Problem of the Week
Problem D and Solution
A Triangle and a Rectangle

Problem

Point $B$ lies on line segment $AC$, in between $A$ and $C$, so that $BC=10\times AB.$ Line segment $DE$ is parallel to $AC$ so that $BCDE$ forms a rectangle and $ABE$ forms a right-angled triangle.

If $AE=25$ and $BD=74$, determine the exact value of the length of $AD.$

Solution

Let $AB=x.$ Then $BC=10\times AB=10x.$ Let $CD=y.$ Then since $BCDE$ is a rectangle, it follows that $BE=CD=y$ and $ED=BC=10x.$ We label our diagram accordingly.

Using the Pythagorean Theorem in $\mathrm{△}ABE$, $$\begin{array}{rl}A{B}^2+B{E}^2& =A{E}^2\\ x^2+y^2& ={25}^2\\ x^2+y^2& =625\\ y^2& =625-x^2\end{array}$$ Since $\mathrm{\angle }BED=90\mathrm{°}$, it follows that $\mathrm{△}BED$ is a right-angled triangle. Using the Pythagorean Theorem in $\mathrm{△}BED$, $$\begin{array}{rl}B{E}^2+E{D}^2& =B{D}^2\\ y^2+(10x{)}^2& ={74}^2\\ y^2+100x^2& =5476\\ y^2& =5476-100x^2\end{array}$$ Then, since $y^2=625-x^2$ and $y^2=5476-100x^2$, it follows that $$\begin{array}{rl}625-x^2& =5476-100x^2\\ 99x^2& =4851\\ x^2& =49\end{array}$$ Thus, $x=\sqrt{49}=7$, since $x>0.$

Then $y^2=625-x^2=625-(7{)}^2=576$. Thus, $y=\sqrt{576}=24$, since $y>0.$

It follows that $CD=y=24$ and $AC=AB+BC=x+10x=11x=11(7)=77.$

Since $\mathrm{\angle }ACD=90\mathrm{°}$, it follows that $\mathrm{△}ACD$ is a right-angled triangle. Using the Pythagorean Theorem in $\mathrm{△}ACD$, $$\begin{array}{rl}A{C}^2+C{D}^2& =A{D}^2\\ {77}^2+{24}^2& =A{D}^2\\ 6505& =A{D}^2\end{array}$$ Thus, since $AD>0$, we can conclude that $AD=\sqrt{6505}.$