Problem 1. The isosceles triangle $ABC$ has a leg that is 3 times longer than the base. If point $D$ is the midpoint of the base, and point $E$ is the midpoint of the leg $AB$, then the perimeter of quadrilateral $AEDC$ is $42 \text{ cm}$ greater than the perimeter of triangle $EBD$. Calculate the perimeter of triangle $ABC$.
Solution. Let the base of triangle $ABC$ have length $a$. Then the length of the leg is $3a$. We have

$$
L_{A E D C}=\frac{3}{2} a+\overline{E D}+\frac{a}{2}+3 a
$$

and

$$
L_{E B D}=\frac{3}{2} a+\overline{E D}+\frac{a}{2}
$$

This means that $3 a=42$ or $a=14 \mathrm{~cm}$. The length of the leg is $42 \mathrm{~cm}$. Therefore, the perimeter of triangle

![](https://cdn.mathpix.com/cropped/2024_06_05_a1a5b9bf3a5d296219b1g-04.jpg?height=613&width=264&top_left_y=732&top_left_x=1202)
$ABC$ is $L=14+42+42=98 \mathrm{~cm}$.