jmc

Question

What is the sum of the lengths of the altitudes of a triangle whose side lengths are 10, 10, and 12? Express your answer as a decimal to the nearest tenth.

Accepted Answer

Let us draw our triangle and altitudes and label our points of interest: We have made

A B = A C = 10

and

B C = 12.

We can notice a few useful things. Since

A B C

is isosceles, it follows that

A D

is a median as well as an altitude, which is useful since that means

B D = D C = \frac{1}{2} \cdot B C = 6.

Now, since

D C = 6

and

A C = 10,

we have a

3 : 4 : 5

Pythagorean triple and

A D = 8

. Now we can find the area of

A B C

by finding

\frac{1}{2} \cdot A D \cdot B C = \frac{1}{2} \cdot 8 \cdot 12 = 48.

Now, we can find

B E

by using the area we just found:

\frac{1}{2} \cdot A C \cdot B E = 48.

Since

A C = 10,

that means

B E = 9.6

By symmetry,

C F

is also

9.6.

Our answer is:

9.6 + 9.6 + 8 = 27.2 .