1978 AHSME Problems/Problem 23

Problem

Vertex $E$ of equilateral $\triangle ABE$ is in the interior of square $ABCD$ , and $F$ is the point of intersection of diagonal $BD$ and line segment $AE$ . If length $AB$ is $\sqrt{1+\sqrt{3}}$ then the area of $\triangle ABF$ is

$\textbf{(A) }1\qquad \textbf{(B) }\frac{\sqrt{2}}{2}\qquad \textbf{(C) }\frac{\sqrt{3}}{2}\qquad \textbf{(D) }4-2\sqrt{3}\qquad \textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4}$

Solution

C

Page

Toolbox

Search

1978 AHSME Problems/Problem 23

Problem

Solution