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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