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**Q.13. In Fig. 12.34, O is the centre of a circle such that diameter AB =13 cm and AC = 12 cm. ISC is joined. Find the area of the shaded region. (Take k = 3.14) [CBSE (AI) 2016] Ans.** In ΔABC, ∠ACB = 90° (Angle in the semicircle)

∴ BC

∴ BC

= 169 - 144 = 25

∴ BC = 5 cm

Area of the shaded region = area of semicircle - area of right AABC

If OP = PQ = 10 cm show that area of shaded region is 25 [CBSE (Delhi) 2016]

Ans.

⇒ APOQ is an equilateral triangle

∴ ∠POQ = 60°

Area of segment PAQM

Area of semicircle with M as centre

Area of shaded region

Ans.

∴ BC = 7 cm

Area of shaded region

Ans.

∴ Area of the circular portion

= area of circle - area of the sector

Now, area of the equilateral triangle OAB

∴ Area of shaded region = area of circular portion + area of equilateral triangle OAB

**Long Answer Type Questions**

**Q.1. 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. [Delhi 2017]Ans.** Radius of the circle = 10 cm

Central angle subtended by chord AB = 60°

Area of minor sector OACB

Area of equilateral triangle OAB formed by radii and chord

∴ Area of minor segment ACBD

= Area of sector OACB - Area of triangle OAB

= (52.38 - 43.30) cm

Area of circle = πr

∴ Area of major segment ADBE

= Area circle - Area of minor segment

= (314.28 - 9.08) cm

Ans.

AB

⇒ (3)

∴ BC = 5 cm

Now, Area of shaded region = Area of semicircle on side AB + area of semicircle on side AC - area of semicircle on side BC + area of ΔABC

Now, Area o f semicircle on side AB

Hence area of the shaded region = 6 cm

[CBSE (AI) 2017]

Ans.

Area of shaded region = area of semicircle - area of ΔCAB + area of quadrant BOD

Ans.

PA = 5√S cm = BP (Tangents from an external point are equal)

Shaded area = 43 .25 - 26 .17 = 17.08 cm

Ans.

In right ΔAOB

Perimeter of shaded region

Q.6. Find the area of the shaded region in Fig. 12.48, where are semicircles of diameter 14 cm, 3.5 cm, 7 cm and 3.5 cm respectively.

Ans.

Ans.

Total area of all sector

[Sum of angles of a quadrilateral is 360°]

From (i) and (ii),

Area of shaded region = 350 - 154 = 196 cm

Ans.

Area of the square = a

= 14

Let radius of a semicircle=x

radius of two semicircles = 2x

side of inner square = diameter of semicircle = 2x

According to figure 2x + 2x = 8

4x = 8 ⇒ x = 2 cm

⇒ Side of inner square = 4 cm

Area of unshaded region = area of inner square + 4 (Area of a semicircle)

∴ Area of shaded region = area of square - area of unshaded region

= (196-16-871) cm

= 180-8 x 3.14 = 180-25.12 = 154.88 cm

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