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Difference between revisions of "1995 AJHSME Problems/Problem 9"

(Created page with "==Problem== Three congruent circles with centers <math>P</math>, <math>Q</math>, and <math>R</math> are tangent to the sides of rectangle <math>ABCD</math> as shown. The circle...")
 
(→‎Solution 1)
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Draw a horizontal line <math>PQR</math> that extends to the sides of the rectangle.  This line is <math>2</math> diameters long, so it has length <math>4\cdot 2 = 8</math>.  It is parallel and congruent to <math>AB</math>, so the width of the rectangle is <math>8</math>.
 
Draw a horizontal line <math>PQR</math> that extends to the sides of the rectangle.  This line is <math>2</math> diameters long, so it has length <math>4\cdot 2 = 8</math>.  It is parallel and congruent to <math>AB</math>, so the width of the rectangle is <math>8</math>.
  
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Thus, the area of the rectangle is <math>lw = 4\cdot 8 = 32</math>, and the answer is <math>\boxed{E}</math>.
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Thus, the area of the rectangle is <math>lw = 4\cdot 8 = 32</math>, and the answer is <math>\boxed{C}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 17:12, 18 September 2011

Problem

Three congruent circles with centers $P$, $Q$, and $R$ are tangent to the sides of rectangle $ABCD$ as shown. The circle centered at $Q$ has diameter $4$ and passes through points $P$ and $R$. The area of the rectangle is

[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,W); label("$Q$",Q,W); label("$R$",R,W); [/asy]

$\text{(A)}\ 16 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$

Solution 1

If circle $Q$ has diameter $4$, then so do congruent circles $P$ and $R$. Draw a diameter through $P$ parallel to $AD$. The diameter will be congruent to $AD$, and thus $AD = 4$, which is the height of the rectangle.

Draw a horizontal line $PQR$ that extends to the sides of the rectangle. This line is $2$ diameters long, so it has length $4\cdot 2 = 8$. It is parallel and congruent to $AB$, so the width of the rectangle is $8$.

Thus, the area of the rectangle is $lw = 4\cdot 8 = 32$, and the answer is $\boxed{C}$.

Solution 2

The area of circle $P$ and circle $R$ almost fill the rectangle. Circle $P$ has radius $\frac{4}{2} = 2$, and so does circle $R$. Thus, the sum of their areas is $\pi \cdot 2^2 + \pi \cdot 2^2 = 8\pi$. Since $\pi \approx 3.14$, the area of the circles is just over $24$. The area of the rectangle is greater than the area of the circles. $64$ would be too high, as the two circles appear to take up much more than half the area of the rectangle. But $24$ is too low, as it is less than the area of the two circles. Thus, the only reasonable answer is $\boxed{C}$, which is $32$.

See Also

1995 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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