Question

The sides of a triangle are $35cm,54cm$ and $61cm$ respectively. The length of its longest altitude

• A. $28cm$
• B. $10\sqrt{5}cm$
• C. $24\sqrt{5}cm$
• D. $16\sqrt{5}cm$

Answer 1

The correct option is C $24\sqrt{5}cm$

Let $ABC$ be a triangle with sides $AB=35cm,BC=54cm,CA=61cm$

If $s$ be its semi-perimeter, then

$s=\frac{35+54+61}{2}=\frac{150}{2}=75cm$

$\therefore$ Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{75(75-35)(75-54)(75-61)}$

$=\sqrt{75\times 40\times 21\times 14}$

$=\sqrt{25\times 3\times 4\times 2\times 5\times 7\times }$
$\sqrt{3\times 7\times 2}$

$=5\times 2\times 2\times 3\times 7\sqrt{5}$

$=420\sqrt{5}c{m}^2$

Lt ${}^{\prime }{x}^{\prime }$ be the longest altitude that will correspond to the shortest side i.e., $AB=35cm$

$\therefore$ Area of $△ABC=\frac{1}{2}\times AB\times \text{Altitude}$

$\Rightarrow \frac{1}{2}\times 35\times x=420\sqrt{5}$

$\Rightarrow x=\frac{420\times 2\sqrt{5}}{35}$

$\Rightarrow x=24\sqrt{5}cm$

Hence, $\left(c\right)$ is the correct answer.