NCERT Solutions for Class 10 Maths Exercise 12.1 Chapter 12 Areas Related To Circles - Free PDF Download (2024)

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

NCERT Solutions involve detailed solutions for all Chapter 12 questions. Get free NCERT Solutions for Maths Chapter 12, Exercise 12.1, at one place, crafted by subject experts according to the NCERT guidelines. Class 10 Mathematics Chapter 12 Area Related to Circles Exercise 12.1 Questions with solutions that allow you to revise the full curriculum and score more.

In order to score good marks in the CBSE Class 10 examination, students must practise the solutions provided. Students will therefore find it extremely easy to understand the questions and how to go about solving the NCERT Solutions.

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.1

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Access Answers to NCERT Class 10 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 1^st circle = 19 cm (given)

∴ Circumference of the 1^st circle = 2π×19 = 38π cm

The radius of the 2^nd circle = 9 cm (given)

∴ Circumference of the 2^nd circle = 2π×9 = 18π cm

So,

The sum of the circumference of two circles = 38π+18π = 56π cm

Now, let the radius of the 3^rd circle = R

∴ The circumference of the 3^rd 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 1^st circle = 8 cm (given)

∴ Area of 1^st circle = π(8)² = 64π

Radius of 2^nd circle = 6 cm (given)

∴ Area of 2^nd circle = π(6)² = 36π

So,

The sum of the 1^st and 2^nd circle will be = 64π+36π = 100π

Now, assume that the radius of 3^rd circle = R

∴ Area of the circle 3^rd circle = πR²

It is given that the area of the circle 3^rd circle = Area of 1^st circle + Area of 2^nd 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 1^st 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 2^nd circle, r₂ = 10.5cm+10.5cm = 21 cm

Thus,

∴ Area of red region = Area of 2^nd 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 3^rd circle, r₃ = 21 cm+10.5 cm = 31.5 cm

The radius of 4^th circle, r₄ = 31.5 cm+10.5 cm = 42 cm

The radius of 5^th circle, r₅ = 42 cm+10.5 cm = 52.5 cm

For the area of n^th 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 a 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 the 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.

Access Other Exercise Solutions of Class 10 Maths Chapter 12 Areas Related to Circles

Exercise 12.2 Solutions : 14 Solved Questions

Exercise 12.3 Solutions: 16 Solved Questions

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.1

NCERT Class 10 Maths Solutions for Chapter 12 Ex 12.1 exercise mainly deals with the perimeter and area of a circle. In this exercise, students learn how to find the perimeter and area of a circle using formulas and relate to day-to-day life. Solutions provide an overview of the main concepts in the chapter and help students to get well-versed in these topics.