# NCERT Solutions for Class 10 Maths Chapter 12

NCERT Solutions for class 10 Maths Chapter 12 Exercise 12.3, Exercise 12.2 & Exercise 12.1 (प्रश्नावली 12.1, प्रश्नावली 12.2 & प्रश्नावली 12.3) of Areas related to Circles in Hindi and English Medium PDF form for UP Board & CBSE Education to free download. Download Offline Apps based on latest NCERT Solutions for class 10 all subjects prepared according to Latest CBSE Syllabus for the academic year 2019-20.

 Class 10: Maths – गणित Chapter 12: Areas Related to Circles

## NCERT Solutions for class 10 Maths Chapter 12

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### Class 10 Maths – Areas Related to Circles – Solutions

#### Previous Years Questions

##### Three marks questions
1. 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 of various figures:

1. Perimeter of a rectangle = 2 (length + breadth)
2. Area of a rectangle = length × breadth
3. Perimeter of a square = 4 × side
4. Area of a square = (side) × (side)
5. Area of a parallelogram = base × corresponding altitude
6. Area of a triangle = 1/2 × base × corresponding altitude
7. Area of a rhombus = 1/2 × product of its diagonals
8. Area of a trapezium = 1/2 × (sum of the two parallel sides) × distance between them
9. Circumference of a circle = 2 × π × radius
11. Area of a sector = angle/360 × π × (radius) × (radius)
12. Length of an arc = angle/360 × 2π × (radius)
13. Perimeter of a sector = angle/360 × 2π × (radius) + 2(radius)
14. 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.

#### 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.

Radius of first circle (r1) = 19 cm,
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.

#### दो वृत्तों की त्रिज्याएँ क्रमशः 8 cm और 6 cm हैं। उस वृत्त की त्रिज्या ज्ञात कीजिए जिसका क्षेत्रफल इन दोनों वृत्तों की क्षेत्रफलों के योग के बराबर है।

पहले वृत्त की त्रिज्या (r1) = 8 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।

#### 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

Let the radius of circle = r, Circumference of circle (perimeter) = 2πr, Area of circle = πr2
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.

#### 15 m भुजा वाले एक वर्गाकार घास के मैदान के एक कोने पर लगे खूंटे से एक घोड़े को 5 m लंबी रस्सी से बाँध दिया गया है। ज्ञात कीजिए: मैदान के उस भाग का क्षेत्रफल जहाँ घोडा चार सकता है।

मैदान का वह भाग जहाँ घोडा चार सकता है, एक त्रिज्यखंड, जिसका कोण 90° है।
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

#### त्रिज्या 21 cm वाले एक वृत्त का एक चाप केंद्र पर 60° का कोण अंतरित करता है। चाप की लंबाई ज्ञात कीजिए।

वृत्त की त्रिज्या = 21 cm
चाप की लंबाई = θ/(360°) × 2πr
= (60°)/(360°) × 2πr
= 1/6 × 2 × π × 21
= 1/6 × 2 × 22/7 × 21
= 22 cm