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

To solve the problem, we need to find the sum of the altitudes in triangle $ABC$ where $AB=AC=50\text{ cm}$ and $BC=80\text{ cm}$.

Triangle $ABC$ is isosceles with $AB=AC$. We first find the area of the triangle using Heron's formula. The semi-perimeter $s$ is:

\( s = \frac{AB + AC + BC}{2} = \frac{50 + 50 + 80}{2} = 90\text{ cm} \) 

Using Heron's formula, the area $A$ is:

\( A = \sqrt{s(s-AB)(s-AC)(s-BC)} \)
\( A = \sqrt{90(90-50)(90-50)(90-80)} \)
\( A = \sqrt{90 \times 40 \times 40 \times 10} = \sqrt{1440000} = 1200\text{ cm}^2 \)

The altitude from $A$ to $BC$ (let's call it \( h_a \)) is:

\( h_a = \frac{2 \times A}{BC} = \frac{2 \times 1200}{80} = 30\text{ cm} \)

Since the triangle is isosceles, the altitudes from $B$ to $AC$ (let's call it \( h_b \)) and from $C$ to $AB$ (let's call it \( h_c \)) are equal. Hence, we only need to calculate one of them. Using the area relation again:

\( h_b = \frac{2 \times A}{AC} = \frac{2 \times 1200}{50} = 48\text{ cm} \)

Since \( h_b = h_c \), we have:

\( h_c = 48\text{ cm} \)

Summing all the altitudes gives:

\( h_a + h_b + h_c = 30 + 48 + 48 = 126\text{ cm} \)

The sum of the altitudes is 126 cm, which matches the given range 126,126.

Answer 2

Step 1: Identify the type of triangle. 
Given: \[ AB = AC = 50 \text{ cm}, \quad BC = 80 \text{ cm}. \] Since two sides are equal, \( \triangle ABC \) is an isosceles triangle with base \( BC \) and equal sides \( AB \) and \( AC \). 

Step 2: Altitude from \( A \) to base \( BC \) (call it \( h_1 \)).
Let \( AD \) be the altitude from vertex \( A \) to side \( BC \). In an isosceles triangle, the altitude from the vertex to the base bisects the base: \[ BD = DC = \frac{BC}{2} = \frac{80}{2} = 40 \text{ cm}. \] Consider right triangle \( \triangle ADC \): \[ AC = 50 \text{ cm (hypotenuse)}, \quad DC = 40 \text{ cm (base)}, \quad AD = h_1 \text{ (height)}. \] Using Pythagoras’ theorem: \[ h_1^2 + 40^2 = 50^2 \] \[ h_1^2 + 1600 = 2500 \] \[ h_1^2 = 2500 - 1600 = 900 \] \[ h_1 = 30 \text{ cm}. \] 

Step 3: Find the area of \( \triangle ABC \).
Using base \( BC \) and altitude \( AD \): \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] \[ = \frac{1}{2} \times 80 \times 30 \] \[ = 40 \times 30 = 1200 \text{ cm}^2. \] 

Step 4: Altitudes to sides \( AB \) and \( AC \) (call them \( h_2 \) and \( h_3 \)).
Let \( h_2 \) be the altitude to side \( AC \) and \( h_3 \) be the altitude to side \( AB \). Since \( AB = AC \), the corresponding altitudes are equal: \[ h_2 = h_3. \] Using the area formula with base \( AC = 50 \) cm and height \( h_2 \): \[ \text{Area} = \frac{1}{2} \times AC \times h_2 \] \[ 1200 = \frac{1}{2} \times 50 \times h_2 \] \[ 1200 = 25 h_2 \] \[ h_2 = \frac{1200}{25} = 48 \text{ cm}. \] Thus: \[ h_3 = h_2 = 48 \text{ cm}. \] 

Step 5: Sum of all three altitudes.
\[ \text{Sum} = h_1 + h_2 + h_3 \] \[ = 30 + 48 + 48 \] \[ = 126 \text{ cm}. \]