1958 AHSME Problems/Problem 31

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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