Class 10th, Area Related Circle Exe-12.3 (Solution3)

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Maths NCERT Class 10th Chapter 12 – Areas Related to Circles

 Class 10th Maths Chapter 12 Exercise: 12.3 (Q9 to Q12)

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9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other

and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:

Radius of larger circle, R = 7 cm

Radius of smaller circle, r = 7/2 cm

Height of ΔBCA = OC = 7 cm

Base of ΔBCA = AB = 14 cm

Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm2

Area of larger circle = πR2 = 22/7 × 72 = 154 cm2

Area of larger semicircle = 154/2 cm2 = 77 cm2

Area of smaller circle = πr2 = 22/7 × 7/2 × 7/2 = 77/2 cm2

Area of the shaded region = Area of larger circle – Area of triangle – Area of larger semicircle + Area of smaller circle

Area of the shaded region = (154 – 49 – 77 + 77/2) cm2

= 133/2 cm2 = 66.5 cm2

10. The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

Solution:

ABC is an equilateral triangle.

∴ ∠A = ∠B = ∠C = 60°

There are three sectors each making 60°.

Area of ΔABC = 17320.5 cm2

⇒ √3/4 × (side)2 = 17320.5

⇒ (side)2 = 17320.5 × 4/1.73205

⇒ (side)2 = 4 × 104

⇒ side = 200 cm

Radius of the circles = 200/2 cm = 100 cm

Area of the sector = (60°/360°) × π r2 cm2

= 1/6 × 3.14 × (100)2 cm2

= 15700/3 cm2

Area of 3 sectors = 3 × 15700/3 = 15700 cm2 =

Area of the shaded region = Area of equilateral triangle ABC – Area of 3 sectors

= 17320.5 – 15700 cm2 = 1620.5 cm2

11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.

Solution:

Number of circular design = 9

Radius of the circular design = 7 cm

There are three circles in one side of square handkerchief.

∴ Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm

Area of the square = 42 × 42 cm2 = 1764 cm2

Area of the circle = π r2 = 22/7 × 7 × 7 = 154 cm2

Total area of the design = 9 × 154 = 1386 cm2

Area of the remaining portion of the handkerchief = Area of the square – Total area of the design

= 1764 – 1386 = 378 cm2

12. In Fig. 12.30, 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:

Radius of the quadrant = 3.5 cm = 7/2 cm

(i) Area of quadrant OACB = (πR2)/4 cm2

= (22/7 × 7/2 × 7/2)/4 cm2

= 77/8 cm2

(ii) Area of triangle BOD = 1/2 × 7/2 × 2 cm2

= 7/2 cm2

Area of shaded region = Area of quadrant – Area of triangle BOD

= (77/8 – 7/2) cm2 = 49/8 cm2

= 6.125 cm2