Track / Stage 3 / 67 of 260 #67 of 2000

Problem 67

AMC 10/12, early questions

Geometry Difficulty 3.3 Find the answer

Rectangle $ABCD$ has sides $CD=3$ and $DA=5$ . A circle of radius $1$ is centered at $A$ , a circle of radius $2$ is centered at $B$ , and a circle of radius $3$ is centered at $C$ . Which of the following is closest to the area of the region inside the rectangle but outside all three circles?

$\text{(A) }3.5\qquad\text{(B) }4.0\qquad\text{(C) }4.5\qquad\text{(D) }5.0\qquad\text{(E) }5.5$

Your answer

Official solution

The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.
The area of the rectangle is $3\cdot5 =15$ . The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$ . Therefore the area in the rectangle but outside the circles is $15-\frac{7\pi}{2}$ . $\pi$ is approximately $\dfrac{22}{7},$ and substituting that in will give $15-11=\boxed{\text{(B) }4.0}$

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