Access Answers for SAT Maths Chapter 12 – Areas Related to Circles
Exercise: 12.1
1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a
circumference equal to the sum of the circumferences of the two circles.
Solution:
The radius of the 1st circle = 19 cm (given)
∴ Circumference of the 1st circle = 2π×19 = 38π cm
The radius of the 2nd circle = 9 cm (given)
∴ Circumference of the 2nd circle = 2π×9 = 18π cm
So,
The sum of the circumference of two circles = 38π+18π = 56π cm
Now, let the radius of the 3rd circle = R
∴ The circumference of the 3rd circle = 2πR
It is given that sum of the circumference of two circles = circumference of the 3 rd circle
Hence, 56π = 2πR
Or, R = 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 of 1st circle = 8 cm (given)
∴ Area of 1st circle = π(8)2 = 64π
Radius of 2nd circle = 6 cm (given)
∴ Area of 2nd circle = π(6)2 = 36π
So,
The sum of 1st and 2nd circle will be = 64π+36π = 100π
Now, assume that the radius of 3rd circle = R
∴ Area of the circle 3rd circle = πR2
It is given that the area of the circle 3rd circle = Area of 1st circle + Area of 2nd circle
Or, πR2 = 100πcm2
R2 = 100cm2
So, R = 10cm
,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:
The radius of 1st circle, r1 = 21/2 cm (as diameter D is given as 21 cm)
So, area of gold region = π r12 = π(10.5)2 = 346.5 cm2
Now, it is given that each of the other bands is 10.5 cm wide,
So, the radius of 2nd circle, r2 = 10.5cm+10.5cm = 21 cm
Thus,
∴ Area of red region = Area of 2nd circle − Area of gold region = (πr22−346.5) cm2
= (π(21)2 − 346.5) cm2
= 1386 − 346.5
= 1039.5 cm2
Similarly,
The radius of 3rd circle, r3 = 21 cm+10.5 cm = 31.5 cm
The radius of 4th circle, r4 = 31.5 cm+10.5 cm = 42 cm
The Radius of 5th circle, r5 = 42 cm+10.5 cm = 52.5 cm
For the area of nth region,
A = Area of circle n – Area of circle (n-1)
∴ Area of blue region (n=3) = Area of third circle – Area of second circle
= π(31.5)2 – 1386 cm2
= 3118.5 – 1386 cm2
= 1732.5 cm2
∴ Area of black region (n=4) = Area of fourth circle – Area of third circle
= π(42)2 – 1386 cm2
= 5544 – 3118.5 cm2
= 2425.5 cm2
, ∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle
= π(52.5)2 – 5544 cm2
= 8662.5 – 5544 cm2
= 3118.5 cm2
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?
Solution:
The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm)
So, the circumference of wheels = 2πr = 80 π cm
Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm
It is given that the distance covered by the car in 1 hr = 66km
Converting km into cm we get,
Distance covered by the car in 1hr = (66×105) cm
In 10 minutes, the distance covered will be = (66×105×10)/60 = 1100000 cm/s
∴ Distance covered by car = 11×105 cm
Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the
wheels)
=( 11×105)/80 π = 4375.
5. Tick the correct Solution: 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
Solution:
Since the perimeter of the circle = area of the circle,
2πr = πr2
Or, r = 2
So, option (A) is correct i.e. the radius of the circle is 2 units.
Exercise: 12.1
1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a
circumference equal to the sum of the circumferences of the two circles.
Solution:
The radius of the 1st circle = 19 cm (given)
∴ Circumference of the 1st circle = 2π×19 = 38π cm
The radius of the 2nd circle = 9 cm (given)
∴ Circumference of the 2nd circle = 2π×9 = 18π cm
So,
The sum of the circumference of two circles = 38π+18π = 56π cm
Now, let the radius of the 3rd circle = R
∴ The circumference of the 3rd circle = 2πR
It is given that sum of the circumference of two circles = circumference of the 3 rd circle
Hence, 56π = 2πR
Or, R = 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 of 1st circle = 8 cm (given)
∴ Area of 1st circle = π(8)2 = 64π
Radius of 2nd circle = 6 cm (given)
∴ Area of 2nd circle = π(6)2 = 36π
So,
The sum of 1st and 2nd circle will be = 64π+36π = 100π
Now, assume that the radius of 3rd circle = R
∴ Area of the circle 3rd circle = πR2
It is given that the area of the circle 3rd circle = Area of 1st circle + Area of 2nd circle
Or, πR2 = 100πcm2
R2 = 100cm2
So, R = 10cm
,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:
The radius of 1st circle, r1 = 21/2 cm (as diameter D is given as 21 cm)
So, area of gold region = π r12 = π(10.5)2 = 346.5 cm2
Now, it is given that each of the other bands is 10.5 cm wide,
So, the radius of 2nd circle, r2 = 10.5cm+10.5cm = 21 cm
Thus,
∴ Area of red region = Area of 2nd circle − Area of gold region = (πr22−346.5) cm2
= (π(21)2 − 346.5) cm2
= 1386 − 346.5
= 1039.5 cm2
Similarly,
The radius of 3rd circle, r3 = 21 cm+10.5 cm = 31.5 cm
The radius of 4th circle, r4 = 31.5 cm+10.5 cm = 42 cm
The Radius of 5th circle, r5 = 42 cm+10.5 cm = 52.5 cm
For the area of nth region,
A = Area of circle n – Area of circle (n-1)
∴ Area of blue region (n=3) = Area of third circle – Area of second circle
= π(31.5)2 – 1386 cm2
= 3118.5 – 1386 cm2
= 1732.5 cm2
∴ Area of black region (n=4) = Area of fourth circle – Area of third circle
= π(42)2 – 1386 cm2
= 5544 – 3118.5 cm2
= 2425.5 cm2
, ∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle
= π(52.5)2 – 5544 cm2
= 8662.5 – 5544 cm2
= 3118.5 cm2
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?
Solution:
The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm)
So, the circumference of wheels = 2πr = 80 π cm
Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm
It is given that the distance covered by the car in 1 hr = 66km
Converting km into cm we get,
Distance covered by the car in 1hr = (66×105) cm
In 10 minutes, the distance covered will be = (66×105×10)/60 = 1100000 cm/s
∴ Distance covered by car = 11×105 cm
Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the
wheels)
=( 11×105)/80 π = 4375.
5. Tick the correct Solution: 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
Solution:
Since the perimeter of the circle = area of the circle,
2πr = πr2
Or, r = 2
So, option (A) is correct i.e. the radius of the circle is 2 units.
