# From the center O of the circle inscribed in the triangle ABC to the plane of this triangle

**From the center O of the circle inscribed in the triangle ABC to the plane of this triangle, a perpendicular OС with a length of 3 cm is drawn. Find the area of the triangle ASB if AB = 14 cm, AC = 15 cm, BC = 13 cm.**

Probably, in the condition there is a typo, the perpendicular to the plane is OS.

The area of the triangle ABS will be found by the formula:

S ABS = 1/2 * AB * SH.

For this formula, we need the height SH.

Consider a right-angled triangle SOH.

OH – the radius of the inscribed circle, we find it by the formula:

OH = √ ((p – AB) * (p – BC) * (p – AC) / p) = √ ((21 – 14) * (21 – 13) * (21 – 15) / 21) = √ ( 7 * 8 * 6/21) = √16 = 4 (cm).

By the Pythagorean theorem, we find the hypotenuse SH:

SH = √ (OS² + OH²) = √ (9 + 16) = √25 = 5 (cm).

Find the area of the triangle ABS:

S ABS = 1/2 * 14 * 5 = 35 (cm²).

Answer: The area of the ABS triangle is 35 cm².