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Class 10 RD Sharma Solutions – Chapter 15 Areas Related to Circles – Exercise 15.4 | Set 3

  • Last Updated : 30 Apr, 2021

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:

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Given that,



Radius of larger circle(OA) = 7 cm

Diameter of smaller circle(OD) = 7 cm

So, the radius of smaller circle = 7/2 cm

Now we find the area of the shaded region = Area of large circle – Area of small circle

= π(7)2 – π(7/2)2

= π × 49 – π × (49/4)

=  22/7[49 – 49/4]

= 115.5cm2

Hence, the area of the shaded region 115.5 cm2

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.  

Solution:

Given that,

Diameter of semicircle PSR = 10 cm

So, radius of semicircle PSR, (r1) = 10/2 = 5 cm

Diameter of semicircle RTQ = 3 cm

So, radius of semicircle RTQ, (r2) = 3 cm = 3/2 cm

Diameter of semicircle PAQ = 7 cm

So, radius of semicircle PAQ, (r3) = 7/2 cm



Now we find the perimeter of the shaded region = Length of the arc PAQ + 

                                                                                      Length of the arc PSR +

                                                                                      Length of the arc RTQ

= πr1 + πr2 + πr3

= π(r1 + r2 + r3)

= π(5 + 3/2 + 7/2)

= π{(10 + 3 + 7)/2}

= π × 20/2  

= 10 π

= 10 × 22/7



= 10 × 3.14

= 31.4 cm

Hence, the perimeter of the shaded region is 31.4 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:

Given that, 

AC = 8 cm & AB = 3 cm

Here, AC is the radius of the bigger circle and BC is the radius of inner circle.

BC = AC – AB  

BC =  8 – 3  

BC = 5 cm

Now we find the area of shaded region = Area of bigger circle – Area of inner circle

= πR2 – πr2

= π(R2 – r2)

= 22/7 (82 – 5)

= 22/7 (64 – 25)

= 22/7 × 39

= (22 × 39)/7

= 858/7  

= 122.57 cm2

Hence, the area of shaded region is 122.57 cm2

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:

Given that,

Side of a square ABCD = 2a  

So, the diameter of incircle = side of a square = 2a  

Radius of a incircle(r) = Diameter of incircle/2  

r = 2a/2  

r = a  

Also, the diameter of circumcircle = diagonal of a square = √2 side



So, the radius of circumcircle(R) = √2 side/2

R = (√2 × 2a) /2  

R = √2a

R = √2a  

(i) Ratio of circumferences of inner circle (C1) and circumcircle (C2)

C1 : C2 = 2πr : 2πR

C1 / C2 = 2πr / 2πR

C1 / C2 = r / R

C1 / C2 = a/√2a

C1 / C2 = 1/√2

C1 : C2 = 1 : √2  

(ii) Ratio of Areas of inner circle (A1) and circumcircle (A2) :  

A1 : A2 = πr2 : πR2

A1 / A2 = πr2 / πR2

A1 / A2 = r2 /R2

A1 / A2 = a2/(√2a)2

A1 / A2 = a2/2a2

A1 / A2 = 1/2

A1 : A2 = 1 : 2  

Hence, the ratio of circumferences of inner circle (A1) and circumcircle (A2) is 1 : √2  

and Ratio of Areas of inner circle (A1) and circumcircle (A2) is 1 : 2  

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:

Given that, 

Three semicircles, A, B and C having diameter 3 cm each, and 

another semicircle E having a circle D with diameter 4.5 cm.

(i) Now area of shaded region = Area of the semicircle with diameter 9 cm – 

                                                      Area of two semicircles with radius 3 cm – 

                                                      Area of the circle with centre D + 

                                                      Area of semicircle with radius 3 cm



= 1/2 π(9/2)2 – 2 × 1/2 π(3/2)2 – π(4.5/2)2 + 1/2 π(3/2)2  

= 1/2 π(4.5)2 – π(1.5)2 – π(2.25)2 + 1/2 π(1.5)2

= 1/2 π(4.5)2 – π(1.5)2 + 1/2 π(1.5)2 – π(2.25)2 

= 1/2 π(4.5)2 –  1/2 π(1.5)2 – π(2.25)2  

= 1/2π(4.52 – 1.52) – π 2.252

= 1/2 π (20.25 – 2.25) – π × 5.0625

= 1/2 π(18) – π × 5.0625

= 9π – π 5.0625

= π(9 – 5.0625)

= π × 3.9375

= 22/7 × 3.9375

= 0.5625 × 22

= 12.375 cm2

(ii) Cost of painting 1 cm² Shaded Region = 25 p

Cost of painting 13.275 cm² Shaded Region = 25 p × 13.275

= 309.375 paise

= ₹ 309.375 /100  

= ₹ 3 (nearest rupee)

Cost of painting the Shaded Region = ₹ 3.

Hence, the area of the shaded region is12.375 cm2 and the cost of painting the Shaded Region is ₹ 3.

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:

Given that,

ABC is a right-angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm.  

AC as diameter a semicircle is drawn and with BC as radius a quarter circle is drawn

So, the area of triangle = 1/2 x 21 x 28 = 294 cm2

Now in ΔABC,

 By using Pythagoras Theorem

AC2 = 282 + 212   

AC = √1225

AC = 35

So, the radius of semi-circle = 35/2 = 17.5

Now, the area of semi-circle

= 1/2 x 3.14 x (17.5)2 = 480.8 cm2

Area of quarter circle = 1/4 x 3.14 x 212 = 346.2 cm2

Now we find the area of the shaded region = Area of Semi-circle + Area of triangle ABC – 

                                                                             Area of quarter circle

 = 294 + 480.8 – 346.2 = 428.75cm2

Hence, the area of shaded region is 428.75cm2

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:



Given that,

∆ACB is a right-angled triangle, in which AC = 12 cm, BC = 16 cm, ∠C = 90°

Now In ∆ACB, 

By using Pythagoras theorem, we get

AB2 = AC2 + BC2         

AB2 = 122 + 162

AB2 = 144 + 256

AB2 = 400

AB = √400

AB = 20 cm

So, the diameter of a semicircle = 20 cm

So, the radius of semicircle, r = 10 cm

Now we find the perimeter of a Shaded region = circumference of semicircle + AC + AB

= πr + 12 + 16

= 3.142 × 10 + 28

= 31.42 + 28

= 59.42 cm

Now we find the area of the shaded region = Area of a semicircle – Area of a right angle triangle  

= 1/2  πr2 – 1/2 × base × height

 = 1/2 × 3.142 × 102 – 1/2 × AC × BC

= 1/2  × 3.142 × 100 – 1/2 × 12 × 16

= 3.142 × 50 – 6 × 16

= 157.1 – 96  

= 61.1 cm2 

Hence, the required Perimeter of a Shaded region is 59.4 cm and area of the shaded region is 61.1 cm2.

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:

Given that,  

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.

(i) Length of the boundary = (boundary of bigger semicircle + 

                                                  boundary of smaller semicircle + 

                                                 2 × boundary of smallest semicircle)

= π(14/2) + π(7/2) + 2× π(3.5/2)

= 7π + 3.5π + 3.5π

= 7π + 7π  

= 14π  

= 14 × 22/7  



= 2 × 22

= 44 cm

Hence, the length of the boundary = 44 cm

(ii) Now we find the area of the shaded region = Area of semicircle with AB as diameter – 

                                                                                   Area of the semicircle with radius AE – 

                                                                                  Area of the semicircle with radius BC + 

                                                                                  Area of semicircle with diameter 7 cm

= 1/2 × π(14/2)2 – 1/2 × π(3.5/2)2 – 1/2 × π(3.5/2)2 + 1/2 × π(7/2)2

= 1/2π [72 – 1.752 – 1.752 + 3.52]

= 1/2 π[49 – 3.0625 – 3.0625 + 12.25]

= 1/2 π[49 – 6.125 + 12.25]

= 1/2 π [42.875 + 12.25]

= 1/2 π [55.125]

= 1/2 × 22/7 × 55.125

= 11 × 7.875

= 86.625 cm2

Hence, the required area of the shaded region is 86.625 cm2

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:

Given that,

AB = 36 cm 

 AM = BM = 1/2 × AB = 1/2 × 36 = 18 cm [M is mid-point of AB]

AM = BM = 18 cm

AP = PM = MQ = QB = 9 cm

Let us considered the radius of circle with centre C be ‘r’ i.e CR = r  

Join P to C and M to C, MC ⊥ AB

MR = AM = 18 cm

CM = MR – CR  

CM = (18 – r )………(1)

PC = PE + CE  

PC = (9 + r)…….(2)

Now In ∆ PCM,

By using Pythagoras theorem, we get

PC2 = PM2 + MC2 

(9 + r)2 = 92 + (18 – r)2

81 + r2 + 18r = 81 + 324 + r2 – 36r         [From eq (1) and (2)]

54r = 324

r = 324/54  

r = 6  

Radius of circle with C as a centre = 6 cm

Now we find the area of shaded region = Area of semicircle with diameter AB –  

                                                                      Area to semicircles with diameter AM and MB –  

                                                                      Area of circle with C as a centre

= 1/2 π(36/2)2 – 2 × 1/2 π(18/2)2 – π(6)2

= 1/2 π(18)2 – π(9)2 – π(6)2

= 1/2 π × 324 – 81π – 36π

= 162π – 81π – 36π

= 162π – 117π

= 45π cm2

Hence, the area of required shaded region is 45π cm2.



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:

Given that,

ABC is a right-angled triangle in which ∠A = 90°, AB = 21 cm and AC = 28 cm. 

To find: the area of the shaded region.

Now In right ΔABC,

By using Pythagoras theorem, we get

 BC2 = AB2 + AC2 

BC2 = 212 + 282

BC2 = 1225

BC = √1225

BC = 35 cm

Diameter BC = 35 cm

Now we find the area of shaded region, A = Area of semicircle with AC as a diameter + 

                                                                           Area of right angle ∆ ABC + 

                                                                           Area of semicircle with AB as a diameter  –  

                                                                           Area of semicircle with BC as diameter

= 1/2 π(21/2)2 + 1/2 π(28/2)2 + 1/2 × 21 × 28 – 1/2 π(35/2)2                   

= 1/2 π(21/2)2 + 1/2 π(28/2)2 – 1/2 π(35/2)² + 1/2 × 21 × 28                            

= 1/2 π [10.52 + 142 – 17.52] + 14 × 21

= 1/2 π [110.25 + 196  – 306.25] × 294

= 1/2 π [306.25 – 306.25] + 294

= 1/2 π × 0 + 294

= 0 + 294

= 294 cm2

Hence, the area of required shaded region is 294 cm2

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:

Given that,

The radius OA of the circular part = 2 m  

∠AOB = 90°

Let OM ⊥ AB.

(i) Now In ∆OAB,  

By using Pythagoras Theorem, we get

AB2 = OA2 + OB2        

AB2 = 22 + 22

AB2 = 8  

AB = √8  

AB = √4×2

AB = 2√2 cm

Here, D b e the mid point so, AD = BD = √2

So, OD2 = OA2 – AD2

= 22 – (√22)

= √2

Let the height of the tunnel to be h.

So,

The area of ∆ OAB = 1/2 × Base × height  

= 1/2 × OA × OB

1/2 × 2 × 2  

= 2  



(i) Height of the tunnel (h) = OC + OD  

h = √2 + 2

h = (2 + √2)m

(ii) Central angle of major arc, θ = 360° – 90° = 270°

Perimeter of cross-section, 

= length of the major Arc AB + AB  

= θ/360° × 2πr + 2√2

= 270°/360° × 2π × 2 + 2√2

= 3/4 × 4π + 2√2

= (3π + 2√2) m

(iii) Area of cross-section, A = θ/360° × Area of circle + area of ∆AOB

= θ/360° × πr2 + 1/2 × base × height  

= 270°/360° × π× 22 + 1/2 × 2 × 2

= 3/4 × π × 4 + 2

= (3π + 2)m

Hence, the height of the tunnel is  (2 + √2)m, 

Perimeter of cross-section is (3π + 2√2) m and 

Area of cross section is (3π + 2)m.

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:

Given :

Radius of a quadrant of a circle, r = 42 cm.  

Equal sides of an isosceles right-angled ∆ = 6 cm  

To find: the area of the shaded region.

Now we find the area of shaded region(A) = Area of quadrant + Area of isosceles ∆

A = 1/4 πr2 + 1/2 × base × height

A = 1/4 × 22/7 × 422 + 1/2 × 6 × 6

A = 1/2 × 11 × 6 × 42 + 18

A = 11 × 3 × 42 + 18

A = 33 × 42 + 18

A = 1386 + 18  

A = 1404 cm2

Hence, the area of shaded region is 1404 cm2

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). 

Solution:

Given,

Area of trapezium ABCD, A = 24.5 cm2

AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle.

Now in trapezium ABCD,

Area of the trapezium, A = 1/2 (sum of parallel sides) × perpendicular distance between the parallel sides(h)

A = 1/2 (AD + BC) × AB  

24.5 = 1/2 (10 + 4) × AB

24.5 × 2 = 14 AB   

AB = 49/14  

AB = 7/2

AB = 3.5 cm

So, the radius of the quadrant of the circle, r = AB = 3.5 cm

Area of the quadrant of the circle = 1/4 ×πr2

= (1/4) (22/7 x 3.5 x 3.5)  

= 9.625 cm2

Now we find the area of the shaded region = Area of the trapezium – Area of the quadrant of the circle

= 24.5 – 9.625

 = 14.875 cm2

Hence, the area of the shaded region is 14.875 cm2

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). 

Solution:

Given,

AB = 18 cm, DC = 32 cm, 

Distance between AB and DC(h)= 14 cm and radius of each circle(r) = 7cm

Since, AB ||DC

So, ∠A + ∠D = 180° & ∠B + ∠C = 180°  

Area of sector = (θ /360) × πr2

Area of sector with ∠A and ∠D = (180 /360) × 22/7 × 72

= 1/2 × 22 × 7 = 11 × 7 = 77 cm2

Similarly,  Area of sector with ∠B & ∠C = (180 /360) × 22/7 × 72

= 1/2 × 22 × 7 = 11 × 7 = 77 cm2

Now in trapezium ABCD,

Area of trapezium = 1/2 (sum of parallel sides) × perpendicular distance between Parallel sides(h)

= 1/2 (AB + DC) × (h)

= 1/2(18 + 32) × 14



= 1/2(50)× 14 

= 25 × 14 = 350 cm2

Now we find the area of shaded region = Area of trapezium – 

                                                                    (Area of sector with ∠B and ∠C + 

                                                                     Area of sector with ∠A and ∠D )

= 350 -(77+77) = 350 – 154 = 196 cm2

Hence, the Area of shaded region is 196 cm2

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:

Given,

In trapezium ABCD 

AB || CD and ∠BCD = 90°  

AB = BC =3.5 cm & DE = 2 cm  

CE = CB = 3.5 cm [CE and BC are the radii of quarter circle BFEC]

So, DC = DE + EC  

DC = 2 cm + 3.5 cm  

DC = 5.5 cm  

Area of remaining piece of the metal sheet (A) = Area of trapezium ABCD – Area of quarter circle BFEC  

A = 1/2(AB + DC) × BC – 1/4 x π x (BC)2

A = 1/2 (3.5 + 5.5) × 3.5  – 1/4 x π(3.5)2             

A = 1/2 × 9  × 3.5 – 1/4 x π(3.5)2

A = 4.5 × 3.5  – 22/7 × 3.5 × 3.5/4

A = 15.75 –   11 × 3.5/4

A = 15.75 – 9.625

A = 6.125 cm2

Hence, the area of remaining piece of the metal sheet (Shaded region) is 6.125 cm2

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 √3 = 1.732).

Solution:

In Equilateral triangle all the angles are each 60°.

The corners form sectors of a circle.

When we join the sectors we form a major sector with the middle angle as (60 × 3) = 180°

Area of the shaded region = Area of the triangle – area of the sector.

Area of the triangle = 1/2 × base × height

As we know that,

Base = 8/2 = 4 cm

Hypotenuse = 8 cm

Height = √82 – 42    

= √48 

= 4√3 

= 4 × 1.732 

= 6.928 cm 

Also, area of the triangle = 1/2 × 6.928 × 8 = 27.712 cm2

Now area of the Sector,

Radius of the sector = 8/2 = 4 cm

= 180/360 × 3.142 × 42 = 25.136 cm2

Now we find the area of the shaded region = 27.712 – 25.136 = 2.576 cm2

Hence, the area of the shaded region is 2.576 cm2

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. 

Solution:

Let ABC be the triangular field with sides AC = 15 m, AB = 16 m and BC = 17 m 

And,

Let the place where the buffalo, the horse and the cow are tied, 

are three sectors i.e. sector BFG, sector CHI and sector ADE 

 Area of triangular field = √s(s – a)(s – b)(s – c)   [by using Heron’s formula]

s = (a + b + c)/2

s = (15 + 16 + 17)/2

s = 48/2

s = 24 m

=√24(24 – 15)(24 – 16)(24 – 17)

=√24 x 9 x 8 x 7



=√12096

=109.98 m2

Area of triangular field = 109.98 m2

Area of the grazed part = Area of the sector ADE + Area of sector BFG + Area of sector CHI

= π x 72 x ∠A/360 + π x 72 x ∠B/360 + π x 72 x ∠C/360

= π x 72(∠ A + ∠ B + ∠ C)/360

= 22/7 x (7)2 x 180/360

= 154/2

= 77 m2

So, the area of the field which cannot be grazed by these animals  

= 109.98 m2 – 77 m2

= 32.98 m2

Hence, the area of the field which cannot be grazed by these animals is 32.98 m2

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. 

Solution:

Given that,

Side of square = 28 cm 

 Radius of each circle is half of the length of the side of the square

So, radius of each circle = 28/2 cm =14 cm

As we know that 

Area of Square = (Side)2

Area of Circle = πr2

Now we find the area of Shaded region = Area of Square +3/4 (Area of Circle) + 3/4(Area of Circle)

= (28)2 + 3/2 x 22/7 × 14 × 14

= 784 cm2 + 924 cm2

= 1708 cm2

Hence, the area of shaded region is 1708 cm2

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? 

Solution:

Given that

Diameter of cylinder (d) = 2 m

Radius of cylinder (r) = 1 m

Height of cylinder (H) = 5 m

Now we know that volume of cylindrical tank is, 

V = πr2H = π × (1)2 × 5 = 5π m

Length of the park (l) = 25 m

Breadth of park (b) = 20 m

Let us considered the height of standing water in the park = h

Volume of water in the park = l x b x h = 25 × 20 × h

Now for irrigation in the park water is used from the tank. So,

Volume of cylindrical tank = Volume of water in the park

⇒ 5π = 25 × 20 × h

⇒ 5π/25 × 20 = h

⇒ h = π/100 m

⇒ h = 0.0314 m

Hence, the height of standing water used for irrigating the park is 0.0314 m




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