Q1
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
Answer: 126
Key idea: Find the area once, then each altitude is \( \frac{2 \times \text{Area}}{\text{base}} \).
1. In an isosceles triangle the altitude from A bisects BC, so BD = 40 cm and the altitude from A is \( \sqrt{50^2 - 40^2} = 30 \) cm.
2. Area \( = \frac{1}{2} \times 80 \times 30 = 1200 \) sq. cm.
3. The altitude to AB (and, by symmetry, to AC) is \( \frac{2 \times 1200}{50} = 48 \) cm.
4. Sum of altitudes \( = 30 + 48 + 48 = 126 \) cm.
Tip: The 50-40-30 split is the 3-4-5 triple scaled by 10; spotting it saves a square root.
Hence, the answer is 126.