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

Let’s denote the height of an isosceles triangle as h, the length of the edge a, and the length of the base b. We know that the height of an isosceles triangle lowered to the base forms a right angle. Thus, by the Pythagorean theorem, we can assert that:

a ^ 2 = h ^ 2 + (b / 2) ^ 2
a ^ 2 = h ^ 2 + b ^ 2/4

Since we need to find the length of the base, that is, b, we can express it from this formula:

b ^ 2/4 = a ^ 2-h ^ 2
b ^ 2 = 4a ^ 2-4h ^ 2

From the condition we know that h = 15 means:

b ^ 2 = 4a ^ 2-4h ^ 2 = 4a ^ 2-4 * 15 ^ 2 = 4a ^ 2-4 * 225 = 4a ^ 2-900

We also know that b = a + 15. Substitute in the resulting equation and get:

(a + 15) ^ 2 = 4a ^ 2-900
a ^ 2 + 2 * 15 * a + 15 ^ 2 = 4a ^ 2-900
a ^ 2 + 30a + 225-4a ^ 2 + 900 = 0
-3a ^ 2 + 30a + 1125 = 0
a ^ 2-10a-375 = 0

This is a quadratic equation and we can find the discriminant:

D = b ^ 2-4ac = (- 10) ^ 2-4 * 1 * (- 375) = 100 + 4 * 375 = 100 + 1500 = 1600
√D = √1600 = 40

a1 = (10 + 40) / 2 = 50/2 = 25
a2 = (10-40) / 2 = -30 / 2 = -15

Since the length cannot be negative, we obtain that the length of the edge is a = 25

Then b = a + 15 = 25 + 15 = 40