The base of an isosceles triangle is 8 and the radius of the inscribed circle is 2. Find the area of the triangle.

Consider a triangle ABC.

The incircle center is the intersection point of the bisectors of triangle ABC.

Since triangle ABC is isosceles, the bisector and height at the base of AC coincide.

By the condition of the problem, the base is AC = 8, and the radius of the inscribed circle is OH = OG = OK = 2.

Obviously, AH = HC = 0.5 * AC.

From the right-angled triangle AOH we have:

OH = AH * tg (angle OAH). We denote the angle OAH = a. Then OH = AH * tg (a) and 2 = 4 * tg (a),

tg (a) = 1/2.

From a right-angled triangle ABH we have:

BH = AH * tg (2a), since AO – divides the BAH angle into hits. Then

BH = 4 * tan (2a) = 4 * sin (2a) / cos (2a) = 4 * 2 * sin (a) * cos (a) / (cos ^ 2 (a) – sin ^ 2 (a)) = 4 * 2 / (cos (a) / sin (a) – sin (a) / cos (a)) = 4 * 2 / (2 – 1/2) = 16/3.

Area of ​​triangle ABC:

S = 0.5 * AC * BH = 0.5 * 8 * 16/3 = 64/3.