57. Cut a line segment of length 14 into two segments, such that the two resulting segments and a line segment of length 10 can form a right triangle. Then the area of this right triangle is $\qquad$ -
Answer: 24 or $\frac{120}{7}$
Solution: Let the lengths of the two segments obtained be $\mathrm{a}$ and $\mathrm{b}$.
If 10 is the hypotenuse of the right triangle, then
$$
\begin{array}{c}
a+b=14, \\
a^{2}+b^{2}=10^{2},
\end{array}
$$
Square both sides of (1), we get
$$
a^{2}+b^{2}+2 a b=196,
$$
Subtract (2) from (3), we get

At this point, the area of the triangle is
$$
\begin{aligned}
2 \mathrm{ab} & =96, \\
\frac{a b}{2} & =24 .
\end{aligned}
$$

If 10 is a leg of the right triangle, then
$$
\begin{array}{c}
\mathrm{a}+\mathrm{b}=14, \\
a^{2}-b^{2}=10^{2},
\end{array}
$$
Divide (5) by (4), we get
$$
\frac{a^{2}-b^{2}}{a+b}=a-b=\frac{100}{14}=\frac{50}{7},
$$

From (4) and (6), we can solve for
$$
\begin{aligned}
a=\frac{74}{7}, b= & \frac{24}{7} \\
\frac{b \times 10}{2}= & \frac{120}{7}
\end{aligned}
$$