1958 AHSME Problems/Problem 31

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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

[asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot("$A$", A, SW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); label("$70^\circ$",C,2*dir(180-35)); [/asy] $\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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