The height drawn to the base of an isosceles triangle is $h$ and is twice as large as its projection on the lateral side. Find the area of the triangle.

#
Find the angle between the given height and the lateral side of the isosceles triangle.

## Solution

Let $B P$ be the projection of the height $B K$ of the isosceles triangle $A B C (A B = B C)$ on the lateral side $B C$.

Since $B P = \frac{1}{2} B K$, then $\angle K B P = 60^{\circ}$. Therefore,

$$
C K = B K \operatorname{tg} 60^{\circ} = h \sqrt{3}, A C = 2 C K = 2 h \sqrt{3} .
$$

Thus,

$$
S_{\triangle \mathrm{ABC}} = \frac{1}{2} A C \cdot B K = h^{2} \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4fce0705b02fa9e106d4g-24.jpg?height=320&width=1106&top_left_y=1455&top_left_x=476)

Answer

$h^{2} \sqrt{3}$.