In triangle ABC, side a = 26 cm, side b = 28 cm, side c = 30 cm.

In triangle ABC, side a = 26 cm, side b = 28 cm, side c = 30 cm. Find the part of the area of this triangle between the height and the bisector of angle B.

We denote the height BP, the bisector of the ВK.
The sides of the triangle are known, we will find its area according to Heron’s formula.
We find the semi-perimeter of the triangle ABC:
p = (AB + BC + AC) / 2 = 42.
We find the area of ​​the triangle ABC:
S ABC = √ (p * (p – AB) * (p – BC) * (p – AC)) = 336.
Knowing the area of ​​the triangle ABC, we find the height BP:
BP = 2 * S ABC / AC = 24.
Let’s use the property of the angle bisector and write the ratio:
AK / KС = AB / BC = 26/30 = 13/15.
AK = 13/28 * AC = 13.
In the right-angled triangle AРВ, we find the AР leg:
AP = √ (AB² – BP²) = √ (676 – 576) = √100 = 10.
PK = AK – AP = 13 – 10 = 3.
Find the area of ​​the ВРK triangle:
S ВРK = 1/2 * РK * ВР = 36.
Answer: The area is 36 square units.