5. An isosceles triangle with sides $A B=B C=3, A C=4$ has an inscribed circle that touches the sides of the triangle at points $K, L$, and $M$. Find the ratio of the areas $S(\triangle A B C): S(\triangle K L M)$.
Answer: $\frac{9}{2}$. Solution. Since the triangle is isosceles, the inscribed circle touches the base at its midpoint, so $A M=M C=$ $A K=C L=2, B K=B L=1$. We will find the areas of the triangles (in relation to the area of $\triangle A B C): S(\triangle A K M)=S(\triangle C L M)=\frac{1}{3} S(\triangle A B C)$,
$S(\triangle B K L)=\frac{1}{9} S(\triangle A B C)$. Therefore, the area of the remaining part $S(\triangle K L M)=\frac{2}{9} S(\triangle A B C)$.

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