Question

The sides of a triangle are 35 cm, 54 cm and 61 cm respectively. The length of its longest altitude is
(a) $16\sqrt{5}\mathrm{cm}$
(b) $10\sqrt{5}\mathrm{cm}$
(c) $24\sqrt{5}\mathrm{cm}$
(d) 28 cm

Answer 1

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



Let CD be the longest altitude of the triangle.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


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

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{75\left(75-35\right)\left(75-54\right)\left(75-61\right)} \\ =\sqrt{75\left(40\right)\left(21\right)\left(14\right)} \\ =\sqrt{\left(5\times 5\times 3\right)\left(5\times 2\times 2\times 2\right)\left(3\times 7\right)\left(2\times 7\right)} \\ =\left(5\times 3\times 7\times 2\times 2\right)\sqrt{5} \\ =420\sqrt{5}{\mathrm{cm}}^2 \end{gathered}$$


We know,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ \Rightarrow 420\sqrt{5}=\frac{1}{2}\times 35\times CD \\ \Rightarrow 12\sqrt{5}=\frac{1}{2}CD \\ \Rightarrow CD=24\sqrt{5} \end{gathered}$$

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

Hence, the correct option is (c).