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Difference between revisions of "2000 AMC 8 Problems/Problem 19"

(Created page with "Three circular arcs of radius <math>5</math> units bound the region shown. Arcs <math>AB</math> and <math>AD</math> are quarter-circles, and arc <math>BCD</math> is a semicircle....")
 
(Solution 1, Solution 2, see also box)
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<math> \text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi </math>
 
<math> \text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi </math>
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==Solution 1==
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Draw two squares:  one that has opposing corners at <math>A</math> and <math>B</math>, and one that has opposing corners at <math>A</math> and <math>D</math>.  These squares share side <math>\overline{AO}</math>, where <math>O</math> is the center of the large semicircle. 
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These two squares have a total area of <math>2 \cdot 5^2</math>, but have two quarter circle "bites" of radius <math>5</math> that must be removed.  Thus, the bottom part of the figure has area
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<math>2\cdot 25 - 2 \cdot \frac{1}{4}\pi \cdot 5^2</math>
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<math>50 - \frac{25\pi}{2}</math>
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This is the area of the part of the figure underneath <math>\overline{BD}</math>.  The part of the figure over <math>\overline{BD}</math> is just a semicircle with radius <math>5</math>, which has area of <math>\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}</math>
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Adding the two areas gives a total area of <math>50</math>, for an answer of <math>\boxed{C}</math>
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==Solution 2==
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Draw line <math>\overline{BD}</math>.  Then draw <math>\overline {CO}</math>, where <math>O</math> 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 <math>\overline {BD}</math> and height <math>\overline {AO}</math>, which are equal to <math>10</math> and <math>5</math>, respectively.  Thus, the total area is <math>50</math>, and the answer is <math>\boxed{C}</math>.
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==See Also==
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{{AMC8 box|year=2000|num-b=18|num-a=20}}

Revision as of 20:12, 30 July 2011

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]

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

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 $\boxed{C}$.

See Also

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