# 1973 AHSME Problems/Problem 4

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Two congruent 30-60-90 are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is $\textbf{(A)}\ 6\sqrt3\qquad\textbf{(B)}\ 8\sqrt3\qquad\textbf{(C)}\ 9\sqrt3\qquad\textbf{(D)}\ 12\sqrt3\qquad\textbf{(E)}\ 24$

## Solution $[asy] fill((0,3.464)--(-6,0)--(6,0)--cycle,yellow); draw((-6,0)--(6,0)--(-3,5.196)--(-6,0)); draw((-6,0)--(6,0)--(3,5.196)--(-6,0)); draw((0,3.464)--(0,0)); label("6",(-3,0),S); label("6",(3,0),S); [/asy]$

Note that the altitude of the shared region bisects the hypotenuse of the original two right triangles (this can be confirmed by using AAS on the two right triangles with that altitude of the shared region). By using 30-60-90 triangles, the altitude’s length is $2\sqrt{3}$. The area of the shared region is $\tfrac12 \cdot 12 \cdot 2\sqrt{3} = \boxed{\textbf{(D)}\ 12\sqrt3}$.