3. Given a triangle $ABC$ with mutually perpendicular medians $t_{a}=12 \mathrm{~cm}$ and $t_{b}=20 \mathrm{~cm}$. Calculate the area of $\triangle ABC$.
Solution. First method. Let $A_{1}, B_{1}, C_{1}$ denote the midpoints of sides $B C, C A$, and $A B$, respectively. We have $\overline{A A_{1}}=t_{a}=12 \mathrm{~cm}$ and $\overline{B B_{1}}=t_{b}=20 \mathrm{~cm}$. We consider the triangles $A C_{1} B_{1}, A_{1} B_{1} C_{1}, C_{1} B A_{1}$, and $B_{1} A_{1} C$. For $\triangle A C_{1} B_{1}$, we have $\overline{A C_{1}}=\frac{\overline{A B}}{2}, \overline{A B_{1}}=\frac{\overline{A C}}{2}$, and $\overline{C_{1} B_{1}}=\frac{\overline{B C}}{2}$ as the midline in $\triangle A B C$. The triangle $A_{1} B_{1} C_{1}$ is a triangle whose

![](https://cdn.mathpix.com/cropped/2024_06_05_b36983d4a97bcdb99a36g-06.jpg?height=514&width=464&top_left_y=1329&top_left_x=1018)
sides are the midlines of $\triangle A B C$, so $\overline{A_{1} C_{1}}=\frac{\overline{A C}}{2}, \overline{A_{1} B_{1}}=\frac{\overline{A B}}{2}$, and $\overline{C_{1} B_{1}}=\frac{\overline{B C}}{2}$. For $\triangle C_{1} B A_{1}$, we have $\overline{C_{1} B}=\frac{\overline{A B}}{2}, \overline{B A_{1}}=\frac{\overline{B C}}{2}$, and $\overline{C_{1} A_{1}}=\frac{\overline{C A}}{2}$ as the midline in $\triangle A B C$. For $\triangle B_{1} A_{1} C$, we have $\overline{B_{1} C}=\frac{\overline{A C}}{2}, \overline{C A_{1}}=\frac{\overline{B C}}{2}$, and $\overline{B_{1} A_{1}}=\frac{\overline{B A}}{2}$ as the midline in $\triangle A B C$.

From the above, it follows that $\triangle A B C$ is composed of four congruent triangles. Let $P^{\prime}$ denote the area of each of the four congruent triangles. Then the area of $\triangle A B C$ is $P=4 P^{\prime}$. Therefore, it is sufficient to calculate the area $P^{\prime}$.

Since $A_{1} B_{1}$ is a midline in $\triangle A B C$, it is parallel to $A B$, which means that $A B A_{1} B_{1}$ is a trapezoid for which the diagonals satisfy $A A_{1} \perp B B_{1}$. From this, the area of the trapezoid is $P_{A B A_{1} B_{1}}=\frac{\overline{A A_{1}} \cdot \overline{B B_{1}}}{2}=\frac{12 \cdot 20}{2}=120 \mathrm{~cm}^{2}$. Since the trapezoid is composed of three congruent triangles $A C_{1} B_{1}, A_{1} B_{1} C_{1}$, and $C_{1} B A_{1}$, each with area $P^{\prime}$, we get $P_{A B A_{1} B_{1}}=3 P^{\prime}$. Therefore, $P^{\prime}=40 \mathrm{~cm}^{2}$ and $P=4 P^{\prime}=160 \mathrm{~cm}^{2}$.

Second method. Let $A_{1}, B_{1}$ denote the midpoints of sides $B C$ and $A C$, respectively, so $\overline{A A_{1}}= t_{a}=12 \mathrm{~cm}$ and $\overline{B B_{1}}=t_{b}=20 \mathrm{~cm}$. Since the centroid divides the medians in the ratio $1: 2$, we get:

$$
\begin{gathered}
\overline{A T}=8 \mathrm{~cm}, \overline{T A_{1}}=4 \mathrm{~cm} \\
\overline{A T}=\frac{40}{3} \mathrm{~cm} \text { and } \overline{T B_{1}}=\frac{20}{3} \mathrm{~cm}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_b36983d4a97bcdb99a36g-07.jpg?height=496&width=571&top_left_y=891&top_left_x=892)

Since the area of $\triangle B T C$ is equal to the area of rectangle $B T A_{1} N$, and the area of $\triangle A T C$ is equal to the area of rectangle $A T B_{1} M$, it follows that the area of $\triangle A B C$ is equal to the sum of the areas of rectangles $B T A_{1} N, A T B_{1} M$ and triangle $A B T$. Therefore, the area of $\triangle A B C$ is

$$
\frac{40}{3} \cdot 4+\frac{20}{3} \cdot 8+\frac{1}{2} \cdot \frac{40}{3} \cdot 8=160 \mathrm{~cm}^{2}
$$