Question : A right-angled isosceles triangle is inscribed in a semi-circle of radius 7 cm. The area enclosed by the semi-circle but exterior to the triangle is:
Option 1: 14 cm2
Option 2: 28 cm2
Option 3: 44 cm2
Option 4: 68 cm2
Correct Answer: 28 cm 2
Solution : Area of the semi-circle = $\frac{1}{2} \pi r^2$ = $\frac{1}{2}\times \frac{22}{7}\times 7^2$ = 77 cm 2 The altitude of the isosceles triangle OC is $r$ = 7 cm. The base of the isosceles triangle AB is the diameter = $2 r$ = 14 cm. Area of the Isosceles triangle ABC = $\frac{1}{2}$ × Base × Height = $\frac{1}{2}\times 14\times 7$ = 49 cm 2 So, the area enclosed by the semi-circle but exterior to the triangle = 77 - 49 = 28 cm 2 Hence, the correct answer is 28 cm 2 .
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