The perimeter of an isosceles triangle is 234 and the base is 104. Find the area of the triangle.

Given is an isosceles triangle ABC with base AB = 104 and perimeter P = 234.

The perimeter of a triangle is the sum of the lengths of all sides. The base AB is 104, which means the sum of the two remaining sides:
AC + CB = P – AB = 234 – 104 = 130
The triangle is isosceles, which means AC = CB. Then the length of each of them is equal to:
AC = CB = 130: 2 = 65

The height of the CO triangle, lowered to the base, divides the base AB into two equal parts:
AB = AO + OB = 2 * AO
AO = AB: 2 = 104: 2 = 52

In a right-angled triangle AOS, by the Pythagorean theorem, we obtain the equality:
AC² = CO² + AO², from where we find the CO height:
CO² = AC² – AO²
CO = √ (AC² – AO²)
CO = √ (65² – 52²)
CO = √1521
CO = 39

The area of ​​the triangle is:
S = (1/2) * AB * CO
S = (1/2) * 52 * 39
S = 1014