Problem D and Solution

Overlapping Shapes 2

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 \(AFGD\).

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 \(224~\text{cm}^2\).