UPTET Paper 1 2016 Solved Paper

Concept:The longest altitude of a triangle lies opposite the smallest side. Use Heron's formula to find area, then equate area with $\frac{1}{2}\times\text{base}\times\text{height}$ to obtain the altitude.Explanation:First, compute the semi‑perimeter: $s = \frac{35+54+61}{2} = 75\ \text{cm}$.Use Heron's formula: Area $= \sqrt{75(75-35)(75-54)(75-61)}$.Simplify: $\sqrt{75\times40\times21\times14} = \sqrt{(3\times5^2)\times(2^3\times5)\times(3\times7)\times(2\times7)}$.Combine: $\sqrt{2^4\times3^2\times5^2\times7^2} = 2^2\times3\times5\times7 = 420\ \text{cm}^2$.But the factors under the square root give $\sqrt{5}$? Let's recalc carefully: $75\times40\times21\times14 = (75\times14)\times(40\times21) = 1050\times840 = 882000$. Factor: $882000 = 882 \times 1000 = (2\times441)\times1000 = 2\times21^2\times10^3 = 2\times21^2\times2^3\times5^3 = 2^4\times3^2\times7^2\times5^3$. Squareroot: $2^2\times3\times7\times5\sqrt{5} = 4\times21\times5\sqrt{5}=420\sqrt{5}$. So area = $420\sqrt{5}\ \text{cm}^2$.Now, area also equals $\frac{1}{2}\times\text{base}\times\text{height}$. To get the longest altitude, use the smallest side (35 cm) as base: $420\sqrt{5} = \frac{1}{2}\times35\times h$.Solve: $h = \frac{2\times420\sqrt{5}}{35} = \frac{840\sqrt{5}}{35} = 24\sqrt{5}\ \text{cm}$.Alternatively, if we used larger sides, the altitude would be smaller.Answer:The longest altitude is $24\sqrt{5}$ cm, which corresponds to option C.