Problem of the Week
Problem C and Solution
Three Triangles

Problem

The right-angled triangle $\mathrm{△}ABC$ has $\mathrm{\angle }ABC=90\mathrm{°}$ and $AB=12$. Point $D$ is on side $BC$ such that $BD=5$ and the area of $\mathrm{△}ADC$ is $80\mathrm{\%}$ of the area of $\mathrm{△}ABD$.

Determine the perimeter of $\mathrm{△}ADC$.

Note: You may find the following useful:

The Pythagorean Theorem states, "In a right-angled triangle, the square of the length of hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides."

In the right-angled triangle shown, $c$ is the hypotenuse, $a$ and $b$ are the lengths of the other two sides, and $c^2=a^2+b^2$.

Solution

We start by labeling our diagram with the given information.

We then notice that since $AB$ is perpendicular to $BC$, it follows that $AB$ is perpendicular to $DC$. Thus, if the base of $\mathrm{△}ADC$ is $DC$, then its height is $AB$.

To find the area of a triangle, multiply the length of the base by the height and divide by $2$. Therefore, $$\text{area of }\mathrm{△}ABD=BD\times AB÷2=5\times 12÷2=30.$$

Since the area of $\mathrm{△}ADC$ is $80\mathrm{\%}$ of the area of $\mathrm{△}ABD$, the area of $\mathrm{△}ADC$ is equal to $0.8\times 30=24$.

We also know the area of $\mathrm{△}ADC$ is equal to $DC\times AB÷2$. Thus, $$\begin{array}{rl}DC\times AB÷2& =24\\ DC\times 12÷2& =24\\ DC\times 6& =24\\ DC& =24÷6=4\end{array}$$ Thus, $BC=BD+DC=5+4=9$. Using the Pythagorean Theorem in $\mathrm{△}ABC$, $$\begin{array}{rl}A{C}^2& =A{B}^2+B{C}^2\\ & ={12}^2+9^2\\ & =225\end{array}$$ Thus, $AC=\sqrt{225}=15$, since $AC>0$.

Using the Pythagorean Theorem in $\mathrm{△}ABD$, $$\begin{array}{rl}A{D}^2& =A{B}^2+B{D}^2\\ & ={12}^2+5^2\\ & =169\end{array}$$ Thus, $AD=\sqrt{169}=13$, since $AD>0$.

Therefore, the perimeter of $\mathrm{△}ADC$ is $AD+DC+AC=13+4+15=32$.