Three sides are given in triangle ABC: AB = 26 cm, BC = 30 cm and AC = 28 cm.

| June 20, 2021 | education

Three sides are given in triangle ABC: AB = 26 cm, BC = 30 cm and AC = 28 cm. Find the part of the area of this triangle, enclosed between the height and the bisector drawn from the vertex B.

We will use the property of the bisector of the angle of a triangle, it divides the opposite side into segments proportional to the adjacent sides.

AB / AM = BC / CM.

Let AM = X cm, then CM = AC – X = 28 – X.

26 / X = 30 / (28 – X).

30 * X = 26 * (28 – X).

56 * X = 728.

X = AM = 13 cm.

CM = 28 – 13 = 15 cm.

Let us determine the area of ​​the triangle ABC through a semiperimeter.

P = (AB + BC + AC) / 2 = (26 + 30 + 28) / 2 = 42.

Savs = √p * (p – AB) * (p – BC) * (p – AC) = √42 * 16 * 12 * 14 = √112896 = 336 cm2.

Also Savs = AC * BH / 2.

336 * 2 = 28 * BH.

BH = 672/28 = 24 cm.

BH is also the height of triangles ABM and BMC.

Sawm = AM * BH / 2 = 13 * 24/2 = 156 cm2.

Svms = CM * BH / 2 = 15 * 24/2 = 180 cm2.

Then Svnm = Svms – Savm = 180 – 156 = 24 cm2.

Answer: Svnm = 24 cm2.



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