Difference between revisions of "1951 AHSME Problems/Problem 3"
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| − | + | == Problem == | |
| + | If the length of a [[diagonal]] of a [[square]] is <math>a + b</math>, then the area of the square is: | ||
| + | |||
| + | <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 == | ||
| + | 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 == | ||
| + | {{AHSME 50p box|year=1951|num-b=2|num-a=4}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:19, 5 July 2013
Problem
If the length of a diagonal of a square is 
, then the area of the square is:
Solution
Let a side be 
; then by the Pythagorean Theorem, 
. The area of a square is 
.
Alternatively, using the area formula for a kite, the area is 
.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
