The figure below represents Dona Idalina's land. This land is divided into two parts by a fence, represented by the segment $A C$. The triangular part $A B C$ has an area equal to $120 \mathrm{~m}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_1c2853cb1112cfb2ee2dg-13.jpg?height=306&width=340&top_left_y=475&top_left_x=961)

a) What is the total area of the land?

b) Dona Idalina wants to build a new fence, represented by the segment $A F$ in the figure below, to divide the land into two parts of equal area. What should be the distance $C F$?

![](https://cdn.mathpix.com/cropped/2024_05_01_1c2853cb1112cfb2ee2dg-13.jpg?height=277&width=306&top_left_y=1061&top_left_x=978)
Solution

a) $1^{st}$ solution: The figure below shows how to decompose the region $A C D E$ into a square $C D E H$ and a triangle $A H E$. Since $C D=D E=10 \mathrm{~m}$ and $A C=20 \mathrm{~m}$, it follows that $A H=10 \mathrm{~m}$. Therefore, the area of the triangle $A H E$ is half the area of a square with a side of $10 \mathrm{~m}$, that is,

$$
\frac{A H \times H E}{2}=\frac{10 \mathrm{~m} \times 10 \mathrm{~m}}{2}=50 \mathrm{~m}^{2}
$$

Since the area of the square $C D E H$ is $10 \mathrm{~m} \times 10 \mathrm{~m}=100 \mathrm{~m}^{2}$, we conclude that the area of the region $A C D E$ is $100 \mathrm{~m}^{2}+50 \mathrm{~m}^{2}=150 \mathrm{~m}^{2}$.

Alternatively, we can calculate the area of $A C D E$ as the difference between the areas of the rectangle $A C D G$ and the triangle $A G E$, that is, $20 \mathrm{~m} \times 10 \mathrm{~m}-\frac{10 \mathrm{~m} \times 10 \mathrm{~m}}{2}=150 \mathrm{~m}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_1c2853cb1112cfb2ee2dg-35.jpg?height=383&width=385&top_left_y=1616&top_left_x=744)

$2^{nd}$ solution: We can calculate the area of the right trapezoid $A C D E$, in square meters, using the usual formula:

$$
\frac{(A C+D E) \times C D}{2}=\frac{(20+10) \times 10}{2}=150
$$

The total area of the land is then area $(A C D E)+$ area $(A B C)=150 \mathrm{~m}^{2}+120 \mathrm{~m}^{2}=270 \mathrm{~m}^{2}$.

b) $1^{st}$ solution:

![](https://cdn.mathpix.com/cropped/2024_05_01_1c2853cb1112cfb2ee2dg-35.jpg?height=320&width=306&top_left_y=2361&top_left_x=778)

Since the land has $270 \mathrm{~m}^{2}$, when dividing it into two equal parts, each part will have an area of

$$
\frac{270 \mathrm{~m}^{2}}{2}=135 \mathrm{~m}^{2}
$$

Thus, we must have

$$
135 \mathrm{~m}^{2}=\operatorname{area}(A B C F)=\operatorname{area}(A B C)+\text { area }(A C F)=120 \mathrm{~m}^{2}+\operatorname{area}(A C F)
$$

and we see that area $(A C F)=15 \mathrm{~m}^{2}$. On the other hand, the area of the triangle $A C F$ is

$$
\frac{A C \times C F}{2}=\frac{20 \mathrm{~m} \times C F}{2}=10 \mathrm{~m} \times C F
$$

Therefore, $10 \mathrm{~m} \times C F=15 \mathrm{~m}^{2}$ and thus $C F=1.5 \mathrm{~m}$.

$2^{nd}$ solution: Since the land has $270 \mathrm{~m}^{2}$, when dividing it into parts of the same area $A B C F$ and $A F D E$, each part will have an area of $135 \mathrm{~m}^{2}$. We note that $A B C F$ is a trapezoid with bases $A B$ and $C F$ and height $A C=20 \mathrm{~m}$; thus

$$
135 \mathrm{~m}^{2}=\operatorname{area}(A B C F)=\frac{(12 \mathrm{~m}+C F) \times 20 \mathrm{~m}}{2}=120 \mathrm{~m}^{2}+10 \mathrm{~m} \times C F
$$

and it follows that $C F=1.5 \mathrm{~m}$.