1955 AHSME Problems/Problem 31

Revision as of 18:17, 18 November 2020 by Angrybird029 (talk | contribs) (Created page with "An equilateral triangle whose side is <math>2</math> is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equa...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

An equilateral triangle whose side is $2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:

$\textbf{(A)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ 2+\sqrt{2}\qquad\textbf{(D)}\ \frac{2+\sqrt{2}}{2}\qquad\textbf{(E)}\ \frac{2\sqrt{3}-\sqrt{6}}{2}$

Solution

The area of the large equilateral triangle is $2^2*\frac{\sqrt{3}}{4}=\sqrt{3}$ square units, so the smaller triangle is $\frac{\sqrt{3}}{2}$. The side length of the smaller length $s$ is $\sqrt{\frac{4}{\sqrt{3}}*\frac{\sqrt{3}}{2}} = \sqrt{2}$. The median of the trapezoid can be determined by the average of the two bases of the trapezoids, so it works out to be $\boxed{\textbf{(D)}\frac{2+\sqrt{2}}{2}}$ units long.

See Also

Go back to the rest of the 1955 AHSME Problems

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png