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

Approach: Every altitude equals \(\dfrac{2\times\text{Area}}{\text{its base}}\). So find the area once, then divide by each side. The \(50\text{-}50\text{-}80\) triangle splits into two clean \(30\text{-}40\text{-}50\) right triangles, so the area comes out instantly.

Step 1: Find the area.
Drop the altitude from \(A\) onto \(BC\). It bisects \(BC\), giving two right triangles with base \(40\) and hypotenuse \(50\). Height \(=\sqrt{50^2-40^2}=\sqrt{2500-1600}=\sqrt{900}=30.\)
Area \(=\dfrac12\times 80\times 30 = 1200\ \text{cm}^2.\)

Step 2: Altitude to \(BC\) (base \(=80\)).
\(h_a = \dfrac{2\times 1200}{80} = 30\ \text{cm}.\) (Same \(30\) we just found — good check.)

Step 3: Altitudes to \(AC\) and \(AB\) (each base \(=50\)).
\(h_b = \dfrac{2\times 1200}{50} = 48\ \text{cm},\) and by symmetry \(h_c = 48\ \text{cm}.\)

Step 4: Add them.
\(h_a+h_b+h_c = 30+48+48 = 126\ \text{cm}.\)

\[\boxed{\text{Sum of altitudes} = 126\text{ cm}}\]

Answer 2

Alternate approach — using Heron's formula for the area:
Sides are \( 50, 50, 80 \). Semi-perimeter \( s = \frac{50+50+80}{2}=90 \).
Area \( = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{90 \times 40 \times 40 \times 10} = \sqrt{1440000} = 1200 \).
Altitude to a side \( = \frac{2 \times \text{Area}}{\text{that side}} \). Altitude to BC (80): \( \frac{2400}{80}=30 \). Altitude to AB (50): \( \frac{2400}{50}=48 \). Altitude to AC (50): \( 48 \) (by symmetry, since \( AB=AC \)).
Sum of altitudes \( = 30+48+48 = \) 126 cm.