Q31 of 49 Page 1

In Fig. 6, ABC is a right-angled triangle right angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.

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fig.6

Area of semicircle with diameter AB = π × (3/2)2 = 2.25π sq. cm

Area of semicircle with diameter AC = π × (4/2)2 = 4π sq. cm


In Δ ABC,


AB2+ AC2 = BC2


[using pythagoras theorem]


∴ BC = √(32+42) = √25 = 5 cm


Area of semicircle with diameter BC = π × (5/2)2 = 6.25π sq. cm


and,


Area of rt. angle Δ ABC = (1/2) × AB×AC = (1/2)×3×4 = 6 sq. cm


Now,


Area of Shaded Region


= (Area of semicircle with diameter AB+


Area of semicircle with diameter AC+


Area of rt. angle Δ ABC)- Area of semicircle with diameter BC


= (2.25π+4π+6)-6.25π


= 6.25π+6-6.25π


= 6 sq. cm


Thus, Area of Shaded Region = 6 sq. cm


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