1973 AHSME Problems/Problem 4

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

1973 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1973 AHSME Problems/Problem 4

Solution

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