1973 AMC 12 Problem 4

Question

Two congruent

30^\circ

-

60^\circ

-

90^\circ

triangles 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

6\sqrt3

8\sqrt3

9\sqrt3

12\sqrt3

24

Place the common hypotenuse on the

x

-axis and reverse which endpoint has the

30^\circ

angle

The upper boundary of the overlap consists of two lines making

30^\circ

angles with the hypotenuse

Put the common hypotenuse from

A=(0,0)

to

B=(12,0).

The two triangles have their third vertices at

(9,3\sqrt3)

and

(3,3\sqrt3).

Their common region is the triangle with base

AB

and apex where

y=\frac{x}{\sqrt3}

and

y=\frac{12-x}{\sqrt3}

meet. This occurs at

x=6,

with height

2\sqrt3.

Hence the common area is

\frac12(12)(2\sqrt3)=12\sqrt3.

Therefore, the correct answer is D.

Problem 4 in other years