Question

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

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

Answer 1

The correct option is C

$24\sqrt{5}cm$

Explanation for the correct option:

Find the area of the triangle

Let the sides of the triangle be $a=35cm,b=54cm,c=61cm$

To find the area of the triangle, we will use Heron's formula.

Heron's formula can be given that: $A=\sqrt{s(s-a)(s-b)(s-c)}$

Where $s=\frac{a+b+c}{2}$ and $a,b,c$ are the sides of the triangle.

Now, the semi-perimeter will be,

$$\begin{gathered} s=\frac{35+54+61}{2} \\ \Rightarrow s=\frac{150}{2} \\ \Rightarrow s=75cm \end{gathered}$$

Now, the area of the triangle will be,

$$\begin{gathered} A=\sqrt{75(75-35)(75-54)(75-61)} \\ \Rightarrow A=\sqrt{75\left(40\right)\left(21\right)\left(14\right)} \\ \Rightarrow A=420\sqrt{5}c{m}^2 \end{gathered}$$

Find the length of the longest altitude

Since the smallest side is $35cm$. Therefore, it will have the longest altitude.

It is known that the area of the triangle can be given as: $\frac{1}{2}\times base\times altitude$

Now, the altitude will be,

$$\begin{gathered} \frac{1}{2}\times 35\times altitude=420\sqrt{5}c{m}^2 \\ \Rightarrow 35\times altitude=840\sqrt{5} \\ \Rightarrow altitude=24\sqrt{5}cm \end{gathered}$$

Hence, the correct option is (c) $24\sqrt{5}cm$.