Question

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

Hint:

In an isosceles triangle, the altitude to the base not only gives the height but also splits the base into two equal parts. Once you know the area from one base–height pair, you can easily find the other altitudes using the same area with different bases.

Answer 1

Correct Answer: 126

Step 1: Drop the altitude from A to BC; it bisects BC (isosceles triangle), so half of BC \( = 40 \). By Pythagoras, altitude \( = \sqrt{50^2-40^2} = \sqrt{900}=30 \).
Step 2: Area of triangle \( = \frac{1}{2}\times 80 \times 30 = 1200 \).
Step 3: Altitude to AC \( = \frac{2\times1200}{50}=48 \), and by symmetry altitude to AB is also 48.
\[ \boxed{30+48+48 = 126 \text{ cm}} \]