The bisector of a triangle is 96 cm, the median perpendicular to it is 96. Find the sides of the triangle.

| August 8, 2021 | education

AR = CM = 96 cm.

The ACM triangle is isosceles, since AO is its height and bisector, then AC = AM, CO = OM = 96/2 = 48 cm.

Let AC = X cm, then AM = BM = X cm, AB = 2 * X cm.

AR is a bisector, then, by its property, AC / CP = AB / PB.

X / CP = 2 * X / BP.

BP = 2 * CP.

The area of ​​the AMC triangle is equal to: Sams = CM * AO / 2 = 96 * AO / 2 = 48 * AO cm2.

The PM segment is the median of the ABP triangle, then Sarm = Svrm = AR * OM / 2 = 96 * 48/2 = 2304 cm2.

The area of ​​the triangle ABC is equal to: Savs = Sarv + Sars = 2 * Sarm + Sarm = 2 * 2304 + 2304 = 6912 cm2.

Sams = Saavs / 2 = 3456 cm2.

48 * AO = 3456.

AO = 3456/48 = 72 cm, then RR = 96 – 72 = 24 cm.

In a right-angled triangle AOC, AC ^ 2 = AO ^ 2 + OC ^ 2 = 5184 + 2304 = 7488.

AC = 24 * √13 cm.

AB = 2 * AC = 48 * √13 cm.

From a right-angled triangle COP, CP ^ 2 = CO ^ 2 + OP ^ 2 = 2304 + 576 = 2880.

СР = 24 * √5 cm, then ВР = 48 * √5 cm, and ВС = 72 * √5 cm.

Answer: The sides of the triangle are 24 * √13 cm, 48 * √13 cm, 72 * √5 cm.



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