NCERT Solutions Class 10 Maths
Chapter – 12 (Areas Related to Circles)
The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 12 Areas Related to Circles Exercise 12.1 has been provided here to help the students in solving the questions from this exercise.
Chapter : 12 Areas Related to Circles
Exercise – 12.1
Unless stated otherwise, take π = 22/7.
1. The radii of the 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 – Let the radius of the third circle be R.
Circumference of the circle with radius R = 2πR
Circumference of the circle with radius 19 cm = 2π × 19 = 38π cm
Circumference of the circle with radius 9 cm = 2π × 9 = 18π cm
Sum of the circumference of two circles = 38π + 18π = 56π cm
Circumference of the third circle = 2πR = 56π
⇒ 2πR = 56π cm
⇒ R = 28 cm
The radius of the circle which has circumference equal to the sum of the circumferences of the two circles is 28 cm.
2. The radii of the two circles are 8 cm and 6 cm, respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.
Solution – Let the radius of the third circle be R.
Area of the circle with radius R = πR2
Area of the circle with radius 8 cm = π × 82 = 64π cm2
Area of the circle with radius 6 cm = π × 62 = 36π cm2
Sum of the area of two circles = 64π cm2 + 36π cm2 = 100π cm2
Area of the third circle = πR2 = 100π cm2
⇒ πR2 = 100π cm2
⇒ R2 = 100 cm2
⇒ R = 10 cm
Thus, 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 the 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 the circle (n-1)
∴ area of the blue region (n=3) = Area of the third circle – Area of the second circle
= π(31.5)2 – 1386 cm2
= 3118.5 – 1386 cm2
= 1732.5 cm2
∴ area of the black region (n=4) = Area of the fourth circle – Area of the third circle
= π(42)2 – 1386 cm2
= 5544 – 3118.5 cm2
= 2425.5 cm2
∴ area of the white region (n=5) = Area of the fifth circle – Area of the 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
r = 2
So, option (A) is correct, i.e., the radius of the circle is 2 units.
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