Track / Stage 3 / 12 of 260 #12 of 2000

Problem 12

AMC 10/12, early questions

Geometry Difficulty 3.0 Find the answer

$\text{ABCD}$ 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

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

Your answer

Official 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 $\boxed{\text{D}}$

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