Question

A triangle with a perimeter of 12 has sides in the ratio 1:3:2 what is the area of this triangle?

Answer 1

Answer:

$perimeter = 12 \\ let \: sides \: be \: x \: \: \: \: \: \: 3x \: \: \: \: \: 2x \\ x + 3x + 2x = 12 \\ 6x = 12 \\ x = 2$

Answer 2

Given,

Perimeter of triangle $\[ = 12\]$

Ratio of sides of the triangle, $\[ = 1:3:2\]$

Solution,

Consider the sides of the triangle are $\[x,3x,2x\]$

$\[\begin{array}{l}x + 3x + 2x = 12\\ \Rightarrow 6x = 12\\ \Rightarrow x = 2\end{array}\]$

Hence the sides of the triangle are $\[a = 2,b = 6,c = 4\]$.

$\[\begin{array}{l}s = \frac{{a + b + c}}{2}\\ \Rightarrow s = \frac{{2 + 6 + 4}}{2}\\ \Rightarrow s = 6\end{array}\]$

Calculate the area of the triangle.

$\[\begin{array}{l} = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \\ = \sqrt {6\left( {6 - 2} \right)\left( {6 - 6} \right)\left( {6 - 4} \right)} \\ = 0\end{array}\]$

It is not possible to create a triangle with these sides.

Hence area of triangle will be zero.