Question

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

$24\sqrt{5}cm$

Here $a=35cm,b=54cm,c=61cm$.
Semi perimeter, $s=\frac{a+b+c}{2}$
$=\frac{35+54+61}{2}$
$=75cm$
Area, $A=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{75(75-35)(75-54)(75-61)}$
$=\sqrt{75\times 40\times 21\times 14}$
$=\sqrt{3\times 5\times 5\times 2\times 2\times 2\times 5\times 7\times 3\times 7\times 2}$
$=7\times 5\times 3\times 2\times 2\sqrt{5}$
$=420\sqrt{5}$ ${\text{cm}}^2$
Now, $A=\frac{1}{2}\times \text{base}\times \text{altitude}$
Area of a triangle is constant.
Hence, the base corresponding to the longest altitude is the shortest.
The shortest side is $35cm$
$\therefore 420\sqrt{5}=\frac{1}{2}\times 35\times \text{altitude}$
$\Rightarrow \text{Shortest altitude}=\frac{2\times 420\sqrt{5}}{35}=24\sqrt{5}cm$