Question

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

Answer 1

Step 1: Find the area of triangle ​

Given : The sides of a triangle are $35$ cm, $54$ cm and $61$ cm

Let $a=35\text{cm},b=54\text{cm}$ and $c=61\text{cm}$

$\text{Semi-perimeter}\left(s\right)=\frac{a+b+c}{2}$

$=\frac{35+54+61}{2}$

$=\frac{150}{2}$


$=75$

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

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

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

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

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

$=420\sqrt{5}{\text{cm}}^2$

Step 2: Find the length of the longest altitude​

The sides of a triangle are $35$ cm, $54$ cm and $61$ cm

Area of triangle $=420\sqrt{5}{\text{cm}}^2$

The length of the longest altitude will be corresponding to the shortest base of the given triangle

$\Rightarrow \text{Base}=35\text{cm}$

$\therefore \text{Area of triangle}=\frac{1}{2}\times \text{Base}\times \text{Height}$

$\Rightarrow \frac{1}{2}\times \text{Base}\times \text{Height}=420\sqrt{5}$

$\Rightarrow \text{Height}=\frac{420\sqrt{5}\times 2}{\text{Base}}$

$\Rightarrow \text{Height}=\frac{840\sqrt{5}}{35}$

$=24\sqrt{5}$

Thus, the length of the longest altitude is $24\sqrt{5}\text{cm}$

Hence, (c) is the correct answer.