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Difference between revisions of "1951 AHSME Problems/Problem 3"

(1951 AHSME Problems/Problem 3 moved to User:Azjps/1951 AHSME Problems/Problem 3: well, this obviously isn't the right question ... move to userspace until we find where it does belong)
 
 
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#REDIRECT [[User:Azjps/1951 AHSME Problems/Problem 3]]
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== Problem ==
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If the length of a [[diagonal]] of a [[square]] is <math>a + b</math>, then the area of the square is:
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<math> \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} }  </math>
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== Solution ==
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Let a side be <math>s</math>; then by the [[Pythagorean Theorem]], <math>s^2 + s^2 = 2s^2 = (a+b)^2</math>. The area of a square is <math>s^2 = \frac{(a+b)^2}{2} \Rightarrow \mathrm{(B)}</math>.
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Alternatively, using the area formula for a [[kite]], the area is <math>\frac{1}{2}d_1d_2 = \frac{1}{2}(a+b)^2</math>.
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== See Also ==
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{{AHSME 50p box|year=1951|num-b=2|num-a=4}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:19, 5 July 2013

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 $\frac{1}{2}d_1d_2 = \frac{1}{2}(a+b)^2$.

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AHSME Problems and Solutions

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