Areas Related to Circles – Exercise 12.1 – Class 10

  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:

Radius (r1) of 1st circle = 19 cm

Radius (r2) or 2nd circle = 9 cm

Let the radius of 3rd circle be r.

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

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

Circumference of 3rd circle = 2πr

Given that, Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle 2πr = 38π + 18π = 56π

r = 56Π/= 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.


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:

Radius (r1) of 1st circle = 8 cm

Radius (r2) of 2nd circle = 6 cm

Let the radius of 3rd circle be r.

Area of 1st circle = πr12 = π(8)2 = 64 π

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

Given that, Area of 3rd circle = Area of 1st circle + Area of 2nd circle

πr2 = πr12 + πr22

πr2 = π64 + π36

r2 = 100

r = ± 10

However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm.


3. Given figure 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[𝑈𝑠𝑒 𝜋 = 22/7 ].

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Solution:

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Radius (r­1) of gold region (i.e., 1st circle) = 21/2 = 10.5 cm

Given that each circle is 10.5cm wider than the precious circle.

Therefore, radius (r2) of 2nd circle = 10.5 + 10.5 = 21 cm

Radius (r3) of 3nd 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 1st circle = πr12 = π(10.5)2 = 346/5 cm2

Area of red region = area of 2nd circle – area of 1st circle

= πr22 – πr12

= π(21)2 – π(10.5)2

= 441π – 110.25 π = 330.75π

= 1039.5 cm2

Area of blue region = area of 3rd circle – area of 2nd circle

= πr32 – πr22

= π(31.5)2 – π(21)2

= 992.25π – 441π = 551.25π

= 1732.5 cm2

Area of black region = area of 4th circle – area of 3nd circle

= πr42 – πr32

= π(42)2 – π(31.5)2

= 1764π – 992.25π = 771.75π

= 2425.5 cm2

Area of white region = area of 5rd circle – area of 4nd circle

= πr52 – πr42

= π(52.5)2 – π(42)2

= 2756.25π – 1764π = 992.25π

= 3118.5 cm2

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


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 speed of 66 km per hour?

Solution:

Diamater oof 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 = (66×100000)/60 cm/min = 1100000 cm/min

Distance travelled by the 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) = Distance travelled in 10 minutes

n x 80π = 1100000

n = (1100000×7)/(80×22) = 35000/8 = 4375

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


  1. 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) 2units

(b) n units

(c ) 4 units

(d) 7 units

Solution:

Let the radius of the circle be r.

Circumference of circle = 2nr

Given that, the circumference of the circle and the area of the circle are equal, then 2πr = πr2

2 = r

Therefore, the radius of the circle is 2 units

Hence (a) is the answer.



 

 

 

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