NCERT Solutions involve detailed solutions of all chapter 12 questions. Get free NCERT Solutions for Maths Chapter 12, Exercise 12.1 at one place crafted by subject expert 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 first term examination, students must practise the NCERT Solution provided. Students will therefore find it extremely easy to understand the questions and how to go about solving them.

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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)

âˆ´ Circumference of the 1^{st} circle = 2Ï€Ã—19 = 38Ï€ cm

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

âˆ´ Circumference of the 2^{nd} circle = 2Ï€Ã—9 = 18Ï€ cm

So,

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

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

âˆ´ The circumference of the 3^{rd} circle = 2Ï€R

It is given that sum of the circumference of two circles = circumference of the 3^{rd} circle

Hence, 56Ï€ = 2Ï€R

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)

âˆ´ Area of 1^{st} circle = Ï€(8)^{2} = 64Ï€

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

âˆ´ Area of 2^{nd} circle = Ï€(6)^{2} = 36Ï€

So,

The sum of 1^{st} and 2^{nd} circle will be = 64Ï€+36Ï€ = 100Ï€

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

âˆ´ Area of the circle 3^{rd} circle = Ï€R^{2}

It is given that the area of the circle 3^{rd} circle = Area of 1^{st} circle + Area of 2^{nd} circle

Or, Ï€R^{2} = 100Ï€cm^{2}

R^{2} = 100cm^{2}

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, r_{1} = 21/2 cm (as diameter D is given as 21 cm)

So, area of gold region = Ï€ r_{1}^{2 }= Ï€(10.5)^{2 }= 346.5 cm^{2}

Now, it is given that each of the other bands is 10.5 cm wide,

So, the radius of 2^{nd} circle, r_{2} = 10.5cm+10.5cm = 21 cm

Thus,

âˆ´ Area of red region = Area of 2^{nd} circle âˆ’ Area of gold region = (Ï€r_{2}^{2}âˆ’346.5) cm^{2}

= (Ï€(21)^{2} âˆ’ 346.5) cm^{2}

= 1386 âˆ’ 346.5

= 1039.5 cm^{2}

Similarly,

The radius of 3^{rd} circle, r_{3} = 21 cm+10.5 cm = 31.5 cm

The radius of 4^{th} circle, r_{4} = 31.5 cm+10.5 cm = 42 cm

The Radius of 5^{th} circle, r_{5} = 42 cm+10.5 cm = 52.5 cm

For the area of n^{th }region,

A = Area of circle n â€“ Area of circle (n-1)

âˆ´ Area of blue region (n=3) = Area of third circle â€“ Area of second circle

= Ï€(31.5)^{2} â€“ 1386 cm^{2}

= 3118.5 â€“ 1386 cm^{2 }

= 1732.5 cm^{2}

âˆ´ Area of black region (n=4) = Area of fourth circle â€“ Area of third circle

= Ï€(42)^{2} â€“ 1386 cm^{2 }

= 5544 â€“ 3118.5 cm^{2 }

= 2425.5 cm^{2}

âˆ´ Area of white region (n=5) = Area of fifth circle â€“ Area of fourth circle

= Ï€(52.5)^{2} â€“ 5544 cm^{2 }

= 8662.5 â€“ 5544 cm^{2 }

= 3118.5 cm^{2}

**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 carâ€™s wheel = 80/2 = 40 cm (as D = 80 cm)

So, the circumference of wheels = 2Ï€r = 80 Ï€ cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 Ï€ 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 = (66Ã—10^{5}) cm

In 10 minutes, the distance covered will be = (66Ã—10^{5}Ã—10)/60 = 1100000 cm/s

âˆ´ Distance covered by car = 11Ã—10^{5} cm

Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels)

=( 11Ã—10^{5})/80 Ï€ = 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) Ï€ units **

**(C) 4 units **

**(D) 7 units**

**Solution:**

Since the perimeter of the circle = area of the circle,

2Ï€r = Ï€r^{2}

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

This 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 with these topics.