Triangle $A B C$ is right-angled with $A B=10$ and $A C=8$. If $B C=3 D C$, then $A D$ equals
(A) 9
(B) $\sqrt{65}$
(C) $\sqrt{80}$
(D) $\sqrt{73}$
(E) $\sqrt{68}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-153.jpg?height=372&width=477&top_left_y=1722&top_left_x=1431)
Solution

Using Pythagoras in $\mathrm{D} A B C$ gives $B C$ to be 6 . Since $B C=3 D C, D C=2$.

$$
\begin{aligned}
A D^{2} & =2^{2}+8^{2} \\
& =68
\end{aligned}
$$

Using Pythagoras again in $\mathrm{D} A D C, A D=\sqrt{68}$

ANSWER: (E)

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