Class 10 NCERT Solutions- Chapter 12 Areas Related to Circles - Exercise 12.3 | Set 2

Question 11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig.). Find the area of the remaining portion of the handkerchief.

Solution:

Side of square=6*Radius

                       =6*7

                      =42 cm

Area of shaded region=Area of square - Area of 9 circles

=side*side - 9πr²

=42*42 - 9*22/7*7*7

=1764-1386

=378 cm²

The area of the remaining portion of the handkerchief =378 cm²

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Question 12. In Fig., OACB is a quadrant of a circle with Centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) quadrant OACB,                       (ii) shaded region.

Solution:

(i) Area of shaded region=Area of quadrant

=1/4πr²

=1/4*22/7*3.5*3.5

=38.5/4

=9.625 cm²

(ii) Area of shaded region=Area of quadrant -Area of ∆BOD

=9.625-1/2*BO*OD

=9.625-1/2*3.5*2

=9.625-3.5

=6.125 cm²

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Question 13. In Fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Solution:

By Pythagoras theorem,

OB²=DA²+AB²

BO²=(20)²+(20)²

OB²=400+400

OB²=800

OB=√800

OB=√(2*2*2*2*5*5)

OB=2*2*5√2

OB=20√2

Area of shaded region=Area of quadrant -Area of square

=1/4πr² - side*side

=1/4*3.14*20√2*20√2-20*20

=1/4*3.14*800-400

=1*3.14

=22cm²

=1/4*3.14  

The area of shaded region is =1/4*3.14  

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Question 14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig.). If ∠AOB = 30°, find the area of the shaded region.

Solution:

Area of shaded region=Area of sector AOB-Area of sector COD

=θ/(360°) πR²-θ/(360°) πr²

=θ/(360°) π[R²-r²]

=30°/360*22/7[(21)²-(7)²]

=1/12*22/7*28*14

=308/3cm²

=102.66cm²  

The area of shaded region 102.66cm²

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Question 15. In Fig., ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution:

Area of segment=Area of quadrant -Area of ∆BAC

=1/4πr²-1/2*AC*AB

=1/4*22/7*14*14-1/2*14*14

=11*14-98

=154-98

=56 cm²

Semicircle R=?

In rt. ∆BAC, By Pythagoras theorem,

BC²=AB²+BC²

BC²= (14)²+(14)²

BC=√((14)²+(14)²)

BC=√((14)²[1+1] )

BC=14√2

∴Diameter of semicircle=14√2cm  

then  radius R of semicircle=14√2/2=7√2cm

Area of semicircle =1/2πR²

=1/2*22/7*7√2*7√2

=22*7

=154 cm²

Area of shaded region=Area of semicircle-Area of segment

                                  =154-56 cm²

                                  =98 cm²

The area of shaded region is =98cm²

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Question 16. Calculate the area of the designed region in Fig. common between the two quadrants of circles of radius 8 cm each.

Solution:

Area of design=Area of 2 quadrant -Area of square

=2*1/4πr²-side*side

=1/2*22/7*8*8-8*8

=704/7-64

=100.57-64

=36.57cm²

Area designed region in figure is 36.57cm²

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Summary

Exercise 8.1 introduces the fundamental concepts of trigonometric ratios in right-angled triangles. It covers the definitions and calculations of sine, cosine, tangent, cosecant, secant, and cotangent for acute angles. Students learn to identify the opposite, adjacent, and hypotenuse sides relative to a given angle in a right triangle, and use these to compute the six trigonometric ratios. The exercise also explores the relationships between these ratios and their values for specific angles (0°, 30°, 45°, 60°, and 90°).