# 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.

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.

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