Question

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

$24\sqrt{5}cm$

Thinking Process
(i) First, determine the semi-perimeter, s and then determine the area of the triangle by using Heron’s formula.

(ii) For the longest altitude, take base as the smallest side. Apply the formula,
$Area=\frac{1}{2}\times Base\times Altitude$

(iii) Equate the area obtained using the two formulae and obtain the required height.

Let ABC be a triangle in which sides AB = 35 cm, BC = 54 cm, CA = 61 cm

Now, semi-perimeter of the triangle,
$s=\frac{a+b+c}{2}=\frac{35+54+61}{2}=\frac{150}{2}=75cm$
$[\because semi-perimeter,s=\frac{a+b+c}{2}]$
$\because Areaof\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{75(75-35)(75-54)(75-61)}$
$=\sqrt{75\times 40\times 21\times 14}$
$=\sqrt{25\times 3\times 4\times 2\times 5\times 7\times 3\times 7\times 2}$
$=5\times 2\times 2\times 3\times 7\sqrt{5}=420\sqrt{5}c{m}^2$
$AlsoAreaof\Delta ABC=\frac{1}{2}\times AB\times Altitude$
$\Rightarrow \frac{1}{2}\times 35\times CD=420\sqrt{5}$
$\Rightarrow CD=\frac{420\times 2\sqrt{5}}{35}$
$\therefore CD=24\sqrt{5}$
Hence, the length of altitude is $24\sqrt{5}cm$