2012/13 AHSMC Part 1 Problem 5

Problem 5 of 16

$ABCD$ is a square. $E$ and $F$ are points on the segment $BC$ such that $BE = EF = FC = 4$ cm. The segments $AF$ and $DE$ intersect at $G$ . What, in cm $^2$ , is the area of triangle $EFG$ ?

Show solution

Solution

Triangles $GAD$ and $GFE$ are similar, with $AD = 3EF$ . Hence the vertical height of triangle $EFG$ is $\dfrac{1}{3}$ of the vertical height of triangle $ADG$ . Hence it is equal to $\dfrac{1}{4}AB = 3$ cm so that the area of triangle $EFG$ is $\dfrac{1}{2} \times 3 \times 4 = 6$ cm $^2$ . The answer is (A).

University of Alberta

Answer choices

Solution