1995 AJHSME Problems/Problem 9


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 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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