#
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 \(\triangle
ABC\).