94. In a right triangle, the hypotenuse is 25 feet, and the area is 150 square feet. Determine the legs.
94. Nipsus's Solution. We will indicate the steps of his calculations using modern notation. Let $b$ and $c$ be the legs, $\Delta$ the area, and $a$ the given hypotenuse. We have:

$$
\begin{aligned}
& a^{2}=b^{2}+c^{2} \\
& 4 \Delta=2 b c
\end{aligned}
$$

Adding and subtracting give:

$$
\begin{aligned}
& a^{2}+4 \Delta=(b+c)^{2} \\
& a^{2}-4 \Delta=(b-c)^{2}
\end{aligned}
$$

from which

$$
\begin{aligned}
& b+c=\sqrt{a^{2}+4 \Delta} \\
& b-c=\sqrt{a^{2}-4 \Delta}
\end{aligned}
$$

Adding leads to

$$
b=\frac{\sqrt{a^{2}+4 \Delta}+\sqrt{a^{2}-4 \Delta}}{2}
$$

Then Nipsus calculates $c$ not by subtraction, as we would do, but by the formula.

$$
c=b-(b-c)
$$

knowing $b$ and $b-c$.

Given numerical values:

$$
\begin{gathered}
b+c=\sqrt{1225}=35 \\
b-c=\sqrt{25}=5
\end{gathered}
$$

from which $b=20$, $c=15$.