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

#
Apply Heron's formula and the property of the angle bisector of a triangle.

## Solution

Let $B P$ and $B Q$ be the height and the angle bisector of the given triangle $A B C$. By Heron's formula,

On the other hand, $S=1 / 2$ AC$\cdot$BP.

Therefore, $B P=\frac{2 S_{\triangle A B C}}{A C}=$

$\frac{2 \cdot 336}{28}=24$

By the property of the angle bisector of a triangle, $\frac{A Q}{Q C}=\frac{A B}{B C}=\frac{26}{30}=$

$\frac{13}{15}$.

Therefore, $A Q=\frac{13}{28} A C=13$. By the Pythagorean theorem in

the right triangle $A P B$, we find that

$$
A P=\sqrt{A B^{2}-B P^{2}}=
$$

$$
\sqrt{26^{2}-24^{2}}=\sqrt{2 \cdot 50}=10
$$

Thus, $P Q=A Q-A P=13$

$-10=3, S_{B P Q}=\frac{1}{2} P Q \cdot B P=\frac{3 \cdot 24}{2}$

$=36$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-16.jpg?height=360&width=646&top_left_y=1415&top_left_x=37)

# Answer

36.