1987 AJHSME Problems/Problem 22

Problem

$\text{ABCD}$ is a rectangle, $\text{D}$ is the center of the circle, and $\text{B}$ is on the circle. If $\text{AD}=4$ and $\text{CD}=3$ , then the area of the shaded region is between

$[asy] pair A,B,C,D; A=(0,4); B=(3,4); C=(3,0); D=origin; draw(circle(D,5)); fill((0,5)..(1.5,4.7697)..B--A--cycle,black); fill(B..(4,3)..(5,0)--C--cycle,black); draw((0,5)--D--(5,0)); label("A",A,NW); label("B",B,NE); label("C",C,S); label("D",D,SW); [/asy]$

$\text{(A)}\ 4\text{ and }5 \qquad \text{(B)}\ 5\text{ and }6 \qquad \text{(C)}\ 6\text{ and }7 \qquad \text{(D)}\ 7\text{ and }8 \qquad \text{(E)}\ 8\text{ and }9$

Solution

The area of the shaded region is equal to the area of the quarter circle with the area of the rectangle taken away. The area of the rectangle is $4\cdot 3=12$ , so we just need the quarter circle.

Applying the Pythagorean theorem to $\triangle ADC$ , we have $(AC)^2=4^2+3^2\Rightarrow AC=5$ Since $ABCD$ is a rectangle, $BD=AC=5$

Clearly $BD$ is a radius of the circle, so the area of the whole circle is $5^2\pi =25\pi$ and the area of the quarter circle is $\frac{25\pi }{4}$ .

Finally, the shaded region is $\frac{25\pi }{4}-12 \approx 7.6$ , so the answer is $\boxed{\text{D}}$

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1987 AJHSME Problems/Problem 22

Problem

Solution

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