Question

All of the triangles in the diagram below are similar to isosceles triangle $ABC,$ in which $AB=AC.$ Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40.$ What is the area of trapezoid $DBCE?$

$16$

$18$

$20$

$22$

$24$

Solution:

We know that the side length of the smaller triangles is $\sqrt{\frac{1}{40}}$ times the length of the larger triangle from similar triangles.

Then the side length of $\triangle ADE$ is $4\sqrt{\frac{1}{40}}$ times the length of the side length of the larger triangle.

This makes the ratio of the areas $\left(4\sqrt{\dfrac{1}{40}}\right)^2 = 16 \cdot \dfrac{1}{40} = \dfrac{2}{5}.$

Therefore, the area of $\triangle ADE$ is $\frac{2}{5} \cdot 40 = 16.$ The area of the trapezoid is then $40 - 16 = 24.$

Thus, E is the correct answer.

2018 AMC 10A Problem 9

Solution:

Problem 9 in Other Years