Difference between revisions of "1973 AHSME Problems/Problem 4"

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Revision as of 23:06, 4 July 2018

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 $\tfrac12 \cdot 12 \cdot 2\sqrt{3} = \boxed{\textbf{(D)}\ 12\sqrt3}$.

See Also

1973 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AHSME Problems and Solutions