**NCERT Solutions Class 10 Maths Chapter 12 Area Related to Circles** – Here are all the NCERT solutions for Class 10 Maths Chapter 12. This solution contains questions, answers, images, explanations of the complete Chapter 12 titled Area Related to Circles of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 12 Area Related to Circles. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 12 Area Related to Circles in one place.

## NCERT Solutions Class 10 Maths Chapter 12 Area related to circles

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Class | 10 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 12 |

Chapter Name |
Area related to circles |

### NCERT Solutions Class 10 Maths chapter 12 Area related to circles

Class 10, Maths chapter 12, Area related to circles solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Area related to circles

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### Question & Answer

Q.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.

Ans :Radius \( \left(r_{1}\right)\) of the 1^{st}circle = 19cm Radius \( \left(r_{2}\right)\) of the 1st circle = 9 cm Let the radius of 3^{rd}circle be r. Circumference of 1^{st}circle = \( 2 \pi r_{1}=2 \pi(19)\) = \( 38 \pi\) Circumference of 2^{nd}= \( 2 \pi r_{1}=2 \pi(9)\) = \( 18 \pi\) Circumference of 3^{rd}circle =\( 2r\pi\) That given , Circumference of 3^{rd}circle = Circumference of 2^{nd}circle = Circumference of 1^{st}circle \( 2 \pi r=38\pi+18\pi=56 \pi\) \( r=\frac{56 \pi}{2 \pi}=28\) Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm.

Q.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.

Ans :Radius \( \left(r_{1}\right)\) of the 1st circle = 8 cm Radius \( \left(r_{2}\right)\) of the 1st circle = 6 cm Let the radius of 3rd circle be r. Area of 1ts circle = \( \pi r_{1}^{2}=\pi(8)^{2}=64 \pi\) Area of 1ts circle = \( \pi r_{1}^{2}=\pi(6)^{2}=36 \pi\) That given, Area of 3rd circle =Area of 2nd circle =Area of 1st circle \( \pi r^{2}=\pi r_{1}^{2}+\pi r_{2}^{2}\) \( \pi r^{2}=64 \pi+36 \pi\) \( \pi r^{2}=100 \pi\) \( r=\pm 10\) 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.

Q.3:Fig. 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.

Ans :Radius \( \left(r_{1}\right)\) of gold region (i.e. 1st circle) = \( \frac{21}{2}=10.5 \mathrm{cm}\) Given that each circle is 10.5 cm wider than the previous circle. Therefore . radius \( r_{2}\) of 2nd circle =10.5 + 10.5 =21 cm radius \( r_{3}\) of 2nd circle =21 + 10.5 =31.5 cm radius \( r_{4}\) of 2nd circle =31.5 + 10.5 =42 cm radius \( r_{5}\) of 2nd circle =42 + 10.5 =52.5 cm Area of gold region = Area of 1st circle =\( \pi r_{1}^{2}=\pi(10.5)^{2}=346.5 \mathrm{cm}^{2}\) Area of red region = Area of 2nd circle - Area of 1st circle =\( \pi r_{2}^{2}-\pi r_{1}^{2}\) =\( \pi(21)^{2}-\pi(10.5)^{2}\) =\( 441 \pi-110.25 \pi=330.75 \pi\) = \( 1039.5 \mathrm{cm}^{2}\) Area of blue region = Area of 3rd circle - Area of 2nd circle =\( \pi r_{3}^{2}-\pi r_{2}^{2}\) =\( \pi(31.5)^{2}-\pi(21)^{2}\) =\( 992.25 \pi-441 \pi=551.25 \pi\) = \( 1732.5\mathrm{cm}^{2}\) Area of black region = Area of 4th circle - Area of 3rd circle =\( \pi r_{3}^{2}-\pi r_{2}^{2}\) =\( \pi(42)^{2}-\pi(31.5)^{2}\) =\( 1764 \pi-992.25 \pi=771.75 \pi\) = \( 2425.5\mathrm{cm}^{2}\) Area of white region = Area of 5th circle - Area of 4th circle =\( \pi r_{3}^{2}-\pi r_{2}^{2}\) =\( \pi(52.5)^{2}-\pi(42)^{2}\) =\( 2756.25\pi-1764 \pi=992.25 \pi\) = \( 3118.5\mathrm{cm}^{2}\) Therefore, areas of gold, red, blue, black, and white regions are \( \ 346.5 \mathrm{cm}^{2}\) , \( 1039.5 \mathrm{cm}^{2}\) , \( 1732.5\mathrm{cm}^{2}\) , \( 2425.5\mathrm{cm}^{2}\) , \( \ 3118.5\mathrm{cm}^{2}\) respectively .

Q.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?

Ans :Diameter of the wheel of the car = 80 cm Radius (r) of the wheel of the car = 40 cm Circumference of wheel =\( 2 \pi r\) \( =2 \pi(40)=80 \pi \mathrm{cm}\) Speed of car = 66 km/hr \( =\frac{66 \times 100000}{60} \mathrm{cm} / \mathrm{min}\) \( =\frac{35000}{8}=4375\) Therefore , each wheel of the car will make 4375 revolutions.

Q.5: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) 2 units (B) π units (C) 4 units (D) 7 units

Ans :Let the radius of the circle be r. Circumference of circle = \( 2 \pi r\) Area of circle =\( \mathrm\pi r^{2}\) Given that, the circumference of the circle and the area of the circle equal. This implies \( 2 \pi r\) = \( \mathrm\pi r^{2}\) r=2 Therefore, the radius of the circle is 2 units. Hence, the correct answer is A.

## NCERT / CBSE Book for Class 10 Maths

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### All NCERT Solutions Class 10

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**Exercise Covered under Maths Book Class 10 Maths Chapter 12 NCERT Solutions**

Here on this page we have covered following exercises from NCERT Maths Book for Class 10 Maths Chapter 12 Area related to circles

- Exercise: 12.1 Class 10 Maths Solutions
- Exercise: 12.2 Class 10 Maths Solutions
- Exercise: 12.3 Class 10 Maths Solutions

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