In the figure, $ABCD$ is a rectangle and $E$ is a point on the segment $\overline{CD}$ such that $|CE|=2|DE|$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-11.jpg?height=362&width=534&top_left_y=506&top_left_x=841)

If the area of triangle $BCE$ is $10 \mathrm{~m}^{2}$, calculate the area of rectangle $ABCD$.

#
Solution

Remember that the area of a triangle is given by (base $\times$ height)/2. For triangle $B E D$, if we consider $E D$ as its base, then $B C$ will be its height. Therefore, the area of triangle $B E D$ is equal to

$$
\frac{|E D| \times|B C|}{2}
$$

On the other hand, the area of triangle $B C E$ can be calculated as

$$
\frac{|C E| \times|B C|}{2}
$$

By comparing quantities $(\square)$ and $(\bullet)$ and using that $C E=2 E D$, we conclude that the area of triangle $B C E$ is twice the area of triangle $B E D$. Therefore, the area of triangle $B E D$ is $5 \mathrm{~m}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-33.jpg?height=391&width=645&top_left_y=1306&top_left_x=587)

Thus, the area of triangle $B C D$ is equal to $(10+5) \mathrm{~m}^{2}$. Finally, the area of rectangle $A B C D$ is equal to twice the area of triangle $B C D$, so the answer is $2 \times(10+5)=30 \mathrm{~m}^{2}$.