Q2 of 52 Page 159

In the figure 7.44, O is the centre of the circle. M (arc PQR) = 60° OP = 10 cm.

Find the area of the shaded region. (π = 3.14, √3 = 1.73)


Since the angle subtended at centre is 60°

And by the property, if two sides of a triangle are equal then their corresponding angles are also equal.


⇒ ∠ ORP = ∠ OPR


As the sum of all internal angles of a triangle is equal to 180°


⇒ ∠ ORP = ∠ OPR = 60°


⇒ Δ OPR is an equilateral triangle.




⇒ AT = 43.25 sq. cm


Area Of Sector (O-PQR), AS is given as:




⇒ AS = 52.33 sq.cm


Area of shaded region, AR = As – AT


⇒ AR = 52.33 – 43.25


⇒ AR = 9.08 sq.cm


∴ Area of shaded region is 9.08 sq.cm


More from this chapter

All 52 →
13

∆ LMN is an equilateral triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centre and radius7 cm. Find,


(1) A (Δ LMN)


(2) Area of any one of the sectors


(3) Total area of all three sectors


(4) Area of shaded region

 1

In figure 7.43, A is the centre of the circle. ∠ ABC = 45° and AC = 7√2 cm. Find the area of segment BXC.

 3

In the figure 7.45, if A is the centre of the circle. ∠ PAR = 30°, AP = 7.5, find the area of the segment PQR. (π = 3.14)

 4

In the figure 7.46, if O is the centre of the circle, PQ is a chord. ∠ POQ = 90°, area of shaded region is 114 cm2, find the radius of the circle. (π = 3.14)