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Difference between revisions of "1958 AHSME Problems/Problem 31"

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== Solution ==
 
== Solution ==
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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>.
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== 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 $8$, and the perimeter $32$. The area of the triangle is:

$\textbf{(A)}\ 56\qquad \textbf{(B)}\ 48\qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 32\qquad \textbf{(E)}\ 24$

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 $x$ and the hypotenuse as $16-x$ and using the Pythagorean Theorem, or by recognizing this $6-8-10$ triangle, we find that the other leg has length 6. So the triangle's total area is 48, and our answer is $\fbox{E}$.

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

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