Q19 of 52 Page 1

In Figure 6, O is the center of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region. (Use π = 3.14)

Given,


AC = 24 cm


AB = 7 cm


∠BOD = 90°


In right triangle BAC,


∠BAC = 90° [Angle in a semi-circle]


By Pythagoras’ theorem,


BC2 = AC2 + AB2


BC2 = 242 + 72


BC2 = 576 + 49 = 625


BC = √625 = 25 cm


∴ Radius = OB = BC/2 = 25/2 = 12.5 cm


Area of the shaded region = Area of the circle – [area of ∆BAC + area of the quadrant]


=


= 490.625 – [84 + 122.656]


= 490.625 – 206.656


= 283.969 = 284 cm2


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 19

In Figure 7, find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

 20

An ice-cream seller sells his ice creams in two ways:

(A) In a cone of r = 5 cm, h = 8 cm


(B) In a cup in shape of cylinder with r = 5 cm, h = 8 cm


He charges the same price for both but prefers to sell his ice-cream in a cone.


(a) Find the volume of the cone and the cup.


(b) Which out of the two has more capacity?


(c) By choosing a cone, which value is not being followed by the ice-cream seller?