Track / Stage 4 / 297 of 340 #557 of 1964

Problem 557

AMC 12 late, AIME early

Geometry Difficulty 5.0 Find the answer

1. As shown in the figure, in $\odot O$ , $\overparen{A D B} = 90^{\circ}$ , chord $A B = a$ , and a circular arc with center $B$ and radius $B A$ intersects $\odot O$ at another point $C$ . Then the area $S$ of the crescent shape (the shaded part in the figure) formed by the two circular arcs is $S=$ . $\qquad$

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

ii. $1 \cdot \frac{1}{2} a^{2}$ .
It is known that $AC$ is the diameter of $\odot O$ , connect $BC$ (figure not shown). Then $\triangle ABC$ is an isosceles right triangle:
$\therefore OA=\frac{1}{2} AC=\frac{1}{2} \sqrt{2} a$ .
$\therefore$ The area of sector $ABC$ $S_{1}=\frac{\pi}{4} \cdot AB^{2}=\frac{\pi}{4} a^{2}$ ,
The upper half area of $\odot O$ $S_{2}=\frac{1}{2} \pi \cdot OA^{2}=\frac{\pi}{4} a^{2}$ .
$\mathrm{F}$ is, $S_{2}=S_{1}$ .
Subtract the common part (the area of the non-shaded segment $AC$ ) from both sides, we get $S=S_{\triangle ABC}=\frac{1}{2} a^{2}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.