# Problem of the Week Problem C and Solution A Bigger Triangle

## Problem

Naveen drew a right-angled triangle, $$\triangle ABC$$, with an area of $$14~\text{cm}^2$$. His brother Anand drew a bigger right-angled triangle, $$\triangle DEF$$, with side lengths four times the lengths of the sides in $$\triangle ABC$$. In particular, $$DE=4\times AB$$, $$EF=4\times BC$$, and $$DF=4\times AC$$.

Calculate the area of $$\triangle DEF$$.

## Solution

In $$\triangle ABC$$, let $$b$$ represent the length of the base, $$BC$$, and $$h$$ represent the length of the height, $$AB$$.

Then the area of $$\triangle ABC$$ is equal to $$\frac{b\times h}{2}$$. We know this area is equal to $$14~\text{cm}^2$$, so it follows that $$14=\frac{b\times h}{2}$$, or $$28=b\times h$$.

$$\triangle DEF$$ is formed by multiplying each of the side lengths of $$\triangle ABC$$ by $$4$$. So the length of the base of $$\triangle DEF$$ is equal to $$4\times b$$ and the length of the height is equal to $$4\times h$$.

We can calculate the area of $$\triangle DEF$$ as follows. \begin{aligned} \text{area of }\triangle DEF &= \frac{(4\times b)\times (4\times h)}{2}\\ &= \frac{16\times b \times h}{2}\\ &= \frac{16 \times 28}{2}, \text{ since } b\times h=28\\ &=224 \end{aligned} Therefore, the area of $$\triangle DEF$$ is $$224~\text{cm}^2$$.

Extension:

Notice that $$\triangle DEF$$ has side lengths that are each $$4$$ times the corresponding side lengths of $$\triangle ABC$$ and that the area of $$\triangle DEF$$ ended up being $$224=16 \times 14=4^2\times \text{ area of }\triangle ABC$$.

Show that if $$\triangle DEF$$ has side lengths that are each $$k$$ times the corresponding side lengths of $$\triangle ABC$$, then the area of $$\triangle DEF$$ will be equal to $$k^2$$ times the area of $$\triangle ABC$$.