# NCERT Solutions for class 10 Maths Chapter 12

NCERT Solutions for class 10 Maths Chapter 12 Areas related to Circles all Exercises in Hindi Medium and English Medium free PDF to download. You can view these NCERT Solutions online. The entire solutions is updated for Students of UP Board and CBSE Board Education for session 2020-21. As teh UP Board Students are now using NCERT Books in current session, so they can download UP Board Solutions for class 10 Maths Chapter 12 all exercise in Hindi Medium. Download NCERT Solutions Offline Apps for 2020-21 based on latest NCERT Textbooks for class 10 all subjects prepared according to Latest CBSE Syllabus for the academic year 2020-2021.

We have updated all the contents of Tiwari Academy for new academic session. Contents are made as per the curriculum changes in CBSE or UP Board. If someone is not getting the contents, please contact us for help. We will help to access NCERT Solutions or required contents.## NCERT Solutions for class 10 Maths Chapter 12

Class: 10 | Maths (English and Hindi Medium) |

Chapter 12: | Areas related to Circles |

### 10th Maths Chapter 12 Solutions

NCERT Solutions for class 10 Maths Chapter 12 Areas related to Circles all exercises are given below updated for new academic year 2020-2021 for all boards who are using NCERT Books as a course books. UP Board Students also can download UP Board Solutions for Class 10 Maths Chapter 12 in Hindi from here.

### 10th Maths Chapter 12 Exercise 12.1

### 10th Maths Chapter 12 Exercise 12.2

### 10th Maths Chapter 12 Exercise 12.3

#### Class 10 Maths Chapter 12 Solutions in Videos

#### Class 10 Maths Exercise 12.3 Solutions in Video

### NCERT Solutions for Class 10 Maths Chapter 12

- NCERT Solutions 10th Maths
- NCERT Solutions 10th Maths
- NCERT Solutions 10th Maths

#### Previous Years Questions

## Three Marks Questions

In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region. [CBSE 2017]

## Four Marks Questions

1. In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where O and O’ are the centres of the circles. Find the area of the shaded region. [CBSE 2017] 2. A chord PQ of a circle of radius 10 cm subtends an angle of 60 at the centre of circle. Find the area of major and minor segments of the circle. [CBSE 2017]

##### Objective of the Area Related to Circles

To find the circumference, area of a circle and circular paths. To derive and understand the formulae for perimeter and area of a sector of a circle. To find the perimeter and the area of a sector, using the above formulae. Finding the areas of some combinations of figures involving circles, sectors as well as triangles, squares, rectangles and solve daily life problems based on perimeters and areas of various plane figures.

###### Formulae for perimeters and areas

- Perimeter of a rectangle = 2 (length + breadth)
- Area of a rectangle = length × breadth
- Perimeter of a square = 4 × side
- Area of a square = (side) × (side)
- Area of a parallelogram = base × corresponding altitude
- Area of a triangle = 1/2 × base × corresponding altitude
- Area of a rhombus = 1/2 × product of its diagonals
- Area of a trapezium = 1/2 × (sum of the two parallel sides) × distance between them
- Circumference of a circle = 2 × π × radius
- Area of a circle = π × (radius) × (radius)
- Area of a sector = angle/360 × π × (radius) × (radius)
- Length of an arc = angle/360 × 2π × (radius)
- Perimeter of a sector = angle/360 × 2π × (radius) + 2(radius)
- Area of segment = Area of sector – area of triangle

###### Historical Facts!

- All the mathematical ideas have emerged out of daily life experiences. The first ever need of human being was counting objects. This gave rise to the idea of numbers. When the man learn to grow crops, following types of problems had to be handled:

Fencing or constructing some kind of a boundary around the field, where the crops were to be grown. - Allotting lands of different sizes for growing different crops.
- Making suitable places for storing different products grown under different crops.
- These problems led to the need of measurement of perimeters (lengths), areas and volumes, which in turn gave rise to a branch of mathematics known as Mensuration. Area related to circle is one of the part of mensuration.

### Important Questions on Class 10 Maths Chapter 12

Radius of second circle (r2) = 9 cm

Let, the radius of the third circle = r

Circumference of the first circle = 2πr1 = 2π (19) = 38π

Circumference of the second circle = 2πr2 = 2π (9) = 18π

Circumference of the third circle = 2πr

According to question,

Circumference of the third circle = Circumference of the first circle + Circumference of the second circle

⇒ 2πr = 38π + 18π

⇒ 2πr = 56π

⇒ r=56π/2π = 28

Hence, the radius of the circle, which has circumference equal to the sum of the circumferences of the two circles, is 28 cm.

दूसरे वृत्त की त्रिज्या (r2) = 6 cm

माना तीसरे वृत्त की त्रिज्या = r

पहले वृत्त का क्षेत्रफल = πr_1^2 = π(64) = 64π

दूसरे वृत्त का क्षेत्रफल = πr_2^2 = π(36) = 36π

तीसरे वृत्त का क्षेत्रफल = πr^2

प्रश्न के अनुसार,

तीसरे वृत्त का क्षेत्रफल = पहले वृत्त का क्षेत्रफल + दूसरे वृत्त का क्षेत्रफल

⇒ πr^2 = 64π + 36π

⇒ πr^2 = 100π

⇒ r = √100 =10 cm

अतः, उस वृत्त की त्रिज्या 10 cm है, जिसका क्षेत्रफल इन दोनों वृत्तों की क्षेत्रफलों के योग के बराबर है।

Given that the perimeter and area are equal in magnitude. Therefore

2πr = πr2

⇒ 2 = r

Therefore, the radius of circle is 2 units.

Hence, the option (A) is correct.

मैदान के उस भाग का क्षेत्रफल जहाँ घोडा चार सकता है

= त्रिज्यखंड OABO का क्षेत्रफल

= (90°)/(360°) × πr^2

= 1/4 × π(5)^2

= 1/4 × 3.14 × 25

= 19.625 m^2

चाप की लंबाई = θ/(360°) × 2πr

= (60°)/(360°) × 2πr

= 1/6 × 2 × π × 21

= 1/6 × 2 × 22/7 × 21

= 22 cm