# Problem of the Week Problem C and Solution Overlapping Shapes 1

## Problem

Omar draws square $$ABCD$$ with side length $$4$$ cm. Jaime then draws $$\triangle AED$$ on top of square $$ABCD$$ so that

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

• $$FG$$ is $$3$$ cm, and

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

Determine the area of $$\triangle FEG$$.

## Solution

Solution 1

In the first solution we will find the area of square $$ABCD$$, the area of $$\triangle AED$$, the area of trapezoid $$AFGD$$, and then use these to calculate the area of $$\triangle FEG$$.

The area of square $$ABCD$$ is $$4\times 4=16~\text{cm}^2$$. Since the area of $$\triangle AED$$ is twice the area of square $$ABCD$$, it follows that the area of $$\triangle AED$$ is $$2 \times 16=32~\text{cm}^2$$.

Recall that to find the area of a trapezoid, we multiply the sum of the lengths of the two parallel sides by the height, and divide the product by $$2$$. In trapezoid $$AFGD$$, the two parallel sides are $$AD$$ and $$FG$$, and the height is the width of square $$ABCD$$, namely $$AB$$. \begin{aligned} \text{Area of trapezoid }AFGD&=(AD+FG)\times AB\div 2\\ &=(4+3)\times 4\div 2\\ &=7\times 4\div 2\\ &=14~\text{cm}^2 \end{aligned} The area of $$\triangle FEG$$ is equal to the area of $$\triangle AED$$ minus the area of trapezoid $$AFGD$$. Thus, the area of $$\triangle FEG$$ is $$32-14=18~\text{cm}^2$$.

Solution 2

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 $$\triangle FEG$$. In this solution we will find the height of $$\triangle FEG$$, that is, the length of $$QE$$, and then use this to calculate the area of $$\triangle FEG$$.

The area of square $$ABCD$$ is $$4\times 4=16~\text{cm}^2$$. Since the area of $$\triangle AED$$ is twice the area of square $$ABCD$$, it follows that the area of $$\triangle AED$$ is $$2 \times 16=32~\text{cm}^2$$.

We also know that \begin{aligned} \text{Area }\triangle AED&=AD\times PE \div 2\\ 32&=4\times PE \div 2\\ 32&=2\times PE\\ PE &=32 \div 2 \\ &= 16~\text{cm} \end{aligned} Since $$\angle APQ = 90\degree$$, we know that $$ABQP$$ is a rectangle, and so $$PQ = AB = 4$$ cm. We also know that $$PE=PQ+QE$$. Since $$PE=16$$ cm and $$PQ=4$$ cm, it follows that $$QE=PE-PQ=16-4=12$$ cm. We can then calculate the area of $$\triangle FEG$$. \begin{aligned} \text{Area }\triangle FEG&=FG\times QE \div 2\\ &=3\times 12 \div 2\\ &=18~\text{cm}^2 \end{aligned} Therefore, the area of $$\triangle FEG$$ is $$18~\text{cm}^2$$.