The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. Findthe length of its longest altitude. (Take √5=2.23)

1. Break Down the Problem

We want to find the longest altitude of a triangle with sides $a = 35 \, \text{cm}$, $b = 54 \, \text{cm}$, and $c = 61 \, \text{cm}$. The altitude corresponding to a side is given by the formula: $h_a = \frac{2A}{a}$ where $A$ is the area of the triangle. We will first calculate the area using Heron's formula.

2. Relevant Concepts

1. Heron's Formula: The area $A$ of a triangle can be calculated as: $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter defined as: $s = \frac{a + b + c}{2}$

2. Altitude Formula: The altitude corresponding to side $a$ (denoted as $h_a$) is: $h_a = \frac{2A}{a}$ Similar equations can be formed for $b$ and $c$:

   • $h_b = \frac{2A}{b}$
   • $h_c = \frac{2A}{c}$

3. Analysis and Detail

1. Calculate the Semi-perimeter: $s = \frac{35 + 54 + 61}{2} = \frac{150}{2} = 75 \, \text{cm}$

2. Calculate the Area $A$: $A = \sqrt{75(75-35)(75-54)(75-61)}$ $= \sqrt{75 \times 40 \times 21 \times 14}$

   Now we calculate each part: $75 \times 40 = 3000$ $21 \times 14 = 294$ Multiplying these: $3000 \times 294 = 882000$ Now find the square root: $A = \sqrt{882000} = 297 \, \text{cm}^2 \quad (\text{using a calculator})$

3. Calculate each altitude:

   • For $h_a$: $h_a = \frac{2A}{a} = \frac{2 \times 297}{35} \approx 17.0 \, \text{cm}$

   • For $h_b$: $h_b = \frac{2A}{b} = \frac{2 \times 297}{54} \approx 11.0 \, \text{cm}$

   • For $h_c$: $h_c = \frac{2A}{c} = \frac{2 \times 297}{61} \approx 9.7 \, \text{cm}$

4. Verify and Summarize

The computed altitudes are:

• $h_a \approx 17.0 \, \text{cm}$
• $h_b \approx 11.0 \, \text{cm}$
• $h_c \approx 9.7 \, \text{cm}$

The longest altitude is $h_a$.

Final Answer

The length of the longest altitude of the triangle is approximately $17.0 \, \text{cm}$.