Let $ABC$ be a triangle. The side lengths of triangle $ABC$ are $AB = 15,$ $BC = 9,$ and $AC = 10.$ Find the length of the shortest altitude in this triangle.

Answer 1

To find the shortest altitude of triangle \( ABC \) with sides \( AB = c = 15 \), \( BC = a = 9 \), and \( AC = b = 10 \), we first compute the area of the triangle using Heron's formula.

1. Calculate the semi-perimeter \( s \):

\[ s = \frac{a + b + c}{2} = \frac{9 + 10 + 15}{2} = 17 \]

2. Compute the area \( K \) using Heron's formula:

\[ K = \sqrt{s(s-a)(s-b)(s-c)} \]

Substituting in the values:

\[ s - a = 17 - 9 = 8, \] \[ s - b = 17 - 10 = 7, \] \[ s - c = 17 - 15 = 2 \]

Now calculating \( K \):

\[ K = \sqrt{17 \cdot 8 \cdot 7 \cdot 2} \]

Calculating step-by-step:

\[ 17 \cdot 8 = 136, \] \[ 7 \cdot 2 = 14, \] \[ 136 \cdot 14 = 1904 \]

Now we find \( K \):

\[ K = \sqrt{1904} \]

To simplify \( \sqrt{1904} \), we factor \( 1904 \):

\[ 1904 = 16 \cdot 119 = 4^2 \cdot 119 \]

Thus,

\[ \sqrt{1904} = 4\sqrt{119} \]

So, the area of triangle \( ABC \) is:

\[ K = 4 \sqrt{119} \]

3. Calculate the altitudes of the triangle:

The formula for the altitude \( h_a \) from side \( a \) is:

\[ h_a = \frac{2K}{a} \]

Calculating \( h_a \):

\[ h_a = \frac{2 \cdot 4 \sqrt{119}}{9} = \frac{8 \sqrt{119}}{9} \]

Now we calculate \( h_b \) from side \( b \):

\[ h_b = \frac{2K}{b} = \frac{2 \cdot 4 \sqrt{119}}{10} = \frac{8 \sqrt{119}}{10} = \frac{4 \sqrt{119}}{5} \]

Finally, we calculate \( h_c \) from side \( c \):

\[ h_c = \frac{2K}{c} = \frac{2 \cdot 4 \sqrt{119}}{15} = \frac{8 \sqrt{119}}{15} \]

4. Compare the altitudes:

We need to compare \( h_a = \frac{8 \sqrt{119}}{9} \), \( h_b = \frac{4 \sqrt{119}}{5} \), and \( h_c = \frac{8 \sqrt{119}}{15} \).

Now we can find a common denominator to compare these values more easily. The least common multiple of \( 9, 5, \) and \( 15 \) is \( 45 \).

Rewriting \( h_a, h_b, \) and \( h_c \) using a denominator of 45:

\[ h_a = \frac{8 \sqrt{119}}{9} = \frac{40 \sqrt{119}}{45} \] \[ h_b = \frac{4 \sqrt{119}}{5} = \frac{36 \sqrt{119}}{45} \] \[ h_c = \frac{8 \sqrt{119}}{15} = \frac{24 \sqrt{119}}{45} \]

Now we can clearly see that:

\[ h_a = \frac{40 \sqrt{119}}{45}, \quad h_b = \frac{36 \sqrt{119}}{45}, \quad h_c = \frac{24 \sqrt{119}}{45} \]

The shortest altitude is \( h_c \).

Thus, the length of the shortest altitude in triangle \( ABC \) is:

\[ \boxed{\frac{8 \sqrt{119}}{15}} \]