In the diagram, $A B=13 \mathrm{~cm}, D C=20 \mathrm{~cm}$, and $A D=5 \mathrm{~cm}$. The length of $A C$, to the nearest tenth of a centimetre, is
(A) 24.2
(B) 20.6
(C) 25.2
(D) 23.4
(E) 24.9

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-112.jpg?height=350&width=531&top_left_y=1690&top_left_x=1141)
We extend $A D$ to the point $E$ where it intersects the perpendicular to $B C$ from $C$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-206.jpg?height=350&width=616&top_left_y=253&top_left_x=814)

By the Pythagorean Theorem in $\triangle A D B, B D^{2}=B A^{2}-A D^{2}=13^{2}-5^{2}=144$, so $B D=12 \mathrm{~cm}$. By the Pythagorean Theorem in $\triangle D B C, B C^{2}=D C^{2}-B D^{2}=20^{2}-12^{2}=256$, so $B C=16 \mathrm{~cm}$.

Since $B C E D$ has three right angles (and in fact, a fourth right angle at $E$ ), then it is a rectangle, so $D E=B C=16 \mathrm{~cm}$ and $C E=B D=12 \mathrm{~cm}$.

Therefore, if we look at $\triangle A E C$, we see that $A E=16+5=21 \mathrm{~cm}$, so by the Pythagorean Theorem, $A C^{2}=21^{2}+12^{2}=585$, so $A C \approx 24.2 \mathrm{~cm}$, to the nearest tenth of a centimetre.

ANSWER: (A)