In the diagram shown, David begins at $A$ and walks in a straight line to $C$, and then walks in a straight line from $C$ to $B$. Cindy also begins at $A$ and walks in a straight line to $B$. How much farther does David walk than Cindy?
(A) $0 \mathrm{~m}$
(B) $2 \mathrm{~m}$
(C) $4 \mathrm{~m}$
(D) $6 \mathrm{~m}$
(E) $7 \mathrm{~m}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-028.jpg?height=250&width=395&top_left_y=563&top_left_x=1293)
Since $\angle A C B=90^{\circ}$, then $\triangle A C B$ is a right-angled triangle.

By the Pythagorean Theorem, $A B^{2}=A C^{2}+C B^{2}=8^{2}+15^{2}=64+225=289 \mathrm{~m}^{2}$.

Since $A B>0$, then $A B=\sqrt{289}=17 \mathrm{~m}$ and so Cindy walks a distance of $17 \mathrm{~m}$.

Walking from $A$ to $C$ to $B$, David walks a total distance of $8+15=23 \mathrm{~m}$.

Thus, David walks $23-17=6 \mathrm{~m}$ farther than Cindy.

ANSwER: (D)