In a triangle ABC AB = 13 BC = 15 AC = 14. Find the perimeter of BMH if BH is the height, AM is the median.

Since, according to the condition AM, the median of the triangle ABC, then BM = CM = BC / 2 = 15/2 = 7.5 cm.

Consider a right-angled triangle BHC, in which HM is the median drawn from the vertex of the right angle and, therefore, is equal to half the length of the hypotenuse. HM = BC / 2 = 15/2 = 7.5 cm.

Let’s define the area of ​​the triangle ABC in two ways.

1) S = √ (p * (p – AB) * (p – BC) * (p – AC), where p is the semiperimeter of the triangle.

2) S = AC * BH / 2 = 7 * BH.

p = (AB + BC + AC) / 2 = (13 + 15 + 14) / 2 = 21 cm.

S = √21 * 8 * 6 * 7 = √7056 = 84 cm2.

84 = 7 * BH.

BH = 12 cm.

Then the perimeter of the triangle BHM is:

Rvnm = 7.5 + 7.5 + 12 = 27 cm.

Answer: The perimeter of the triangle BHM is 27 cm.