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1951 AHSME Problems/Problem 3

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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 {...

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.

See also

1951 AHSME (Problems)
Preceded by
Problem 2 		Followed by
Problem 4
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 • 26 • 27 • 28 • 29 • 30
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