2000 AMC 8 Problems/Problem 19

Problem

Three circular arcs of radius $5$ units bound the region shown. Arcs $AB$ and $AD$ are quarter-circles, and arc $BCD$ is a semicircle. What is the area, in square units, of the region?

$[asy] pair A,B,C,D; A = (0,0); B = (-5,5); C = (0,10); D = (5,5); draw(arc((-5,0),A,B,CCW)); draw(arc((0,5),B,D,CW)); draw(arc((5,0),D,A,CCW)); label("$A$",A,S); label("$B$",B,W); label("$C$",C,N); label("$D$",D,E);[/asy]$

$\text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi$

Solutions

Solution 1

Draw two squares: one that has opposing corners at $A$ and $B$ , and one that has opposing corners at $A$ and $D$ . These squares share side $\overline{AO}$ , where $O$ is the center of the large semicircle.

These two squares have a total area of $2 \cdot 5^2$ , but have two quarter circle "bites" of radius $5$ that must be removed. Thus, the bottom part of the figure has area

$2\cdot 25 - 2 \cdot \frac{1}{4}\pi \cdot 5^2$

$50 - \frac{25\pi}{2}$

This is the area of the part of the figure underneath $\overline{BD}$ . The part of the figure over $\overline{BD}$ is just a semicircle with radius $5$ , which has area of $\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}$

Adding the two areas gives a total area of $50$ , for an answer of $\boxed{C}$

Solution 2

Draw line $\overline{BD}$ . Then draw $\overline {CO}$ , where $O$ is the center of the semicircle. You have two quarter circles on top, and two quarter circle-sized "bites" on the bottom. Move the pieces from the top to fit in the bottom like a jigsaw puzzle. You now have a rectangle with length $\overline {BD}$ and height $\overline {AO}$ , which are equal to $10$ and $5$ , respectively. Thus, the total area is $50$ , and the answer is $\boxed{C}$ .

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2000 AMC 8 Problems/Problem 19

Contents

Problem

Solutions

Solution 1

Solution 2

See Also