The figure represents Mr. Arlindo's land. This land is divided 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_4fd1b602786ad9981479g-02.jpg?height=690&width=785&top_left_y=570&top_left_x=506)

a) What is the total area of the land?

b) As this land will be donated to Mr. Arlindo's two sons, the eldest son, Marlindo, proposes that the fence be rebuilt, represented by the segment $A F$ in the new drawing, with $F$ belonging to the segment $C D$, in such a way that it divides the land into two equal areas. What should be the measure of $C F$?

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-02.jpg?height=690&width=790&top_left_y=1730&top_left_x=541)
c) The younger son, Solindo, agrees with his brother regarding the equal division of the areas, but thinks the fence should connect points $C$ and $G$, with $G$ being a point on the segment $A E$. What is the measure of $A G$?

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-03.jpg?height=682&width=785&top_left_y=627&top_left_x=739)
Solution

a) Let $A$ be the total area, then $A=120+\frac{(20+10) \cdot 10}{2}=120+150=270 \mathrm{~m}^{2}$.

b) Since the fence must divide it into two equal areas, the area of the quadrilateral $A B C F$ must be half of the total area of the land, that is, $135 \mathrm{~m}^{2}$, and consequently, the area of the triangle $A C F$ must be $135-120=15 \mathrm{~m}^{2}$. For this, we must have $\frac{20 \cdot C F}{2}=15$ and thus it follows that $C F=1.5 \mathrm{~m}$.
c) The perpendiculars to the segment $A C$ from points $G$ and $E$ intersect this segment at $H$ and $I$, respectively. Since $C D E I$ is a square with side $10 \mathrm{~m}$, $I$ is the midpoint of $A C$. Therefore, the triangle $A E I$ is isosceles and right-angled. The same applies to the triangle $A G H$, which results in $A H=H G$. As in the previous item, the area of the triangle $A C G$ must be $15 \mathrm{~m}^{2}$, and thus we have $\frac{20 \cdot H G}{2}=15$. It follows that $H G=1.5 \mathrm{~m}$. Finally, applying the Pythagorean Theorem to the triangle $A G H$, we arrive at $A G=1.5 \sqrt{2} \mathrm{~m}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-04.jpg?height=764&width=873&top_left_y=649&top_left_x=502)