In the diagram, $AB = 13\text{ cm},$ $DC = 20\text{ cm},$ and $AD = 5\text{ cm}.$ What is the length of $AC,$ to the nearest tenth of a centimeter?

[asy]
draw((0,0)--(5,12)--(21,12)--(5,0)--cycle,black+linewidth(1));
draw((5,12)--(5,0),black+linewidth(1));
draw((0,0)--(21,12),black+linewidth(1));
draw((5,0)--(5,0.5)--(4.5,0.5)--(4.5,0)--cycle,black+linewidth(1));
draw((5,12)--(5.5,12)--(5.5,11.5)--(5,11.5)--cycle,black+linewidth(1));
label("$A$",(0,0),NW);
label("$B$",(5,12),NW);
label("$C$",(21,12),E);
label("$D$",(5,0),SE);
label("13 cm",(0,0)--(5,12),NW);
label("5 cm",(0,0)--(5,0),S);
label("20 cm",(5,0)--(21,12),SE);
[/asy]
We extend $AD$ to the point $E$ where it intersects the perpendicular to $BC$ from $C.$

[asy]
draw((0,0)--(5,12)--(21,12)--(5,0)--cycle,black+linewidth(1));
draw((5,12)--(5,0),black+linewidth(1));
draw((0,0)--(21,12),black+linewidth(1));
draw((5,0)--(5,0.5)--(4.5,0.5)--(4.5,0)--cycle,black+linewidth(1));
draw((5,12)--(5.5,12)--(5.5,11.5)--(5,11.5)--cycle,black+linewidth(1));
label("$A$",(0,0),NW);
label("$B$",(5,12),NW);
label("$C$",(21,12),E);
label("$D$",(5,0),SE);
label("13 cm",(0,0)--(5,12),NW);
label("5 cm",(0,0)--(5,0),S);
label("20 cm",(5,0)--(21,12),SE);
draw((5,0)--(21,0),black+linewidth(1)+dashed);
draw((21,0)--(21,12),black+linewidth(1)+dashed);
draw((21,0)--(21,0.5)--(20.5,0.5)--(20.5,0)--cycle,black+linewidth(1));
label("$E$",(21,0),SE);
label("16 cm",(5,0)--(21,0),S);
label("12 cm",(21,0)--(21,12),E);
[/asy]

By the Pythagorean Theorem in $\triangle ADB,$ $BD^2 = BA^2 - AD^2 = 13^2 - 5^2 = 144,$ so $BD=12\text{ cm}.$

By the Pythagorean Theorem in $\triangle DBC,$ $BC^2 = DC^2 - BD^2 = 20^2 - 12^2 = 256,$ so $BC=16\text{ cm}.$

Since $BCED$ has three right angles (and in fact, a fourth right angle at $E$), then it is a rectangle, so $DE=BC=16\text{ cm}$ and $CE=BD=12\text{ cm}.$

Therefore, if we look at $\triangle AEC,$ we see that $AE = 16+5=21\text{ cm},$ so by the Pythagorean Theorem, $AC^2 = 21^2 + 12^2 = 585,$ so $AC \approx \boxed{24.2}\text{ cm},$ to the nearest tenth of a centimeter.
24.2
Level 5
Prealgebra
Problem: In the diagram, $AB = 13\text{ cm},$ $DC = 20\text{ cm},$ and $AD = 5\text{ cm}.$ What is the length of $AC,$ to the nearest tenth of a centimeter?

[asy]
draw((0,0)--(5,12)--(21,12)--(5,0)--cycle,black+linewidth(1));
draw((5,12)--(5,0),black+linewidth(1));
draw((0,0)--(21,12),black+linewidth(1));
draw((5,0)--(5,0.5)--(4.5,0.5)--(4.5,0)--cycle,black+linewidth(1));
draw((5,12)--(5.5,12)--(5.5,11.5)--(5,11.5)--cycle,black+linewidth(1));
label("$A$",(0,0),NW);
label("$B$",(5,12),NW);
label("$C$",(21,12),E);
label("$D$",(5,0),SE);
label("13 cm",(0,0)--(5,12),NW);
label("5 cm",(0,0)--(5,0),S);
label("20 cm",(5,0)--(21,12),SE);
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the length of segment AC, we can apply the Pythagorean theorem. First, we need to identify the coordinates of points A, B, C, and D based on the given lengths. 

Let point A be at the origin (0, 0). Since AB = 13 cm and is vertical, point B will be at (5, 12) because the height from A to B is 12 cm. Point D, which is directly below B, will be at (5, 0) since AD = 5 cm. 

Point C is horizontally aligned with point B and is 20 cm away from point D. Therefore, the coordinates of point C will be (21, 12).

Next, we calculate the length of AC using the distance formula. The distance formula states that the distance between two points (x1, y1) and (x2, y2) is given by:

d = √((x2 - x1)² + (y2 - y1)²)

Substituting the coordinates of points A (0, 0) and C (21, 12) into the formula:

AC = √((21 - 0)² + (12 - 0)²)
   = √(21² + 12²)
   = √(441 + 144)
   = √585

Calculating √585 gives approximately 24.2 when rounded to the nearest tenth. Thus, the length of AC is 24.2 cm.<|END_OF_LATENT|> The final answer is $24.2$.