Question

In right angled triangle $\Delta ACE$ above, $\overline{BD}$ is parallel to $\overline{AE}$, and $\overline{BD}$ is perpendicular to $\overline{EC}$ at $D$. The length of $\overline{AC}$ is $20$ feet, the length of $\overline{BD}$ is $3$ feet and the length of $\overline{CD}$ is $4$ feet. Find the length of $\overline{AE}$.

• A. 10
• B. 12
• C. 15
• D. 16
• E. 17

Answer 1

The correct option is B 12

In the given triangle $ACE$, it is given that $\overline{BD}=3$ and $\overline{CD}=4$,

therefore, $\overline{BC}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

Now to find the length of $\overline{CE}$, consider,

$\frac{\overline{BC}}{\overline{AC}}=\frac{\overline{CD}}{\overline{CE}}$

$\Rightarrow \frac{5}{20}=\frac{4}{\overline{CE}}$

$\Rightarrow \frac{1}{4}=\frac{4}{\overline{CE}}$

$\Rightarrow \overline{CE}=16$

Now we find $\overline{AE}$ as follows:

$\overline{AE}=\sqrt{{20}^2-{16}^2}=\sqrt{400-256}=\sqrt{144}=12$.

Therefore, the length of $\overline{AE}$ is $12$ feet.