Question

If the area of a triangle is 6$\sqrt{6}{cm}^2$ and its two sides are $5$ cm and $6$ cm. Find the perimeter of the triangle.

Answer 1

Let the third side of the triangle is x.

Semi perimeter of the triangle = $\frac{(5+6+x)}{2}$ = $\frac{(11+x)}{2}$

area of the triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

$\sqrt{\left(\frac{(11+x)}{2}\right)(\frac{(11+x)}{2}-x)(\frac{(11+x)}{2}-5)(\frac{(11+x)}{2}-6)}$ = 6$\sqrt{6}{cm}^2$
$\Rightarrow \left(\frac{(11+x)}{2}\right)\left(\frac{(11-x)}{2}\right)\left(\frac{1+x}{2}\right)\left(\frac{x-1}{2}\right)=216$
$\Rightarrow \left(\frac{121-x^2}{2}\right)\left(\frac{x^2-1}{2}\right)=216$
$\Rightarrow \left(\frac{121-p}{2}\right)\left(\frac{p-1}{2}\right)=216$
$\Rightarrow 121p-121-p^2+p-3456=0$
$\Rightarrow -p^2+122p-3577=0$
$\Rightarrow p^2-122p+3577=0$
$\Rightarrow p=\frac{122\pm \sqrt{14884-14308}}{2}$
$=\frac{122\pm \sqrt{576}}{2}$
$=\frac{122\pm 24}{2}$
$=\frac{122+24}{2}$ or $\frac{122-24}{2}$
$=\frac{146}{2}$ or $\frac{98}{2}$
$P=73$ or $49$
$\Rightarrow x^2=73$ or $49$
$\Rightarrow x=\sqrt{73}$ or $\sqrt{49}$

$\Rightarrow x=\sqrt{73}cm\text{or}7cm$

Case 1: If $x=7cm$
Therefore, perimeter = $5cm+6cm+7cm=18cm$
Case 2: If $x=\sqrt{73}cm$
Perimeter=$5cm+6cm+\sqrt{73}cm=(11+\sqrt{73})cm$