CAT 2025 Slot 3 · Quant · Question 1

The triangle is isosceles, so the altitude from A bisects BC, cutting off a right triangle with legs 40 and the altitude, and hypotenuse 50.

That altitude is \sqrt{{50}^{2}-{40}^{2}}=\sqrt{2500-1600}=\sqrt{900}=30 cm.

The area is therefore \frac{1}{2}\times 80\times 30=1200 sq cm.

Every altitude satisfies \text{area}=\frac{1}{2}\times \text{side}\times \text{altitude}, so the altitude to a side of length s is \frac{2400}{s}.

The altitude to AB is \frac{2400}{50}=48 cm, and by symmetry the altitude to AC is also 48 cm.

The sum is 30+48+48=126 cm.

Note that \angle A is obtuse here, so those two altitudes meet the extensions of AC and AB rather than the sides themselves — the lengths are unaffected.

Shortcut. Find the area once, then every altitude is \frac{2\times \text{area}}{\text{side}}. With sides 50, 50 and 80 the three altitudes are \frac{2400}{50}, \frac{2400}{50} and \frac{2400}{80}, so the sum is 48+48+30 without any further geometry.