1987 AJHSME Problems/Problem 22

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$\text{ABCD}$ 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]

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


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 $\boxed{\text{D}}$

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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