Question

Let $ABC$ be a right-angled isosceles triangle with hypotenuse $BC$. Let $BQC$ be a semi-circle, away from $A$, with diameter $BC$. Let $BPC$ be an arc of a circle centered at $A$ and lying between $BC$ and $BQC$. If $AB$ has length $6	ext{ cm}$ then the area, in sq. cm, of the region enclosed by $BPC$ and $BQC$ is:

Accepted Answer

B. Radius $AB = AC = 6$. Hypotenuse $BC = \sqrt{6^2 + 6^2} = 6\sqrt{2}$. Radius of semicircle $BQC = R = 3\sqrt{2}$. Area of semicircle $BQC = \frac{\pi R^2}{2} = \frac{\pi (18)}{2} = 9\pi$. Area of sector $ABPC = \frac{1}{4} \pi (6^2) = 9\pi$. Area of segment $BPC = 	ext{Sector}(ABPC) - 	ext{Area}(	riangle ABC) = 9\pi - \frac{1}{2}(6)(6) = 9\pi - 18$. Area enclosed by $BPC$ and $BQC = 	ext{Area}(BQC) - 	ext{Area}(BPC) = 9\pi - (9\pi - 18) = 18$.

Area Enclosed by Arcs

Explanation

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