The height of an isosceles triangle is 15. The side is 15 less than the base. Find the base of this triangle.

Let the length of the base of the triangle ABC be equal to X cm, then, by condition, the length of the lateral side is equal to: AB = BC = (X – 15) cm.

Since triangle ABC is isosceles, its height BN divides the base of the AC into two equal segments. AH = CH = AC / 2 = X / 2.

From the right-angled triangle BCH, according to the Pythagorean theorem,

BC ^ 2 = BH ^ 2 + CH ^ 2.

(X – 15) ^ 2 = 15 ^ 2 + (X / 2) ^ 2.

X ^ 2 – 30 * X +225 = 225 + X ^ 2/4.

3 * X ^ 2/4 – 30 * X = 0.

3 * X ^ 2 – 120 * X = 0.

X ^ 2 – 40 * X = 0.

X * (X – 40) = 0.

X1 = 0. (The root does not fit, since it does not make sense).

X2 = AC = 40 cm.

Answer: The length of the base is 40 cm.