Question

$△ABC$ is a triangle with vertices $A\equiv (x,x+3),B\equiv (2,1)$ and $C\equiv (3,-2)$. Area of the triangle is $5$. Then what is the length of its largest side ?

• A. $\sqrt{10}$
• B. $\frac{\sqrt{65}}{2}$
• C. $\frac{\sqrt{290}}{2}$
• D. none of these

Answer 1

Answer: (C)

$\frac{\sqrt{290}}{2}$

Given, vertices of triangle $A\equiv (x,x+3),B\equiv (2,1)$ and $C\equiv (3,-2)$
Therefore, $\frac{1}{2}\left|\begin{array}{ccc}x& x+3& 1\\ 2& 1& 1\\ 3& -2& 1\end{array}\right|=5$

$\Rightarrow |x\left(3\right)-(x+3)(-1)+1(-7)|=10$

$\Rightarrow |3x+x+3-7|=10$

$\Rightarrow 4x=14$

$\Rightarrow x=\frac{7}{2}$

Now, $AB=\sqrt{{(2-\frac{7}{2})}^2+{(1+\frac{13}{2})}^2}=\frac{\sqrt{234}}{2}$

$BC=\sqrt{{(3-2)}^2+{(-2-1)}^2}=\sqrt{10}$

$AC=\sqrt{{(3-\frac{7}{2})}^2+{(-2-\frac{13}{2})}^2}=\sqrt{\frac{290}{4}}=\frac{\sqrt{290}}{2}$

$AC>AB>BC$

Hence, $AC$ is the largest side.