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Area of a triangle (using two sides and the angle between them). [Ratio of areas of triangles with a common base or common height] Class

Given triangle $A B C$, where angle $B$ is $30^{\circ}, A B=4, B C=6$. The bisector of angle $B$ intersects side $A C$ at point $D$. Find the area of triangle $A B D$.

#
Apply the property of the bisector of a triangle.

## Solution

Note that

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot B C \cdot \sin \angle B=6
$$

By the property of the bisector of a triangle,

$$
\frac{A D}{C D}=\frac{A B}{B C}=\frac{2}{3}
$$

Therefore, $\frac{A D}{A C}=\frac{2}{5}$. Hence,

$$
S_{\triangle \mathrm{ABD}}=\frac{A D}{A C} S_{\Delta \mathrm{ABC}}=\frac{2}{5} \cdot 6=2.4
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-22.jpg?height=277&width=924&top_left_y=2217&top_left_x=567)

Answer

2.4 .

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