In the triangle ABC AB = 39, BC = 42, CA = 45. Find the area of the triangle formed by the side AC

In the triangle ABC AB = 39, BC = 42, CA = 45. Find the area of the triangle formed by the side AC, the bisector BK and the median BM.

We apply Heron’s theorem and determine the area of ​​the triangle ABC.

p = (AB + BC + AC) / 2 = (39 + 42 + 45) / 2 = 126/2 = 63 cm.

Saws = √p * (p – AB) * (p – BC) * (p – AC) = √63 * (63 – 39) * (63 – 42) * (63 – 45) = √63 * 24 * 21 * 18 = √ 571536 = 756 cm2.

The median BM divides the ABC triangle into two equal triangles, then Svcm = Sawm = Savs / 2 = 756/2 = 378 cm2.

BK is the bisector of the angle ABC, then the angle ABK = SBK = ABC / 2.

Then:

Savk = AB * BK * Sin (ABC / 2) / 2 = 39 * BK * Sin (ABC / 2) / 2.

Scvk = SV * BK * Sin (ABC / 2) / 2 = 42 * BK * Sin (ABC / 2) / 2.

Savk / Ssvk = 39/42.

Savk = 39 * Ssvk / 42.

Savs = Ssvk + 39 * Ssvk / 42.

756 = Ssvk * (81/42).

Scvk = 756 * 42/81 = 392 cm2.

Then Smvk = Ssvk – Ssvk = 392 – 378 = 14 cm2.

Answer: The area of ​​the triangle is 14 cm2.