10.207. A rectangle with a perimeter of 24 cm is inscribed in a triangle with sides 10, 17, and 21 cm, such that one of its sides lies on the largest side of the triangle. Find the sides of the rectangle.
Solution.

In $\triangle ABC$, we have $AC=21 \text{ cm}, AB=10 \text{ cm}, BC=17$ cm, vertices $L$ and $Q$ of rectangle $MNQL$ lie on $AC$, $M$ on $AB$, and $N$ on $BC$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0721.jpg?height=587&width=1181&top_left_y=74&top_left_x=84)

Fig. 10.17

Fig. 10.18

(Fig. 10.17). Let $ML = x$ cm. Then the perimeter of the rectangle is 24 cm and $MN = (12 - x)$ cm. $BD$ is the height of $\triangle ABC$, $K$ is the intersection point of $BD$ and $MN$, and $BK$ is the height of $\triangle MBN$. Using Heron's formula, we find the area $S$ of triangle $\triangle ABC: S = 84 \text{ cm}^2$. Then $BD = \frac{2S}{AC} = 8 \text{ cm}, BK = (8 - x) \text{ cm}$. Since $MN \parallel AC$, $\triangle MBN \sim \triangle ABC$. Therefore, $\frac{BK}{BD} = \frac{MN}{AC}; \frac{8 - x}{8} = \frac{12 - x}{21}$, $x = 5 \frac{7}{13}$. Thus, $ML = 5 \frac{7}{13}$ cm, $MN = 6 \frac{6}{13} \text{ cm}$.

Answer: $5 \frac{7}{13}$ cm, $6 \frac{6}{13} \text{ cm}$.