In the given figure, $ABCD$ is a rectangle and $\triangle ABE$ and $\triangle CDF$ are right triangles. The area of triangle $\triangle ABE$ is $150 \mathrm{~cm}^{2}$ and the segments $AE$ and $DF$ measure, respectively, 15 and $24 \mathrm{~cm}$. What is the length of the segment $CF$?

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-006.jpg?height=302&width=417&top_left_y=1134&top_left_x=1416)
The segment $C F$, whose length we want to calculate, is a leg of the right triangle $\triangle C D F$. The Pythagorean Theorem, applied to this triangle, states that $(C D)^{2}=(C F)^{2}+(F D)^{2}=(C F)^{2}+24^{2}$, and from this, we get $(C F)^{2}=(C D)^{2}-24^{2}$. In other words, to find $C F$, it is enough to know $C D$. Since the opposite sides of a rectangle (and more generally, of a parallelogram) are equal, we have $C D=A B$. Our goal, then, becomes the calculation of $A B$. For this, let's look at the triangle $\triangle A B E$. Its area is

$$
\frac{1}{2}(A E \times B E)=\frac{1}{2}(15 \times B E)=150
$$

from which we get $B E=20$. The Pythagorean Theorem applied to this triangle gives us $(A B)^{2}=(A E)^{2}+(B E)^{2}=15^{2}+20^{2}=625=25^{2}$, from which $A B=25$. Therefore, $C D=$ $A B=25$ and, according to our previous observation, we obtain

$$
(C F)^{2}=(C D)^{2}-24^{2}=25^{2}-24^{2}=(25+24)(25-24)=49
$$

Thus, $C F=7$.

Notice that the solution is independent of the measures of the sides $A D$ and $B E$.