10.258. The distances from a point $M$, lying inside triangle $A B C$, to its sides $A C$ and $B C$ are 2 cm and 4 cm, respectively. Calculate the distance from point $M$ to line $A B$, if $A B=10 \mathrm{~cm}, B C=17 \mathrm{~cm}$, $A C=21 \mathrm{~cm}$.
## Solution.

Let the required distance be $x$ (Fig. 10.63). Then $S_{\triangle A M B}=0.5 \cdot 10 x=5 x$. $S_{\triangle A M B}=S_{\triangle A B C}-S_{\triangle A M C}-S_{\triangle B M C}$. Let's find these areas: $S_{\triangle A B C}=\sqrt{24 \cdot 14 \cdot 3 \cdot 7}=$ $=84 \text{ cm}^{2}, S_{\triangle A M C}=0.5 \cdot 21 \cdot 2=21 \text{ cm}^{2}, S_{\triangle B M C}=0.5 \cdot 17 \cdot 4=34 \text{ cm}^{2}$. Therefore, $S_{\triangle A M B}=84-55=29 \text{ cm}^{2}$ and $x=5.8$ cm.

Answer: 5.8 cm.