Question

The sides of a triangle are 34 cm, 55 cm and 61 cm. The length of its longest altitude is

• A.
  $\frac{10\sqrt{8610}}{17}cm$
• B.
  $\frac{\sqrt{8610}}{17}cm$
• C.
  $10\sqrt{8610}cm$
• D.
  $17\sqrt{8610}cm$

Answer 1

The correct option is A

$\frac{10\sqrt{8610}}{17}cm$
Let the sides of the triangle be a, b and c.

$\therefore a=34cm,b=55cm,c=61cm$
$\therefore \text{Semi-perimeter, (S)}=\frac{a+b+c}{2}=\frac{34+55+61}{2}=75cm\therefore \text{Area of triangle}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{75\times 41\times 20\times 14}\sqrt{25\times 3\times 41\times 5\times 4\times 14}=5\times 2\sqrt{3\times 41\times 5\times 14}=10\sqrt{8610}c{m}^2$
Let h be the height of the longest altitude corresponding to the smallest side, i.e., 34 cm.

Now, area of a triangle $=\frac{1}{2}\times Base\times Height\Rightarrow 10\sqrt{8610}=\frac{1}{2}times34\times h\Rightarrow 10\sqrt{8610}=\frac{34}{2}\times h\Rightarrow h=\frac{20}{34}\sqrt{8610}=\frac{10}{17}\sqrt{8610}cm$

Hence, the correct answer is option (a).