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Question

Question 15
In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.


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Solution

Radius of the the quadrant ABC of circle = 14 cm
AB = AC = 14 cm
BC is diameter of thesemicircle.
ABC is a right angled triangle.
By Pythagoras theorem in ΔABC,
BC2=AB2+AC2
BC2=142+142
BC=142 cm

Radius of semicircle
=1422cm=72 cm

Area of Δ ABC
= 12×(base)×(height)
=12×14×14=98 cm2

Area of quadrant
=14×227×14×14=154 cm2
Area of the semicircle
=12×227×72×72=154 cm2
Area of the shaded region = Area of the semicircle + Area of ΔABC - Area of quadrant
=154+98154 cm2=98 cm2


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