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

To find the longest altitude of triangle with sides \( 35 \, \text{cm} \), \( 54 \, \text{cm} \), and \( 61 \, \text{cm} \), we first need to calculate the area of the triangle using Heron's formula. The semi-perimeter \( s \) is calculated as \( s = \frac{35 + 54 + 61}{2} = 75 \, \text{cm} \). Using Heron's formula, the area \( A \) of the triangle can be calculated as: \[ A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{75(75-35)(75-54)(75-61)} \] Calculating that gives us: \[ A = \sqrt{75 \times 40 \times 21 \times 14} = 420 \sqrt{5} \, \text{cm}^2 \] The altitudes corresponding to each side can be computed using the formula \( h = \frac{2A}{\text{base}} \). The lengths of the altitudes are: 1. For side 61 cm: \( h_a = \frac{2A}{61} = \frac{2 \times 420\sqrt{5}}{61} = \frac{840\sqrt{5}}{61} \), 2. For side 54 cm: \( h_b = \frac{2A}{54} = \frac{840\sqrt{5}}{54} = \frac{140\sqrt{5}}{9} \), 3. For side 35 cm: \( h_c = \frac{2A}{35} = \frac{840\sqrt{5}}{35} = 24\sqrt{5} \). Thus, the longest altitude is \( 24\sqrt{5} \, \text{cm} \). Answer: a) \( 24 \sqrt{5} \, \text{cm} \) Now for some fun facts: Did you know that triangles are one of the simplest and most foundational shapes in geometry? They were studied as far back as ancient Babylon, where mathematicians explored their properties and connections to other geometric figures! Additionally, in real-world applications, understanding the properties of triangles is crucial in fields like architecture and engineering. For instance, triangular structures provide a strong framework for buildings and bridges, ensuring stability and durability against forces like wind and earthquakes. Triangles really do hold the key to strong design!