Question

The triangles $ABC$ and $DFG$ are similar and the ratio of their corresponding sides is $6:5$.

The area of the triangle $ABC$ is greater than the area of the triangle $DFG$ by $77{\mathrm{cm}}^2$.

Find the areas of these triangles.

Answer 1

Determine the areas.

The ratio of the sides of the triangles $ABC$ and $DFG$ is $6:5$.

The ratio of the areas of similar polygons is equal to the square of the ratio of their sides.

So, the ratio of the areas will be ${\left(\frac{6}{5}\right)}^2=\frac{36}{25}$.

Assume that the area of $ABC$ is $x{\mathrm{cm}}^2$.

So, the area of $DFG$ will be $\left(x-77\right){\mathrm{cm}}^2$

Determine the ratio of the areas:

$\frac{x}{x-77}$

Equate the ratios of areas obtained and solve for $x$:

$\begin{array}{rcl}\frac{x}{x-77}& =& \frac{36}{25}\\ & \Rightarrow & 25x=36x-2772\\ & \Rightarrow & 11x=2772\\ & \Rightarrow & x=252\end{array}$

Determine the area of $DFG$:

$\begin{array}{rcl}x-77& =& 252-77\\ & \Rightarrow & x-77=175\end{array}$

Hence, the areas of the triangles are $252{\mathrm{cm}}^{2}and175{\mathrm{cm}}^2$.