The triangle ABC gives: AB = 3, AC = 5 and BC = 6. find the distance from the vertex C

The triangle ABC gives: AB = 3, AC = 5 and BC = 6. find the distance from the vertex C to the height lowered from the vertex B to the AC side.

By Heron’s theorem, we determine the area of the triangle ABC.

The semi-perimeter of the triangle is: p = (AB + BC + AC) / 2 = (3 + 6 + 5) / 2 = 7 cm.

Then Sav = √7 * (7 – 3) * (7 – 5) * (7 – 6) = √56 = 2 * √14 cm2.

Also Savs = AC * ВН / 2.

ВН = 2 * Savs / AC = 4 * √14 / 5 cm.

The BCН triangle is rectangular, in which, according to the Pythagorean theorem, we determine the length of the CH leg.

CH ^ 2 = BC ^ 2 – BH ^ 2 = 36 – 8.96 = 27.04.

CH = 5.2 cm.

Since the CH is greater than the AС, the HВ height lies on the continuation of the AС.

Answer: From top C to a height of 5.2 cm.