A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the sum of the lengths, in cm, of all three altitudes of the triangle ABC is

The correct option is 126.

Let the vertices of the isosceles triangle be $A$, $B$, and $C$ with side lengths:
$AB=AC=50\text{cm}$
$BC=80\text{cm}$

First, let us find the altitude from the vertex $A$ to the base $BC$. Let this altitude be $h_a$, meeting $BC$ at its midpoint $D$. Since $\Delta ABC$ is isosceles with $AB=AC$, the altitude $AD$ bisects the base $BC$. Therefore, we have:
$BD=DC=\frac{BC}{2}=\frac{80}{2}=40\text{cm}$

Using the Pythagorean theorem in the right-angled triangle $ABD$:
$A{D}^2+B{D}^2=A{B}^2$
$h_a^2+{40}^2={50}^2$
$h_a^2+1600=2500$
$h_a^2=900$
$h_a=30\text{cm}$

Now, we calculate the area $\left(\Delta \right)$ of the triangle $ABC$:
$\Delta =\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times BC\times AD$
$\Delta =\frac{1}{2}\times 80\times 30=1200{\text{cm}}^2$

Next, let us find the altitudes to the other two equal sides $AB$ and $AC$. Let $h_b$ be the altitude to side $AC$ and $h_c$ be the altitude to side $AB$. Since $AB=AC$, these two altitudes are equal in length:
$h_b=h_c$

We can express the area of the triangle using the base $AC$ and its corresponding altitude $h_b$:
$\Delta =\frac{1}{2}\times AC\times h_b$
$1200=\frac{1}{2}\times 50\times h_b$
$1200=25\times h_b$
$h_b=\frac{1200}{25}=48\text{cm}$

Since the triangle is symmetric with respect to the equal sides:
$h_c=48\text{cm}$

Finally, we sum the lengths of all three altitudes:
$\text{Sum}=h_a+h_b+h_c$
$\text{Sum}=30+48+48=126\text{cm}$