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

Draw this isosceles triangle with its base being the side = 12.

Then, the left side is 10, and the right side is also 10.

Drop a line from the top vertex down to the middle of the base; it will be perpendicular to the base.

The left-hand triangle is a right triangle with a base of 6 and a hypotenuse of 12.

Using the Pythagorean Theorem:  a² + b²  =  c²:

   a = 6     c = 10    --->                      6² + b²  =  10²     --->     36 + b²  =  100     --->   b²  =  64

                               --->     b  =  8, which is the altitude to the base = 12

The area of a triangle can be found using the formula:  A  =  ½ · b · h

The base of the triangle = 12 and the height = 8   --->   A  =  ½ · 12 · 8  =  48

Now, the area of the triangle on the left side of the triangle is one-half the total area =  24.

Using the formula for the area of the triangle for the triangle on the left side:

     the base = 10  and  the area = 24     --->     A  =  ½ · b · h     --->     48  =  ½ · 10 · h

                                       --->     h = 9.6, which is the altitude to the left side.

Since the right-hand side triangle is congruent, the altitude to the right side is also 9.6.

Summing these three values:  12 + 9.6 + 9.6  =  31.2

Here is a pic of the triangle :

The three altitudes are DC = 8 , AE and BF  = 9.6

And the sum of these = 8 + 2(9.6)  =  27.2