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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^{2} + AC^{2} = AB^{2}
∴ BC^{2} = AB^{2} AC^{2}
= 169  144 = 25
∴ BC = 5 cm
Area of the shaded region = area of semicircle  area of right AABC
Q.14. In Fig. 12.35, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre 0 and radius OP while arc PBQ is a semicircle drawn on PQ as diameter with centre M.
If OP = PQ = 10 cm show that area of shaded region is 25 [CBSE (Delhi) 2016]
Ans. Since OP = PQ = QO
⇒ APOQ is an equilateral triangle
∴ ∠POQ = 60°
Area of segment PAQM
Area of semicircle with M as centre
Area of shaded region
Q.15. In figure, the boundary of shaded region consists of four semicircular arcs, two smallest being equal. If diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the area of the shaded region. [Foreign 2016]
Ans. Given AD = 14 cm, AB = CD = 3.5 cm
∴ BC = 7 cm
Area of shaded region
Q.16. Find the area of the shaded region in Fig. 12.23, where a circular arc of radius 6 cm has been drawn with vertex 0 of an equilateral triangle ΔOAB of side 12 cm as centre. [NCERT, CBSE (F) 2016 ]
Ans. We have, radius of circular region = 6 cm and each side of ΔOAB = 12 cm.
∴ 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^{2} = 9.08 cm^{2}
Area of circle = πr^{2 }
∴ Area of major segment ADBE
= Area circle  Area of minor segment
= (314.28  9.08) cm^{2} = 305.20 cm^{2}
Q.2. In the given figure, ΔABC is a rightangled triangle in which ∠A is 90°. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region. [Al 2017]
Ans. In right triangle ABC.
AB^{2} + AC^{2} = BC^{2}
⇒ (3)^{2} + (4)^{2} = BC^{2} ⇒ 9 + 16 = BC^{2} ⇒ 25 = BC^{2}
∴ 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^{2}
Q.3. In Fig. 12.51, O is the centre of the circle with AC = 24 cm , AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.
[CBSE (AI) 2017]
Ans. In right angle triangle ABC
Area of shaded region = area of semicircle  area of ΔCAB + area of quadrant BOD
Q.4. An elastic belt is placed around the rim of a pulley of radius 5 cm. (Fig. 12.46). From one point C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from the point O. Find the length o f the belt that is still in contact with the pulley. Also find the shaded area. (Use π = 3.14 and √3 = 1.73) [CBSE Delhi 2016]
Ans.
PA = 5√S cm = BP (Tangents from an external point are equal)
Shaded area = 43 .25  26 .17 = 17.08 cm^{2}
Q.5. In Fig. 12.47, a sector OAP of a circle with centre O, containing angle 0. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is [CBSE (AI) 2016]
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. [CBSE (E) 2016]
Ans. Area of shaded region =
Q.7. In figure, ABCD is a trapezium with AB  DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region. [Foreign 2015]
Ans. Area of shaded region = area of trapezium  (area of 4 sectors)
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^{2}
Q.8. Find the area of the shaded region given in Fig. [Delhi 2015]
Ans. Side of the square = 14 cm
Area of the square = a^{2}
= 14^{2}  196 cm^{2}
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
= (19616871) cm^{2} = (1808π) cm^{2}
= 1808 x 3.14 = 18025.12 = 154.88 cm^{2}
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