In an isosceles triangle ABC, angle A is twice the angle B, and the lengths of the sides

In an isosceles triangle ABC, angle A is twice the angle B, and the lengths of the sides opposite these angles are 16 and 4 cm. Find the area of ABC.

In triangle ABC: AB = BC – lateral sides, AC – base, angle A = 2 * angle B.
1. Since AC is the base, then the angles A and C (angles at the base) are equal, then:
angle A = angle C = 2 * angle B.
Let’s designate angle B as x, then angle A = angle C = 2x.
By the theorem on the sum of the angles of a triangle:
angle A + angle B + angle C = 180 degrees;
2x + x + 2x = 180;
5x = 180;
x = 180/5;
x = 36.
Thus, angle B = x = 36 degrees, angle A = angle C = 2x = 2 * 36 = 72 degrees.
It is known that opposite a larger angle lies a large side, then:
AB = BC = 16 cm;
AC = 4 cm.
2. You can find the area using Heron’s formula for an isosceles triangle:
S = (p – b) * √p (p – a),
where p is a semi-perimeter, a is the length of the base, b is the length of the lateral side.
Semi-perimeter:
p = (a + 2b) / 2 = (4 + 2 * 16) / 2 = (4 + 32) / 2 = 36/2 = 18.
S = (18 – 16) * √18 (18 – 4) = 2 * √18 * 14 = 2 * √2 * 9 * 2 * 7 = 2 * 2 * 3 * √7 = 12√7 (cm ^ 2 ).
Answer: S = 12√7 cm ^ 2.