The sides of a triangle have lengths of $15$, $20$, and $25$. Find the length of the shortest altitude.
First notice that this is a right triangle, so two of the altitudes are the legs, whose lengths are $15$ and $20$.  The third altitude, whose length is  $x$, is the one drawn to the hypotenuse. The area of the triangle is $\frac{1}{2}(15)(20) = 150$. Using 25 as the base and $x$ as the altitude, we have $$
\frac{1}{2}(25)x = 150, \quad \text{so} \quad
x = \frac{300}{25} = \boxed{12}.
$$


[asy]
draw((0,0)--(15,0)--(0,20)--cycle);
draw((0,0)--(9.6,7.2),dashed);
label("15",(7.5,0),S);
label("20",(0,10),W);
label("25",(7.5,10),NE);
label("$x$",(4.8,3.6),N);
[/asy]

$$
\text{OR}
$$

Since the three right triangles in the figure are similar, $$
\frac{x}{15} = \frac{20}{25}, \quad \text{so} \quad
x=\frac{300}{25} = \boxed{12}.
$$
12
Level 4
Geometry
Problem: The sides of a triangle have lengths of $15$, $20$, and $25$. Find the length of the shortest altitude.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the shortest altitude of the triangle, we first identify the triangle's type. The sides 15, 20, and 25 suggest it is a right triangle, as they satisfy the Pythagorean theorem: \(15^2 + 20^2 = 225 + 400 = 625 = 25^2\). 

Next, we calculate the area of the triangle using the formula for the area of a right triangle: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). We can take the legs of the triangle as the base and height. Thus, the area is:

\[
\text{Area} = \frac{1}{2} \times 15 \times 20 = 150.
\]

The altitude corresponding to each side can be calculated using the area formula \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude} \). We will find the altitudes corresponding to each side.

1. For the side of length 25 (hypotenuse):
   \[
   150 = \frac{1}{2} \times 25 \times h_{25} \implies 150 = 12.5 \times h_{25} \implies h_{25} = \frac{150}{12.5} = 12.
   \]

2. For the side of length 20:
   \[
   150 = \frac{1}{2} \times 20 \times h_{20} \implies 150 = 10 \times h_{20} \implies h_{20} = \frac{150}{10} = 15.
   \]

3. For the side of length 15:
   \[
   150 = \frac{1}{2} \times 15 \times h_{15} \implies 150 = 7.5 \times h_{15} \implies h_{15} = \frac{150}{7.5} = 20.
   \]

Now we compare the altitudes calculated: \( h_{25} = 12 \), \( h_{20} = 15 \), and \( h_{15} = 20 \). The shortest altitude is \( h_{25} \), which is \( 12 \).

Thus, the final answer is \( 12 \).<|END_OF_LATENT|> The final answer is $12$.