3. Calculate the sum of the squares of the heights in a triangle with the given data: $c=6 \mathrm{~cm}, v_{c}=4 \mathrm{~cm}$, $a=5 \mathrm{~cm}$. Draw a diagram.
3. If we draw a picture, we observe that we have two triangles with the given data: $\triangle A B C_{1}$ and $\triangle A B C_{2}$. Let's first look at $\triangle A B C_{1}$. From $S=$ $\frac{c \cdot v_{c}}{2}=\frac{a \cdot v_{a}}{2}$, it follows that $v_{a}=\frac{c \cdot v_{c}}{a}=4.8 \mathrm{~cm}$. Let $x_{1}$ be the length of the perpendicular projection of side $a$ onto side $c$. We calculate it using the Pythagorean theorem: $x_{1}=3 \mathrm{~cm}$. Since side $c$ is $6 \mathrm{~cm}$ long, $c-x_{1}=3 \mathrm{~cm}$. Triangle $A B C_{1}$ is isosceles, so $v_{b}=v_{a}=4.8 \mathrm{~cm}$. The sum of the squares of the heights is $v_{a}^{2}+v_{b}^{2}+v_{c}^{2}=62.08 \mathrm{~cm}^{2}$.

Now let's look at $\triangle A B C_{2}$. Since the lengths of sides $c$ and $a$ and the height $v_{c}$ are the same as in triangle $\triangle A B C_{1}$, $v_{a}=4.8 \mathrm{~cm}$. Let $x_{2}$ be the length of the perpendicular projection of side $a$ onto the extension of side $c$. We calculate it using the Pythagorean theorem: $x_{2}=3 \mathrm{~cm}$. The length of side $b$ is also calculated using the Pythagorean theorem: $b=\sqrt{\left(c+x_{2}\right)^{2}+v_{c}^{2}}=\sqrt{81+16}=\sqrt{97}$. Then, from $S=\frac{c \cdot v_{c}}{2}=\frac{b \cdot v_{b}}{2}$, we calculate $v_{b}=\frac{c \cdot v_{c}}{b}=\frac{24}{\sqrt{97}} \mathrm{~cm}$. Finally, we have $v_{a}^{2}+v_{b}^{2}+v_{c}^{2}=$ $44.98 \mathrm{~cm}^{2}$.

Sketch with labeled data 1 point

Calculated $v_{a}=\frac{2 S}{a}=4.8 \mathrm{~cm}$. .1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_40efadbcfacda9b65264g-09.jpg?height=48&width=1642&top_left_y=1324&top_left_x=207)

![](https://cdn.mathpix.com/cropped/2024_06_07_40efadbcfacda9b65264g-09.jpg?height=66&width=1639&top_left_y=1366&top_left_x=206)

Calculation of the length of $b$ and $v_{b}$ for the second triangle....................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_40efadbcfacda9b65264g-09.jpg?height=60&width=1642&top_left_y=1466&top_left_x=204)