Question

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

• A. $24\sqrt{5}cm$
• B. $48\sqrt{5}cm$
• C. $56\sqrt{5}cm$
• D. $108cm$

Answer 1

The correct option is A

$24\sqrt{5}cm$

Given, $a=35\text{cm};b=54\text{cm};c=61\text{cm}.$

Semi-perimeter
$s=\frac{a+b+c}{2}=\frac{35+54+61}{2}=75\text{cm}$

$A\left(△\right)=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{75(75-35)(75-54)(75-61)}=\sqrt{75\times 40\times 21\times 14}$
$A\left(△\right)=420\sqrt{5}{\text{cm}}^2.$

Now, $A\left(△\right)=\frac{1}{2}\times \text{base}\times \text{altitude}$

The altitude corresponding to shortest base is longest.
The shortest side = 35 cm.

$\therefore 420\sqrt{5}=\frac{1}{2}\times 35\times \text{altitude}\text{altitude}=\frac{2\times 420\sqrt{5}}{35}=24\sqrt{5}\text{cm}.$