Question

Question 7
The sides of a triangle are 35cm, 54cm, and 61cm respectively. The length of its longest altitude.

Answer 1

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

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