Question

In the figure 7.44, O is the centre of the circle. m(arcPQR)=60° OP=10 cm. Find the area of the shaded region. (π=3.14, √3=1.73)

Answer 1

given , m(arcPQR)=60°

so , angle POR = 60° ----- central angle of corresponding arc

now , PO = OR ---- radius of same circle

so , angle OPR = angle ORP

so , in triangle POR ,

angle POR = 60°

and sum of all angles = 180°

so , angle OPR = angle ORP = 60°

so , triangle POR is equilateral triangle

area of arc O-PQR = area of triangle POR + area of shaded region

¢/360° × πr^2 = √3/4 × side^2 + area of shaded region

60/360 × 3.14× 10^2 = √3/4 × 10^2 + area of shaded region

314 /6 = 173/4 + area of shaded region

area of shaded region = 52.33 - 42.25


area of shaded region = 10.08 sq cm

Answer 2

Answer:

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