1952 AHSME Problems/Problem 33

Problem

A circle and a square have the same perimeter. Then:

$\text{(A) their areas are equal}\qquad\\ \text{(B) the area of the circle is the greater} \qquad\\ \text{(C) the area of the square is the greater} \qquad\\ \text{(D) the area of the circle is } \pi \text{ times the area of the square}\qquad\\ \text{(E) none of these}$

Solution 1

Assume that the side length of the square is 4 units. Then the perimeter for both shapes is is 4 * 4 = 16 units. Since the perimeter of a circle is 2πr, then 2πr =16 and r = 8/π, which is about 2.5. The area of a circle is πr^2 so (2.5^2)*π = 6.25π or about 19.6. This is more than the area of the square, which is 4^2 =16. So the answer is $\fbox{B}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 32	Followed by Problem 34
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1952 AHSME Problems/Problem 33

Problem

Solution 1

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