Difference between revisions of "1958 AHSME Problems/Problem 31"
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\textbf{(D)}\ 32\qquad | \textbf{(D)}\ 32\qquad | ||
\textbf{(E)}\ 24</math> | \textbf{(E)}\ 24</math> | ||
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| + | == Solution == | ||
| + | Consider the half of the triangle that is left of the altitude. Using the information given in the problem, we can determine that this is a right triangle with one leg of length 8 whose other two sides sum to 16. Through either setting the other leg as <math>x</math> and the hypotenuse as <math>16-x</math> and using the Pythagorean Theorem, or by recognizing this <math>6-8-10</math> triangle, we find that the other leg has length 6. So the triangle's total area is 48, and our answer is <math>\fbox{E}</math>. | ||
== See Also == | == See Also == | ||
Latest revision as of 20:27, 20 August 2020
Problem
The altitude drawn to the base of an isosceles triangle is 
, and the perimeter 
. The area of the triangle is:
Solution
Consider the half of the triangle that is left of the altitude. Using the information given in the problem, we can determine that this is a right triangle with one leg of length 8 whose other two sides sum to 16. Through either setting the other leg as 
and the hypotenuse as 
and using the Pythagorean Theorem, or by recognizing this 
triangle, we find that the other leg has length 6. So the triangle's total area is 48, and our answer is 
.
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 30 |
Followed by Problem 32 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
