RD Sharma Maths Solutions Class 10

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 13 Areas Related to Circles Ex 13.3

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 13.3

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.
AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm and divides the circle into two segments. Find the area of the minor segment.
Solution:
Radius of the circle (r) = 4 cm
Length of the chord AB = 4 cm
∴  In ΔOAB
OA = OB = AB    (each = 4 cm)

Question 2.
A chord PQ of length 12 cm subtends an angle of 120° at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.
Solution:
Length of chord PQ = 12 cm
Angle at the centre (θ) = 120°
∵  Draw OD ⊥ DQ
which bisects PQ at D and also bisects ∠POQ

Question 3.
A chord of a circle of radius 14 cm makes a right angle at the centre. Find the areas of the minor and major segments of the circle.
Solution:
Radius of the circle (r) = 14 cm
Angle at the centre (θ) = 90°

Question 4.
A ehord 10 cm long is drawn in a circle whose radius is 5\(\sqrt { 2 } \)
cm. Find area of both the segments. (Take π = 3.14).
Solution:
Radius of the circle (r) = 5\(\sqrt { 2 } \)  cm
And length of chord AB = 10 cm

Question 5.
A chord AB of a circle, of radius 14 cm makes an angle of 60° at the centre of the circle. Find the area of the minor segment of the circle. (Use π = 22/7)
Solution:
Radius of the circle (r) – 14 cm
Angle at the centre subtended in the fnui
AB = 60°

Question 6.
Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°. [NCERT Exemplar]
Solution:
Given that, radius of circle (r) = 14 cm

Question 7.
A chord of a circle of radius 20 cm subtends an angle of 90° at the centre. Find the area of the corresponding major segment of the circle. (Use π = 3.14) [NCERT Exemplar]
Solution:
Let AB be the chord of a circle of radius 10 cm,
with O as the centre of the circle.

Question 8.
The radius of a circle with centre O is 5 cm (see figure). Two radii OA and OB are drawn at right angles to each other. Find the areas of the segments made by the chord AB (Take π = 3.14).

Solution:
Radius of the circle (r) = 5 cm
∵   OA and OB are at right angle
∴ ∠AOB = 90°
∵  Chord AB makes two segments which are minor segment and major segment Now area of minor segment

Question 9.
AB is the diameter of a circle, centre O. C is a point on the circumference such that ∠COB = 0. The area of the minor segment cut off by AC is equal to twice the area of the sector BOC. Prove that

Solution:

Question 10.
A chord of a circle subtends an angle of 0 at the centre of the circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle. Prove that

Solution:
Let chord AB subtends angle 0 at the centre
of a circle with radius r
Now area of the circle = nr¹
and area of the minor segment ACB

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1

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

Other Exercises

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

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

Solution:
∴ Perpendicular BC – 2 units and
Hypotenuse AC = 3 units
By Phythagoras Theorem, in AABC,
(Hypotenuse)² = (Base)² + (Perpendicular)²
AC² = AB² + BC²
⇒ (3)² = (AB)² + (2)²
⇒ 9 = AB² + 4 ⇒ AB² = 9-4 = 5
AB = √5 units

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

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

Solution:

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

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

Question 6.
In ΔPQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R.
Solution:

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

Solution:

Question 8.
If 3 cot A = 4, check whether \(\frac { 1-{ tan }^{ 2 }A }{ 1+{ tan }^{ 2 }A }\) = cos² A – sin² A or not.
Solution:

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

Question 10.
If 3 tan θ = 4, find the value of 4cos θ – sin θ \(\frac { 4cos\theta -sin\theta }{ 2cos\theta +sin\theta }\).
Solution:

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

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

Solution:

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

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

Question 15.

Solution:

Question 16.

Solution:

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

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

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

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

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

Question 24.

Solution:

Question 25.

Solution:

Question 26.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
∠A and ∠B are acute angles and cos A = cos B
Draw a right angle AABC, in which ∠C – 90°

Question 27.
In a ∆ABC, right angled at A, if tan C =√3 , find the value of sin B cos C + cos B sin C. (C.B.S.E. 2008)
Solution:

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

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

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

Solution:

Question 33.

Solution:

Question 34.

Solution:

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

Question 36.
If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P.
Solution:
∠A and ∠P are acute angles and tan A = tan P Draw a right angled AAPB in which ∠B = 90°

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

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

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

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

Other Exercises

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

Question 1.
Find the distance between the following pair of points :
(i) (-6, 7) and (-1, -5)
(ii) (a + b, b + c) and (a – b, c – b)
(iii) (a sin α, -b cos α) and (-a cos α, -b sin α)
(iv) (a, 0) and (0, b)
Solution:

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

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

Question 4.
Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.
Solution:
Distance between (x, y) and (-3, 0) is

Question 5.
The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.
Solution:
Let the ordinate of other end by y, then The distance between (2, -3) and (10, y) is

Question 6.
Show that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices points of a rectangle. (C.B.S.E. 2006C)
Solution:

AB = CD and AD = BC
and diagonal AC = BD
ABCD is a rectangle

Question 7.
Show that the points A (1, -2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.
Solution:

Question 8.
Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square. [NCERT]
Solution:
Vertices A (1, 7), B (4, 2), C (-1,-1), D (-4, 4)
If these are the vertices of a square, then its diagonals and sides are equal

Question 9.
Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle. (C.B.S.E. 2006C)
Solution:

Question 10.
Prove that (2, -2), (-2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.
Solution:

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

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

Question 13.
The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC. [NCERT Exemplar]
Solution:
Given that, the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ∆ABC right angled at B.
By Pythagoras theorem, AC² = AB² + BC² ………(i)
Now, by distance formula,

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

Question 15.
Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.
Solution:
Two vertices of an isosceles ∆ABC are A (2, 0) and B (2, 5). Let co-ordinates of third vertex C be (x, y)

Question 16.
Which point on x-axis is equidistant from (5, 9) and (-4, 6) ?
Solution:
Let co-ordinates of two points are A (5, 9), B (-4, 6)
The required point is on x-axis
Its ordinates or y-co-ordinates will be 0
Let the co-ordinates of the point C be (x, 0)
AC = CB

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

Now AB + BC = 2√5 +2√5
and CA = 4√5
AB + BC = CA
A, B and C are collinear

Question 18.
The co-ordinates of the point P are (-3,2). Find the co-ordinates of the point Q which lies on the line joining P and origin such that OP = OQ.
Solution:
Co-ordinates of P are (-3, 2) and origin O are (0, 0)
Let co-ordinates of Q be (x, y)
O is the mid point of PQ


= 9 + 4 = (±3)² + (±2)²
The point will be in fourth quadrant
Its y-coordinates will be negative
and x-coordinates will be positive
Now comparing the equation
x² = (±3)² => x = ±3
y² = (±2)² => y = ±2
x = 3, y = -2
Co-ordinates of the point Q are (3, -2)

Question 19.
Which point on y-axis is equidistant from (2, 3) and (-4, 1) ?
Solution:
The required point lies on y-axis
Its abscissa will be zero
Let the point be C (0, y) and A (2, 3), B (-4, 1)
Now,

Question 20.
The three vertices of a parallelogram are (3, 4), (3, 8) and (9, 8). Find the fourth vertex.
Solution:
Let ABCD be a parallelogram and vertices will be A (3, 4), B (3, 8), C (9, 8)

Question 21.
Find a point which is equidistant from the point A (-5, 4) and B (-1, 6). How many such points are there? [NCERT Exemplar]
Solution:
Let P (h, k) be the point which is equidistant from the points A (-5, 4) and B (-1, 6).
PA = PB


So, the mid-point of AB satisfy the Eq. (i).
Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the point A and B.
Replacing h, k, by x, y in above equation, we have 2x + y + 1 = 0

Question 22.
The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10√2 units. [NCERT Exemplar]
Solution:
By given condition,
Distance between the centre C (2a, a-1) and the point P (11, -9), which lie on the circle = Radius of circle

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

Question 24.
Find the value of k, if the point P (0, 2) is equidistant from (3, k) and (k, 5).
Solution:
Let P (0, 2) is equidistant from A (3, k) and B (k, 5)
PA = PB
=> PA² = PB²

Question 25.
If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the
(i) interior,
(ii) exterior of the triangle. [NCERT Exemplar]
Solution:
Let the third vertex of an equilateral triangle be (x, y).
Let A (-4, 3), B (4,3) and C (x, y).
We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal.



But given that, the origin lies in the interior of the ∆ABC and the x-coordinate of third vertex is zero.
Then, y-coordinate of third vertex should be negative.
Hence, the require coordinate of third vertex,
C = (0, 3 – 4√3). [C ≠ (0, 3 + 4√3)]

Question 26.
Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.
Solution:
Let the co-ordinates of the vertices A, B, C and D of a rhombus are A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4)

Question 27.
Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.
Solution:
Let ABC is a triangle whose vertices are A (3, 0), B (-1, -6) and C (4, -1)

Question 28.
Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4). [CBSE 2005]
Solution:
The required point is on x-axis
Its ordinate will be O
Let the co-ordinates of the required point P (x, 0)
Let the point P is equidistant from the points A (7, 6) and B (-3, 4)

Question 29.
(i) Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square. [CBSE 2004]
(ii) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD. [CBSE 2013]
(iii) Name the type of triangle PQR formed by the point P(√2 , √2), Q(- √2, – √2) and R (-√6 , √6 ). [NCERT Exemplar]
Solution:

Question 30.
Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3). [CBSE 2004]
Solution:
The point P lies on x-axis
The ordinates of P will be 0 Let the point P be (x, 0)
Let P is equidistant from A (-2, 5) and B (2, -3)

Question 31.
Find the value of x such that PQ = QR where the co-ordinates of P, Q and R are (6, -1) (1, 3) and (x, 8) respectively. [CBSE 2005]
Solution:

Question 32.
Prove that the points (0, 0), (5, 5) and (-5, 5) are the vertices of a right isosceles triangle. [CBSE 2005]
Solution:
Let the vertices of a triangle be A (0, 0), B (5, 5) and C (-5, 5)

Question 33.
If the points P (x, y) is equidistant from the points A (5, 1) and B (1,5), prove that x = y. [CBSE 2005]
Solution:

Question 34.
If Q (0, 1) is equidistant from P (5, -3) and R (x, 6) find the values of x. Also find the distances QR and PR. [NCERT]
Solution:
Q (0, 1) is equidistant from P (5, -3) and R (x, 6)
PQ = RQ

Question 35.
Find the values ofy for which the distance between the points P (2, -3) and Q (10, y) is 10 units. [NCERT]
Solution:
Distance between P (2, -3) and Q (10, y) = 10

Question 36.
If the point P (k – 1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k. [CBSE 2014]
Solution:
Point P (k – 1, 2) is equidistant from A (3, k) and B (k, 5)
PA= PB

Question 37.
If the point A (0, 2) is equidistant from the point B (3, p) and C (p, 5), find p. Also, find the length of AB. [CBSE 2014]
Solution:
Point A (0, 2) is equidistant from B (3, p) and C (p, 5)
AB = AC

Question 38.
Name the quadrilateral formed, if any, by the following points, and give reasons for your answers :
(i) A (-1, -2), B (1, 0), C (-1, 2), D (-3, 0)
(ii) A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)
(iii) A (4, 5), B (7, 6), C (4, 3), D (1, 2)
Solution:

Question 39.
Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3, 5).
Solution:
Let the given points are A (7, 1) and B (3, 5) and mid point be M

Question 40.
Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order, form a rhombus. Also find its area.
Solution:

Question 41.
In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at A (3, 1), B (6, 4) and C (8, 6). Do you think they are seated in a line ?
Solution:
A (3, 1), B (6, 4) and C (8, 6)

Question 42.
Find a point ony-axis which is equidistant from the points (5, -2) and (-3, 2).
Solution:
The point lies on y-axis
Its x = 0
Let the required point be (0, y) and let A (5, -2) and B (-3, 2)

Question 43.
Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4). [NCERT]
Solution:

Question 44.
If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p. [CBSE 2012]
Solution:
Point A (0, 2) is equidistant from the points B (3, p) and C (p, 5)
AB = AC

Question 45.
Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle. [CBSE 2013]
Solution:
Let points are A (7, 10), B (-2, 5) and C (3, -4)

Question 46.
If the point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8), find the value of x and And the distance AP. [CBSE 2014]
Solution:
Point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8)
PA = PB

Question 47.
If A (3, y) is equidistant from points P (8, -3) and Q (7, 6), find the value of y and find the distance AQ. [CBSE 2014]
Solution:
Point A (3, y) is equidistant from P (8, -3) and Q (7, 6)
i.e., AP = AQ

Question 48.
If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex. [CBSE 2014]
Solution:
Let A (0, -3) and B (0, 3) are vertices of an equilateral triangle
Let the coordinates of the third vertex be C (x, y)

Question 49.
If the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3), find k. Also, find the length of AP.
Solution:
Point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3)
AP = BP
Now,

Question 50.
Show that ∆ABC, where A (-2, 0), B (2, 0), C (0, 2) and ∆PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar.
Solution:

Question 51.
An equilateral triangle has two vertices at the points (3, 4), and (-2, 3). Find the co-ordinates of the third vertex.
Solution:
Let two vertices of an equilateral triangle are A (3,4), and B (-2,3) and let the third vertex be C (x, y)

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

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

Question 54.
Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).
Solution:

Question 55.
If two opposite vertices of a square are (5, 4) and (1, -6), find the coordinates of its remaining two vertices.
Solution:
Two opposite points of a square are (5, 4) and (1, -6)
Let ABCD be a square and A (5, 4) and C (1, -6) are the opposite points
Let the co-ordinates of B be (x, y). Join AC

Question 56.
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3).
Solution:
Let O is the centre of the circle is (x, 7) Join OA, OB and OC

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

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2 are helpful to complete your math homework.

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