# Problem of the Week Problem D and Solution Overlapping Shapes 2

## Problem

Selena draws square $$ABCD$$ with side length $$16$$ cm. Endre then draws $$\triangle AED$$ on top of square $$ABCD$$ so that

• sides $$AE$$ and $$DE$$ meet $$BC$$ at $$F$$ and $$G$$, respectively, and

• the area of $$\triangle AED$$ is twice the area of square $$ABCD$$.

Determine the area of trapezoid $$AFGD$$.

## Solution

We construct an altitude of $$\triangle AED$$ from $$E$$, intersecting $$AD$$ at $$P$$ and $$BC$$ at $$Q$$. Since $$ABCD$$ is a square, we know that $$AD$$ is parallel to $$BC$$. Therefore, since $$PE$$ is perpendicular to $$AD$$, $$QE$$ is perpendicular to $$FG$$ and thus an altitude of $$\triangle FEG$$.

The area of square $$ABCD$$ is $$16\times 16=256~\text{cm}^2$$. Since the area of $$\triangle AED$$ is twice the area of square $$ABCD$$, it follows that the area of $$\triangle AED$$ is $$2 \times 256=512~\text{cm}^2$$.

We also know that \begin{aligned} \text{Area }\triangle AED&=AD\times PE \div 2\\ 512&=16\times PE \div 2\\ 512 &=8\times PE\\ PE &=512 \div 8 \\ &= 64~\text{cm} \end{aligned} Since $$\angle APQ=90\degree$$, we know that $$ABQP$$ is a rectangle, and so $$PQ = AB = 16$$ cm. We also know that $$PE=PQ+QE$$. Since $$PE=64$$ cm and $$PQ=16$$ cm, it follows that $$QE=PE-PQ=64-16=48$$ cm.

From here we proceed with two different solutions.

Solution 1

We will use the relationships between the areas of the shapes to determine the length of $$FG$$. \begin{aligned} \text{Area of trapezoid }AFGD+\text{Area }\triangle FEG &= \text{Area }\triangle AED\\ (AD+FG) \times AB \div 2 + FG \times QE \div 2 &= 512\\ (16+FG)\times 16 \div 2 + FG \times 48 \div 2 &= 512\\ (16+FG)\times 8 + 24\times FG &= 512\\ 128 + 8 FG + 24 FG &= 512\\ 32 FG &= 384\\ FG &=12~\text{cm} \end{aligned} Now we can use $$FG$$ to calculate the area of trapezoid $$AFGD$$. \begin{aligned} \text{Area of trapezoid }AFGD &= (AD+FG) \times AB \div 2\\ &= (16+12) \times 16 \div 2\\ &=28 \times 8 = 224~\text{cm}^2 \end{aligned} Therefore, the area of trapezoid $$AFGD$$ is $$224~\text{cm}^2$$.

Solution 2

We will use similar triangles to determine the length of $$FG$$. We know that $$\angle AED=\angle FEG$$. Also, since $$AD$$ is parallel to $$FG$$ it follows that $$\angle EAD$$ and $$\angle EFG$$ are corresponding angles, so are equal. Thus, by angle-angle similarity, $$\triangle AED \sim \triangle FEG$$. Therefore, \begin{aligned} \frac{AD}{PE}&=\frac{FG}{QE}\\ \frac{16}{64}&=\frac{FG}{48}\\ \frac{1}{4}&=\frac{FG}{48}\\ FG&=48\times \frac{1}{4}=12~\text{cm} \end{aligned} Now we can use $$FG$$ to calculate the area of trapezoid $$AFGD$$. \begin{aligned} \text{Area of trapezoid }AFGD &= (AD+FG) \times AB \div 2\\ &= (16+12) \times 16 \div 2\\ &=28 \times 8 = 224~\text{cm}^2 \end{aligned} Therefore, the area of trapezoid $$AFGD$$ is $$224~\text{cm}^2$$.