Areas Related To Circles Class 10th Mathematics CBSE Solution

Class 10th Mathematics CBSE Solution
Exercise 12.1
  1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of…
  2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the…
  3. Fig. 12.3 depicts an archery target marked with its five scoring areas from the…
  4. The wheels of a car are of diameter 80 cm each. How many complete revolutions…
  5. Tick the correct answer in the following and justify your choice: If the…
Exercise 12.2
  1. Find the area of a sector of a circle with radius 6 cm if angle of the sector…
  2. Find the area of a quadrant of a circle whose circumference is 22 cm.…
  3. The length of the minute hand of a clock is 14 cm. Find the area swept by the…
  4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find…
  5. In a circle of radius 21 cm, an arc subtends an angle of 60 at the centre.…
  6. A chord of a circle of radius 15 cm subtends an angle of 60 at the centre. Find…
  7. A chord of a circle of radius 12 cm subtends an angle of 120 at the centre.…
  8. A horse is tied to a peg at one corner of a square shaped grass field of side…
  9. A brooch is made with silver wire in the form of a circle with diameter 35 mm.…
  10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming…
  11. A car has two wipers which do not overlap. Each wiper has a blade of length 25…
  12. To warn ships for underwater rocks, a lighthouse spreads a red colored light…
  13. A round table cover has six equal designs as shown in Fig. 12.14. If the…
  14. Tick the correct answer in the following: Area of a sector of angle p (in…
Exercise 12.3
  1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and…
  2. Find the area of the shaded region in Fig. 12.20, if radii of the two…
  3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side…
  4. Find the area of the shaded region in Fig. 12.22, where a circular arc of…
  5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm…
  6. In a circular table cover of radius 32 cm, a design is formed leaving an…
  7. In Fig. 12.25, ABCD is a square of side 14 cm. With centers A, B, C and D, four…
  8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.…
  9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O)…
  10. The area of an equilateral triangle ABC is 17320.5cm2. With each vertex of the…
  11. On a square handkerchief, nine circular designs each of radius 7 cm are made…
  12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm.…
  13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm,…
  14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7…
  15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle…
  16. Calculate the area of the designed region in Fig. 12.34 common between the two…

Exercise 12.1
Question 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.


Answer:

Given: Radius (r1) of 1st circle = 19 cm

Radius (r2) or 2nd circle = 9 cm


Now,


Let the radius of 3rd circle be r


Circumference of 1st circle = 2πr1 = 2π (19) = 38π cm


Circumference of 2nd circle = 2πr2 = 2π (9) = 18π cm


Circumference of 3rd circle = 2πr


Given:


Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle


r = 38π + 18π = 56π cm


r = 


= 28


Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm


Question 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.


Answer:

Radius (r1) of 1st circle = 8 cm

Radius (r2) of 2nd circle = 6 cm


Let the radius of 3rd circle be r


Area of 1stcircle = r12 = π (8)2 = 64π


Area of 2nd circle = πr22 = π (6)2 = 36 π


Given that,


Area of 3rd circle = Area of 1st circle + Area of 2nd circle


πr2 = πr12 + πr22


πr2 = 64 π + 36 π


πr2 = 100 π


r2 = 100


r = 


But we know that the radius cannot be negative. Hence, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm



Question 3.

Fig. 12.3 depicts an archery target marked with its five scoring areas 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.



Answer:

Radius (r1) of gold region (i.e., 1st circle) =  = 10.5 cm

Given:


Each circle is 10.5 cm wider than the previous circle.


Hence,


Radius (r2) of 2nd circle = 10.5 + 10.5


21 cm


Radius (r3) of 3rd circle = 21 + 10.5= 31.5 cm


Radius (r4) of 4th circle = 31.5 + 10.5= 42 cm


Radius (r5) of 5th circle = 42 + 10.5= 52.5 cm


Area of gold region = Area of 1stcircle


= πr12


= π (10.5)2


= 346.5 cm2


Area of red region = Area of 2nd circle - Area of 1stcircle


= πr22 – πr12


= π (21)2 – π (10.5)2


= 441π – 110.25π


= 330.75π


= 1039.5 cm2


Area of blue region = Area of 3rdcircle - Area of 2ndcircle


= πr32 – πr22


= π (31.5)2 – π (21)2


= 992.25 π - 441 π


= 551.25 π


= 1732.5 cm2


Area of black region = Area of 4thcircle - Area of 3rd circle


= πr42 – πr32


= π (42)2 – π (31.5)2


= 1764 π – 992.25 π


= 771.75 π


= 2425.5 cm2


Area of white region = Area of 5thcircle - Area of 4thcircle


= πr52 – πr42


= π (52.5)2 – π (42)2


= 2756.25 π - 1764 π


= 992.25 π


= 3118.5 cm2


Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm2, 1039.5 cm2, 1732.5 cm2, 2425.5 cm2, and 3118.5 cm2 respectively


Question 4.

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


Answer:


The diameter of the wheel of the car = 80 cm

Radius (r) of the wheel of the car = 40 cm

Circumference of wheel = 2πr

⇒ 2π (40) = 80π cm

Speed of car = 66 km/hour

1 km = 1000 m and 1m = 100 cm 
⇒ 1 km = 100000 cm

= 110000 cm/min
As distance= speed × time

Distance traveled by car in 10 minutes

= 110000 × 10 
= 1100000 cm

Let the number of revolutions of the wheel of the car be "n".

n × Distance travelled in 1 revolution (i.e., circumference)
(In one revolution of the wheel, a wheel covers the distance equal to its circumference)


= Distance travelled in 10 minutes


n × 80 π = 1100000


n = 



= 4375


Therefore, each wheel of the car will make 4375 revolutions.


Question 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. π units

C. 4 units

D. 7 units


Answer:

Let the radius of the circle be r

Circumference of circle = 2πr


Area of circle = πr2


Given that, the circumference of the circle and the area of the circle are equal.


This implies 2πr = πr2


2 = r


Therefore, the radius of the circle is 2 units


Hence, the correct answer is A




Exercise 12.2
Question 1.

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


Answer:

Let OACB be a sector of the circle making 60° angle at centre O of the circle

Area of sector of angle θ =  ×πr2

Area of sector OACB = × (6)2

 cm2

Hence,

The area of the sector of the circle making 60° at the centre of the circle is cm2


Question 2.

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


Answer:

To find: Area of the quadrant of the circle
Given: Circumference of the circle 
Let the radius of the circle be r

Circumference = 22 cm

Circumference = 2πr

Therefore, 2πr = 22


Quadrant of circle means 1/4 th of the circle.


Area of such quadrant of the circle =





 cm2
Hence, The area of the quadrant of the circle is cm2


Question 3.

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


Answer:


So 15 minutes = 90º

5 min = 5 × 6°
= 30°
So 5 minutes subtend an angle of 30º.

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area swept by min hand = Area of sector




Question 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 π = 3.14)


Answer:


Let AB be the chord of the circle subtending 90° angle at centre O of the circle.



= 3.14 × 25
= 78.5 cm2

OAB is a right angles triangle, 
Area of triangle OAB = 1/2 (base x height)
Area of ΔOAB = x OA x OB


 x 10 x 10


= 50 cm2


Area of minor segment ACB = Area of minor sector OACB -Area of ΔOAB


= 78.5 - 50


= 28.5 cm2
ii) Let AB be the chord of the circle subtending 90° angle at centre O of the circle.

Area of major sector OACB =

=3/4 x 3.14 x 10 x 10

=235.5cm2


Question 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.


Answer:


Radius (r) of circle = 21 cm

The angle subtended by the given arc = 60°


Length of an arc of a sector of angle θ =


(i) Length of arc ACB = 

 × 2 × 22 × 3


= 22 cm


(ii) Area of sector OACB =


=


 ×  × 21 × 21


= 231 cm2


(iii) In ΔOAB,

As OA = OB

⇒ ∠OAB = ∠OBA (Angles opposite to equal sides are equal)


⇒ ∠OAB + ∠AOB + ∠OBA = 180°


⇒ 2∠OAB + 60° = 180°

⇒ 2∠OAB = 180° - 60° 

⇒ ∠OAB = 60°

Hence,

ΔOAB is an equilateral triangle


Area of equilateral triangle = (Side)2

⇒ Area of ΔOAB = (Side)2


 × (21)2


 cm2


Area of segment ACB = Area of sector OACB - Area of ΔOAB


= (231 - ) cm2


Question 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 π = 3.14 and = 1.73)


Answer:


Radius (r) of circle = 15 cm

Area of sector OPRQ = (60/360)×Πr2

= 1/6 × 3.14 × (15)2

= 117.75 cm2

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

2 ∠OPQ = 120°

∠OPQ = 60°

ΔOPQ is an equilateral triangle.

Area of ΔOPQ = ×(Side)2

 × (15)2

= 56.25 × 

= 97.3125 cm2

Area of segment PRQ = Area of sector OPRQ - Area of ΔOPQ

= 117.75 - 97.3125

= 20.4375 cm2

Area of major segment PSQ = Area of circle - Area of segment PRQ

= Π(15)2 – 20.4375

= 3.14 ×225 – 20.4375

= 706.5 – 20.4375

= 686.0625 cm2


Question 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 π = 3.14 and √3 = 1.73)


Answer:

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST and the angle O.

SV = VT

In ΔOVS,
As, 

 = Cos 60o

 = 

OV = 6 cm

 = Sin 60o

 = 

SV = 6√3 cm

ST = 2 × SV

= 2 × 6√3

= 12√3 cm

Area of ΔOST = × 12√3 × 6

= 36√3

= 36 × 1.73

= 62.28 cm2

Area of sector OSUT = × π × (12)2

 × 3.14 × 144

= 150.72 cm2

Area of segment SUT = Area of sector OSUT - Area of ΔOST

= 150.72 - 62.28

= 88.44 cm2


Question 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 π = 3.14)


Answer:

From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius

(i) Area that can be grazed by horse = Area of sector


 × r2


 × 3.14 × 5 × 5


= 19.625 m2


The area that can be grazed by the horse when the length of rope is 10 m long =  ×  (10)2


 × 3.14 × 100


= 78.5 m2


(ii) Increase in grazing area = Area grazed by a horse now - Area grazed previously
(78.5 - 19.625)


= 58.875 m2


Question 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



Answer:

(i) To Calculate total length of wire required to make brooch, we need to find Circumference of Brooch and the length of 5 diameters of Brooch.

Diameter of Circle = 35 mm



Radius of Circle = 35/2 mm

Circumference of Circle = 2 π r



Total Length of wire required = Circumference of Circle + 5 x Diameter of Circle 

Total Length of wire required = 110 + 5 x 35

Total Length of wire required = 285 mm

(ii)

A complete Circle subtends an angle of 360º

10 sectors = 360º

1 sector will subtend = 36º





Question 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.



Answer:



There are 8 ribs in the given umbrella. The area between two consecutive ribs is subtending = 45at the centre of the assumed flat circle

Now we know that area of a sector of a circle subtending an angle x is given by


Question 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.


Answer:


It can be seen from the figure that each blade of wiper would sweep an area of a sector of 115° in a circle with 25 cm radius

Area of sector = 



Area swept by 2 blades :


= 1254.96 cm2

=1255 cm2 (approx)


Question 12.

To warn ships for underwater rocks, a lighthouse spreads a red colored 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 π = 3.14)


Answer:


It can be seen from the figure that the lighthouse spreads light across a sector of 80° in a circle of 16.5 km radius

Area of sector OACB = × r2


 × 3.14 × 16.5 × 16.5


= 189.97 km2


Question 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 cm2. (Use  = 1.7)


Answer:


It can be concluded that these designs are segments of the circle.

Let us take segment APB.


Chord AB is a side of the regular hexagon.


And,


Each chord will substitute = 60o at the centre of the circle


In ΔOAB,


∠OAB = ∠OBA (As OA = OB)


∠AOB = 60°


∠OAB + ∠OBA + ∠AOB = 180°


2∠OAB = 180° - 60°
= 120°


∠OAB = 60°


Hence,


ΔOAB is an equilateral triangle.


Area of ΔOAB = (Side)2


 × (28)2


= 196 


= 333.2 cm2


Area of sector OAPB = × r2


 ×  × 28 × 28


 cm2


Now,


Area of segment APB = Area of sector OAPB - Area of ΔOAB


= ( - 333.2) cm2


Area of design = 6 × ( - 333.2)


= 2464 – 1999.2


= 464.8 cm2


Cost of making 1 cm2 designs = Rs 0.35


Cost of making 464.76 cm2 designs = 464.8 × 0.35


= Rs 162.68


Hence, the cost of making such designs would be Rs 162.68.


Question 14.

Tick the correct answer in the following:

Area of a sector of angle p (in degrees) of a circle with radius R is

A. 

B. 

C. 

D. 


Answer:

Area of sector of angle θ = ×R2

Where θ = angle , r = radius of circle

Here θ = p and radius = R

Area of sector of angle p = (R2)

Multiply and divide by 2,

Area of sector of angle p = ()(2R2)


Hence, (D) is the correct answer



Exercise 12.3
Question 1.

Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.


Answer:


As, ΔPQR is in the semicircle and we know angle in the semicircle is a right angle, therefore PQR is a right-angled triangle.
[ If an angle is inscribed in a semi-circle, that angle measures 90 degrees.]
Here, QR is the hypotenuse of ΔPQR as lines from two ends of diameter always make a right angle when they meet at the circumference of that circle.


Hence by Pythagoras theorem we get,
Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

QR2 = PQ2 + PR2


QR2 = 242 + 72


QR2 = 576 + 49


QR = 25 cm

Diameter = 25 cm

Or, radius = 12.5cm

Area of right angled triangle triangle = 



= 12 × 7
= 84 cm2

= 245.3125 sq cm


Hence,


Area of shaded region = Area of semicircle - area of ΔPQR
= 245.3125 – 84


Area of shaded region = 161.3125 sq cm


Question 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 ∠ AOC = 40°.



Answer:



Area of the shaded region = Area of Bigger sector – Area of Smaller Sector

If r2 and r1 are the two radii of Bigger and Smaller circles respectively, then:

Area of bigger sector = 40/360 πr22

Area of smaller sector = 40/360 πr12


Question 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.



Answer:

It can be observed from the figure that the radius of each semi-circle is 7 cm

Area of each semi-circle =r2


 *  * (7)2


= 77 cm2


Area of square ABCD = (Side)2


= (14)2


= 196 cm2


Area of the shaded region= Area of square ABCD - Area of semi-circle APD - Area of semi-circle BPC


= 196 - 77 - 77


= 196 - 154


= 42 cm2



Question 4.

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



Answer:

We know that each interior angle of an equilateral triangle is of measure 60°

Area of sector = r2


 x  x 6 x 6


 cm2


Area of triangle OAB =  x (12)2

= 36  cm2


Area of circle = πr2


 × 6 × 6


 cm2


Area of shaded region = Area of ΔOAB + Area of circle - Area of sector


 - 


= (36 + ) cm2


Question 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.


Answer:

Each quadrant is a sector of 90° in a circle of 1 cm radius

Area of each quadrant =  * r2


 *  * (1)2


 cm2


Area of square = (Side)2


= (4)2


= 16 cm2


Area of circle = πr2 = π (1)2


 cm2


Area of the shaded region = Area of square - Area of circle - 4 × Area of quadrant


= 16 -  - 4 * 


= 16 -  - 


= 16 - 



 cm2



Question 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 (shaded region).



Answer:

Radius of the circle "R" = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2

Since, AD is the median of the triangle

∴ AO = Radius of the circle = 2/3 AD [ By the property of equilateral triangle inscribed in a circle]

⇒ 2/3 AD = 32 cm

⇒ AD = 48 cm
In ΔADB,

By Pythagoras theorem,

AB= AD+ BD2

⇒ AB= 48+ (AB/2)2

⇒ AB= 2304 + AB2/4
⇒ 3/4 (AB2)= 2304

⇒ AB= 3072

⇒ AB= 32√3 cm
Area of ΔABC = √3/4 × (32√3)cm2
= 768√3 cm2

Area of circle = π R2
= 22/7 × 32 × 32 
= 22528/7 cm2

Area of the design = Area of circle - Area of ΔABC

= (22528/7 - 768√3) cm2


Question 7.

In Fig. 12.25, ABCD is a square of side 14 cm. With centers 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.



Answer:

To Find: Area of shaded region

Given: Side of square ABCD = 14 cm

Radius of circles with centers A, B, C and D = 14/2 = 7 cm


Area of shaded region = Area of square - Area of four sectors subtending right angle


Area of each of the 4 sectors is equal to each other and is a sector of 90° in a circle of 7 cm radius. So, Area of four sectors will be equal to Area of one complete circle

Area of 4 sectors = Πr2


Area of square ABCD = (Side)2


Area of square ABCD = (14)2


Area of square ABCD = 196 cm2


Area of shaded portion = Area of square ABCD - 4 × Area of each sector

= 196 – 154

= 42 cm2


Therefore, the area of shaded portion is 42 cm2


Question 8.

Fig. 12.26 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.


Answer: