# A circle inscribed in a triangle divides one of the sides into segments equal to 3 and 4

**A circle inscribed in a triangle divides one of the sides into segments equal to 3 and 4, and the angle opposite to this side is equal to 120. Find the area of the triangle.**

Let’s draw the radii OK, ОМ, ОН to the points of tangency. By the property of a tangent drawn from one point, AK = AH = 3 cm, CM = CH = 4 cm.Let BK = BM = X cm, then AB = (3 + X) cm, BC = (4 + X) cm.

By the cosine theorem: AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 * AB * BC * Cos120.

49 = X ^ 2 + 6 * X + 9 + X ^ 2 + 8 * X + 16 + X ^ 2 + 3 * X + 4 * X + 12.

3 * X ^ 2 + 21 * X – 12 = 0.

X ^ 2 + 7 * X – 4 = 0.

Let’s solve the quadratic equation.

X = (√65 – 7) / 2.

Let’s define the semiperimeter of the triangle ABC. P = (7 + 3 + 4 + (√65 – 7)) / 2 = (7 + √65) / 2 cm.

The triangle BОК and BОМ is equal in leg and hypotenuse, then the angle ODK = 60.

In a right-angled triangle BOK, OK = R = BK * tg60 = ((√65 – 7) * √3 / 2) see.

Then Sас = р * R = ((7 + √65) / 2) * ((√65 – 7) * √3 / 2) = (65 – 49) * √3 / 4 = 4 * √3 cm2.

Answer: The area of the triangle is 4 * √3 cm2.