$_{\text {[Ratio of areas of triangles with a common base or common height] }]}$

Point $M$ divides side $A B$ of triangle $A B C$ in the ratio 2 : 5. In what ratio does segment $C M$ divide the area of triangle $A B C$?
Triangles $A C M$ and $B C M$ have a common height drawn from vertex $C$.

## Solution

Let $C H$ be the height of triangle $A B C$. Then

$$
S_{\triangle \mathrm{ACM}}=\frac{1}{2} A M \cdot C H, S_{\triangle \mathrm{BCM}}=\frac{1}{2} B M \cdot C H .
$$

Therefore,

$$
\frac{S_{\triangle A C M}}{S_{\triangle B C M}}=\frac{\frac{1}{2} A M \cdot C H}{\frac{1}{2} B M \cdot C H}=\frac{A M}{M B}=\frac{2}{5} \text {. }
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_67482594d9a3488694ecg-14.jpg?height=297&width=717&top_left_y=2310&top_left_x=676)

## Answer

 $\qquad$![](https://cdn.mathpix.com/cropped/2024_05_06_67482594d9a3488694ecg-14.jpg?height=56&width=58&top_left_y=140&top_left_x=2014)$\qquad$

![](https://cdn.mathpix.com/cropped/2024_05_06_67482594d9a3488694ecg-14.jpg?height=58&width=52&top_left_y=248&top_left_x=2009)

In a triangle, one side is 16 and forms a $60^{\circ}$ angle with the base; the other side is 14. Find the base.

## Hint

Apply the Law of Cosines.

## Solution

Let in triangle $A B C$ it is known that $\angle C=60^{\circ}, A C=16, A B=14$. Denote $B C=x$. By the Law of Cosines

$$
A B^{2}=A C^{2}+B C^{2}-2 A C \cdot B C \cos \angle C
$$

or

$$
196=x^{2}+256-16 x \text {, or } x^{2}-16 x+60=0 .
$$

From this, we find that $x=6$ or $x=10$.

![](https://cdn.mathpix.com/cropped/2024_05_06_67482594d9a3488694ecg-15.jpg?height=277&width=468&top_left_y=839&top_left_x=795)

## Answer

10 or 6.