# The perimeter of an isosceles triangle is 90 cm, and the height dropped to the base is 15 cm.

**The perimeter of an isosceles triangle is 90 cm, and the height dropped to the base is 15 cm. Find the area of the triangle.**

Let us introduce the designations: b – lateral side, and – half of the base of this isosceles triangle. Then, knowing that its perimeter is 90 cm, we can write:

b + b + a + a = 90;

2 * b + 2 * a = 90;

b + a = 90/2 = 45.

From a right-angled triangle formed by the side, half of the base and the height h, drawn to this base, we can write:

a ^ 2 + h ^ 2 = b ^ 2;

a ^ 2 + 225 = b ^ 2.

We get the system of equations:

1) b + a = 45;

2) a ^ 2 + 225 = b ^ 2.

From the first equation, we express b through a: b = 45 – a.

Substitute this expression into the second equation of the system and solve it with respect to a:

a ^ 2 + 225 = (45 – a) ^ 2;

a ^ 2 + 225 = 2025 – 90a + a ^ 2;

90a = 1800;

a = 1800/90 = 20 cm – half of the base.

We find the area of the triangle as the product of the height and half the base:

S = h * a = 15 * 20 = 300 cm2.