2. The sides of a triangle are $$ 35 \mathrm{~cm}, 54 \mathrm{~cm} $$ and 61 cm , respectively. The length of its longest altitude $$ \begin{array}{ll} ext { a) } 24 \sqrt{5} \mathrm{~cm} & ext { b) } 28 \mathrm{~cm} \ ext { c) } 10 \sqrt{5} \mathrm{~cm} & ext { d) } 16 \sqrt{5} \mathrm{~cm}\end{array} $$

Question

2. The sides of a triangle are $$ 35 \mathrm{~cm}, 54 \mathrm{~cm} $$ and 61 cm , respectively. The length of its longest altitude $$ \begin{array}{ll}	ext { a) } 24 \sqrt{5} \mathrm{~cm} & 	ext { b) } 28 \mathrm{~cm} \ 	ext { c) } 10 \sqrt{5} \mathrm{~cm} & 	ext { d) } 16 \sqrt{5} \mathrm{~cm}\end{array} $$

UpStudy AI · Accepted Answer

To determine the longest altitude of a triangle with sides $$35 \mathrm{~cm}$$, $$54 \mathrm{~cm}$$, and $$61 \mathrm{~cm}$$, follow these steps:

1. **Identify the Longest Side:**
   The longest side of the triangle is $$61 \mathrm{~cm}$$. The altitude corresponding to this side will be the shortest altitude. Therefore, the longest altitude will correspond to the shortest side, which is $$35 \mathrm{~cm}$$.

2. **Calculate the Area Using Heron's Formula:**
   - **Semi-perimeter (s):**
     $$
     s = \frac{35 + 54 + 61}{2} = 75 \mathrm{~cm}
     $$
   - **Area (A):**
     $$
     A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{75 \times 40 \times 21 \times 14} = \sqrt{882000} = 420\sqrt{5} \mathrm{~cm}^2
     $$

3. **Determine the Longest Altitude:**
   The altitude corresponding to the side $$35 \mathrm{~cm}$$ is calculated as:
   $$
   h = \frac{2A}{a} = \frac{2 \times 420\sqrt{5}}{35} = \frac{840\sqrt{5}}{35} = 24\sqrt{5} \mathrm{~cm}
   $$

**Answer:**  
**a) $$24 \sqrt{5} \mathrm{~cm}$$**
The longest altitude is $$24 \sqrt{5} \mathrm{~cm}$$.

The sides of a triangle are and 61 cm , respectively. The length of its longest altitude

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Geometry Questions