No.39

The altitude, equal sides, and nonequal sides of an isosceles triangle have lengths that are, in the order listed, consecutive even numbers of centimeters.
What is the area of the triangle?

The (even) altitude = 2k for some integer k.
The slanted side = 2k + 2
The base is 2k + 4, so half of it is k + 2.

The altitude of an isosceles triangle splits it into two right triangles,
so from the Pythagorean Theorem we have:
(k + 2)^2 + (2k)^2 = (2k + 2)^2

k^2 + 4k + 4 + 4k^2 = 4k^2 + 8k + 4
k^2 = 4k

k = 0 or 4

0 would not be a "triangle", but if you allow for
the degenerate case of 0 altitude, 2 on the sides and 4 on the base, the numbers work,
and the area is 0.

For k = 4, we have:
altitude 8, sides 10, base 12 (two 6-8-10 right triangles, which also have consecutive even sides)
and the area is 8 * 12/2 = 48