Chapter 12. AREAS RELATED TO CIRCLES

Chapter 12 . Areas related to circles

Exercise 12.1 solution complete  Exercise 12.2 solution complete  Exercise 12.3 solution complete

1. Circumference of a circle . 2. Area of a circle  . 3. The sector of a circle :  Here, OAPB is the minor sector of the circle  and OAQB is the major sector of the circle . 4. The segment of a circle :  Here, APB is the minor segment of the circle  and AQB is the major segment of the circle . 5. Length of an arc of a sector of a circle with radius r and angle with degree measure θ is  ⋅ 6. Area of a sector of a circle with radius r and angle with degree measure θ is 7. Area of segment of a circle = Area of the corresponding sector – Area of the corresponding triangle . OR Area of the segment APB = Area of the sector OAPB – Area of ∆ OAB =    – area of OAB  8. In figure :  Area of the major sector OAQB =  – Area of the minor sector OAPB and   In figure :    Area of major segment AQB =  – Area of the minor segment APB .

Class 10 Maths Chapter 12. AREAS RELATED TO CIRCLES Exercise 12.1 Solutions

Unless stated otherwise , use 

1. The radii of two circles are 19 cm and 9 cm respectively . Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles .

Solution: Let  be the radius of new circle .

 Here,  cm and cm  .

  A/Q , 

       

       

       cm 

        cm

Therefore , the radius of the new circle is 28 cm .

2. The radii of two circles are 8 cm and 6 cm respectively .Find the radius of the circle having area equal to the sum of the areas of the two circles .

Solution: Let  be the radius of new circle .

 Here,  cm and  cm  .

 A/Q,  

      

      

      

       cm

Therefore, the radius of the new circle is 10 cm .

3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold , Red , Blue , Black and White . The diameter of the region representing Gold score is 21 cm and each of the

other bands is 10.5 cm wide . Find the area of each of the five scoring regions .

   

Solution: For Gold region :

Here, ,

Area of the gold region

For Red region : Here,

Area of the red region

 

For blue region : Here,

Area of the red region

 

  

For Black region : Here,

Area of the black region

  

 

For White region : Here,

Area of the white region

 

 

4. The wheels of a car are of diameter 80 cm each . How many complete revolution does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour ?

Solution: Here,  and

The perimeter of the wheels of a car

The speed of the car

Time = 10 minutes .

The distances travel by the car = Speed of the car × time 

The number of complete revolution of the wheels of the car

5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is :

 (A) 2 units    (B)  units       (C) 4 units        (D) 7 units

Solution: Let  be the radius of the circle .

A/Q ,   

 

 

 

  or   units

(A) 2 units

Class 10 Maths Chapter 12. AREAS RELATED TO CIRCLES Exercise 12.2 Solutions

Unless stated otherwise , use  .

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60° .

Solution : Here, cm and

        

 The area of a sector AOB 

2. Find the area of a quadrant of a circle whose circumference is 22 cm .

Solution : We have, 

The area of a quadrant of a circle

3. The length of the minute hand of a clock is 14 cm . Find the area swept by the minute hand in 5 minutes .

Solution:   Here,

The angle of swept in 5 minutes

Therefore, the area swept by the minute hand in 5 minutes

4. A chord of a circle of radius 10 cm subtends a right angle at the centre . Find the area of the corresponding : (i) minor segment   (ii) major sector . ( Use )

Solution: Here,  

     

The area of the minor segment ACB

(ii)  Here,OA = OB = and

    

Area of major sector OADB

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre . Find : (i) the length of the arc   (ii) area of the sector formed by the arc    (iii) area of the segment formed by the corresponding chord .

Solution:  Here, Radius  cm and  

(i) the length of an arc ACB

(ii)  The area of the sector OACB

(iii)  Here, OA = OB = 21 cm  and

So, OAB be an equilateral  triangle .

     Area of triangle OAB

The area of the segment ACB = Area of sector  – Area of triangle

(6)  A chord of a circle of radius 15 cm subtends an angle of 60° at the centre . Find the areas of the corresponding minor and major segments of the circle . (Use and  )

Solution:  Here , radius cm ,  

      

Area of the minor segment ACB =  Area of the sector OACB – 

 

Area of major segment of a circle = Area of the circle – Area of the minor segment of a circle 

 

  

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre . Find the area of the corresponding segment of the circle .  (Use and  )

Solution: Here,  and

         

The area of the segment ACB

8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 12.11) . Find : (i) the area of that part of the field in which the horse can graze . (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m . (Use )

Solution:   Since , ABCD is a square .

So, AB = BC = CD = AD = 15 cm

     

(i)  Here, and

Therefore, the area of the field in which the horse can graze

 

(ii) Here,  ,  and

The increasing area of the field

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm . The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12 . Find : (i) the total length of the silver wire required . (ii) the area of each sector of the brooch .

Solution: (i) Here, diameter   and Radius

The circumference

 

The length of the wire required to make 5 diameters  

Therefore, the total length of the silver wire required   

(ii)  The angle of each brooch

So, the area of each sector of the brooch

10 . An umbrella  has 8 ribs which are equally spaced (see Fig. 12.13) . Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella .

Solution :  Here ,  cm

 The area between the two consecutive ribs of the umbrella

11. A car has two wipers which do not overlap . Each wiper has a blade of length 25 cm sweeping through an angle of 115° . Find the total area cleaned at each sweep of the blades .

Solution : Here ,  

The  area cleaned at each sweep of the blades

Therefore, the total area cleaned at each sweep of the blades

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km . Find the area of the sea over which the ships are warned . (Use ) .

Solution:  Here , ,  

The area of the sea over which the ships are warn

13. A round table cover has six equal designs as shown in Fig. 12.14 . If the radius of the cover is 28 cm , find the cost of making the designs at the rate of Rs 0.35 per  . (Use  )

    

Solution:  Here, and

 

Area of a design part of the round table cover

Area of six design part of the round table cover

Therefore, the cost of making the designs

14.Tick the correct answer in the following :  Area of a sector of angle  (in degrees) of a circle with radius R is :

(A)      (B)     (C)      (D) 

Solution: (D) 

Class 10 Maths Chapter 12. AREAS RELATED TO CIRCLES Exercise 12.3 Solutions 

Unless stated otherwise , use  .

1. Find the area of the shaded region in Fig.12.19, if  ,  and O is the centre of the circle .

Solution:

Solution:  Here, and

In  we have ,

Radius

Area of

Area of semicircle QPR

2. Find the area of the shaded region in Fig. 12.20 , if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and .

Solution:  Here,  , and

The area of the shaded region

3. Find the area of the shaded region in Fig. 12.21 , if ABCD is a square of side 14 cm and APD and BPC are semicircles .

     

Solution:   For square :   Since, ABCD is a square .

So, AB = BC = CD = AD = 14 cm

Area of the square ABCD

For semicircle : Here, and

Area of the two semicircle

The area of the shaded region  Area of the square ABCD  – Area of 2 semicircle

4. Find the area of the shaded region in Fig. 12.22 , where a circular arc of radius 6 cm has been draw with vertex O of an equilateral triangle OAB of side 12 cm as centre .

    

Solution:  For equilateral triangle OAB :

Here, OA = AB = OB = 12 cm and

Area of an equilateral triangle OAB

For circle : Here, and  the major angle of the sector

The area of the major sector

The area of the shaded region  area of major sector  +  area of an equilateral triangle OAB

5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23 . Find the area of the remaining portion of the square .

Solution:  For square ABCD : Here,  AB = BC = CD = AD = 4 cm ,  and

The area of the remaining portion of the square = Area of the square ABCD  – Area of the circle  – Area of 4 quadrant

6. In a circular table cover of radius 32 cm , a design is formed leaving  an equilateral triangle ABC in the middle as shown in Fig. 12.24 . Find the area of the design .

Solution:  In given figure :

 

Here,  and

In  , we have

Since,

and

Therefore, the area of the design = Area of the circular cover  – Area of an equilateral triangle ABC

7. In Fig. 12.25, ABCD is a square of side 14 cm . With centres A , B , C and D , four circles are drawn such that each circle touch externally two of the remaining three circles . Find the area of the shaded region .

 

Solution:  Since , AB = BC = CD = AD = 14 cm

 ,  and

Therefore, the area of the shaded region = Area of the square ABCD – Area of four quadrant

8.In Fig. 12.36 depicts a racing track whose left and right ends are semicircular . The distance between the two inner parallel line segments is 60 m and they are each 106 m long . If the track is 10 m wide , find : (i) the distance around the track along its inner edge . (ii) the area of the track .

Solution:  In given figure :

(i)  Here,  AD = BC = EH = FG = 106 m , AH = DE = 60 m , Diameter  and Radius

The distance around the track along its inner edge = AD + HE + 2 × circumference of the semicircle

(ii) Here, AB = CD = EF = HG = 10 m ,  and

The area of the track = 2 × area of rectangle ABCD (= EFGH) + 2 × Area of the semicircular part

9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle . If   , find the area of the shaded region .

Solution:  Here, , ,   and

Therefore. the area of the shaded region = Area of circle with radius   + Area of semicircle with radius 7 cm  – Area of  triangle ACB

10. The area of an equilateral triangle ABC is 17320.5 cm² . With each vertex of the triangle as centre , a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28) . Find the area of the shaded region . (Use )

Solution:  Since, ABC is an equilateral triangle .

So,

We know that, Area of an equilateral triangle

A/Q,  

Here,  and

The area of the shaded region = Area of an equilateral triangle ABC  – Area of 3 sector

11. On a square handkerchief , nine circular designs each of radius 7 cm are made (see Fig. 12.29) . Find the area of the remaining portion of the handkerchief .

    

Solution:   Here, Radius  and Diameter

Since, ABCD is a square .

So, AB = BC = CD = AD = 3 × 14 = 42 cm

The area of the remaining portion of the handkerchief  = Area of the square ABCD  –  Area of 9 circular designs .

12. In Fig. 12.30 , OACB is a quadrant of a circle with centre O and radius 3.5 cm . If  cm , find the area of the (i) quadrant OACB ,  (ii) shaded region.

   

Solution:   Here,  ,

(i) The area of the quadrant OACB

(ii) Area of the shaded region = Area of the quadrant OACB  –  Area of the triangle OBD

13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ . If  cm, find the area of the shaded region . (Use )  

Solution:  Given,  be a square .

So ,  cm

Area of  square OABC  cm²

In  , we have

 cm

Radius  cm

Area of the quadrant OPBQ

Therefore, the area of shaded region

14. AB and CD are respectively arcs of two concentric circles of radii 21cm and 7 cm and centre O (see Fig. 12.32) . If , find the area of the shaded region.

Solution:  In given figure :

   

Here,  , and

The area of the shaded region

15. In Fig. 12.33 , ABC is a quadrant of a  circle of radius 14 cm and a semicircle is drawn with BC as diameter . Find the area of the shaded region .

    

Solution:   Here,

In  , we have

Again, ,

The area of the shaded region = Area of the triangle ABC + Area of the semicircle  – Area of the quadrant

16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each .

     

Solution:  Here, AB = BC = CD = AD = 8 cm

Radius

The area of the designed region = Area of 2 quadrant  – Area of the square ABCD