1951 AHSME Problems/Problem 3

Problem

If the length of a diagonal of a square is $a + b$ , then the area of the square is:

$\mathrm{(A) \ (a+b)^2 } \qquad \mathrm{(B) \ \frac{1}{2}(a+b)^2 } \qquad \mathrm{(C) \ a^2+b^2 } \qquad \mathrm{(D) \ \frac {1}{2}(a^2+b^2) } \qquad \mathrm{(E) \ \text{none of these} }$

Solution

Let a side be $s$ ; then by the Pythagorean Theorem, $s^2 + s^2 = 2s^2 = (a+b)^2$ . The area of a square is $s^2 = \frac{(a+b)^2}{2} \Rightarrow \mathrm{(B)}$ .

Alternatively, using the area formula for a kite, the area is $\frac{1}{2}d_1d_2 = \frac{1}{2}(a+b)^2$ .

1951 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1951 AHSME Problems/Problem 3

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