CAT 2025 Slot3 Quant: the sum of the lengths, in cm, of all three altitudes of the triangle ABC is:

Step 1: Find altitude to the base $BC$
Let $AD$ be the altitude to $BC$. Since $AB=AC$, $D$ is the midpoint of $BC$.
$BD = \frac{80}{2} = 40$ cm.
In $\triangle ABD$: $AD^2 + BD^2 = AB^2 \implies AD^2 + 40^2 = 50^2$.
$AD^2 = 2500 - 1600 = 900 \implies AD = \mathbf{30}$ cm.

Step 2: Calculate the Area of $\triangle ABC$
$Area = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 80 \times 30 = \mathbf{1200}$ sq. cm.

Step 3: Find altitudes to the equal sides $AB$ and $AC$
Let $h_2$ be the altitude to side $AB$ (where $AB = 50$).
$Area = \frac{1}{2} \times AB \times h_2 \implies 1200 = \frac{1}{2} \times 50 \times h_2$.
$1200 = 25 \times h_2 \implies h_2 = \frac{1200}{25} = \mathbf{48}$ cm.
Since the triangle is isosceles, the altitude to side $AC$ is also $48$ cm.

Step 4: Calculate the total sum
Sum of altitudes $= 30 + 48 + 48 = \mathbf{126}$ cm.