Radius of the circular part of the tunnel = 2 m
i.e. OA = OB = 2 m
and ∠AOB = 90°
OD ⊥ AB
∴ D in mid-point of AB
∴ In right ΔAOB,

Question 46.
In the figure, shows a kite in which BCD is the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and ΔCEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.

Solution:
ABCD is a square with side = 42 cm
BCD is a quadrant in which ∠BCD = 90° and radius = 42 cm
ΔCEF is an isosceles right triangle in which CE = CF = 6 cm

Question 47.
In the figure, ABCD is a trapezium of area 24.5 cm2. In it, AD || BC, ∠DAB = 90°, AD = 10 cm and BC = 4 cm. If ABE is a quadrant of a circle, find the area of the shaded region. (Take π – (22/7). [ cbse2014]

Solution:
In the figure, ABCD is a trapezium
Area = 24.5 cm2, AD || BC
∠DAB = 90°, AD = 10 cm BC = 4 cm

Question 48.
In the figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and the distance between AB and DC is 14 cm. Circles of equal radii 7 cm with centres A, B, C and D have been drawn. Then, find the area of the shaded region of the figure. (Use π  = 22/7). [CBSE 2014]

Solution:
In trapezium ABCD
AB || DC
AB = 18 cm, DC = 32 cm Height = 14 cm
Radius of each at the corner of trapezium = 7 cm

Question 49.
From a thin metallic piece, in the shape of a trapezium ABCD, in which AB || CD and ∠BCD = 90°, a quarter circle BEFC is removed (see figure). Given AB = BC = 3.5 cm and DE = 2 cm, calculate the area of the remaining piece of the metal sheet.

Solution:
ABCD is a trapezium in shape in which
AB || CD and ∠BCD = 90°
AB = BC = 3.5 cm, DE = 2 cm
∴ DC = DE + EC = DE + BC = 2 + 3.5 = 5.5 cm
Now area of trapezium ABCD = \(\frac { 1 }{ 2 }\) (AB + CD) x BC (∵ BC is height)

Question 50.
In the figure, ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of radius 4 cm. Find the area of the shaded region correct upto 2 decimal places. (Take π = 3.142
and \(\sqrt { 3 } \)  = 1.732).

Solution:
In the figure, ABC is an equilateral triangle with 8 cm as side with centres A, B and C,
circular arcs drawn of radius 4 cm
Each side of ΔABC = 8 cm

Question 51.
Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by three animals. (NCERT Exemplar)
Solution:
Given that, a triangular field with the three comers of the field a cow, a buffalo and a horse are tied separately with ropes. So, each animal grazed the field in each corner of triangular field as a sectorial form.
Given, radius of each sector (r) = 7m

Question 52.
In the given figure, the side of a square is 28 cm, and radius of each circle is half of the length of the side of the square where O and O’ are centres of the circles. Find the area of shaded region. [CBSE 2017]

Solution:

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

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

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

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

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.

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

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.