The side of an isosceles triangle is 10, and the base is 12. Find the area of this triangle

The area of the triangle is a * h / 2, where h is the height of the triangle, and is the side of the triangle to which the height is drawn.
SABC = AC * BD / 2
AD = DC = AC / 2 = 12/2 = 6 (by the property of an isosceles triangle, the height is the median)
Then, by the Pythagorean theorem:
AB2 = BD2 + AD2
10 ^ 2 = BD ^ 2 + 62
100 = BD ^ 2 + 36
BD ^ 2 = 64
BD = 8
SABC = AC * BD / 2 = 12 * 8/2 = 48
Answer: SABC = 48