10.278. Given a square with side length $a$. Determine the sides of an equal-area isosceles triangle, where the sum of the lengths of the base and the height dropped to it is equal to the sum of the lengths of the two legs.
Solution.

Let $x$ be the base, $z$ be the lateral side, and $y$ be the height of the sought triangle drawn to the base. According to the problem, we have $\left\{\begin{array}{l}\frac{1}{2} x y=a^{2}, \\ x+y=2 z,\end{array}\right.$ By the Pythagorean theorem (Fig. 10.73) $y^{2}+\frac{x^{2}}{4}=z^{2}$. Thus, we have the system of equations:

$$
\begin{aligned}
& \left\{\begin{array}{l}
x y=2 a^{2}, \\
x+y=2 z, \\
y^{2}+\frac{x^{2}}{4}=z^{2}
\end{array},\left\{\begin{array}{l}
z=\frac{1}{2}(x+y), \\
x y=2 a^{2}, \\
4 y^{2}+x^{2}=4 \cdot\left(\frac{1}{2}(x+y)\right)^{2}, \\
x y=2 a^{2}, \\
4 y^{2}+x^{2}=x^{2}+2 x y+y^{2},
\end{array}\right.\right. \\
& \left\{\begin{array} { l } 
{ z = \frac { 1 } { 2 } ( x + y ) , } \\
{ x y = 2 a ^ { 2 } , } \\
{ 3 y ^ { 2 } = 4 a ^ { 2 } , }
\end{array} \left\{\begin{array}{l}
z=\frac{1}{2}(x+y), \\
x=\frac{2 a \sqrt{3}}{3}, \\
z=\frac{1}{2}\left(\frac{2 a \sqrt{3}}{3}+a \sqrt{3}\right)=\frac{5 a \sqrt{3}}{6} .
\end{array}\right.\right.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0764.jpg?height=510&width=521&top_left_y=64&top_left_x=161)

Fig. 10.73

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0764.jpg?height=670&width=424&top_left_y=65&top_left_x=727)

Fig. 10.74

Answer: $a \sqrt{3}, \frac{5 a \sqrt{3}}{6}, \frac{5 a \sqrt{3}}{6}$.