Question

The length of the sides of a triangle are $2x+\frac{y}{2},5\frac{x}{3}+y+\frac{1}{2},2\frac{x}{3}+2y+\frac{5}{2}$. If the triangle is equilateral, then find its perimeter.

• A. $\frac{13}{2}$
• B. $\frac{26}{2}$
• C. $\frac{39}{2}$
• D. $\frac{42}{2}$

Answer 1

The correct option is C $\frac{39}{2}$

The given triangle is equilateral then,

$2x+\frac{y}{2}=\frac{5x}{3}+y+\frac{1}{2}\Rightarrow 2x+\frac{y}{2}-\frac{5x}{3}-y=\frac{1}{2}\Rightarrow \frac{x}{3}-\frac{y}{2}=\frac{1}{2}\Rightarrow \frac{2x-3y}{6}=\frac{1}{2}\Rightarrow 2x-3y=3\left(1\right)\frac{5x}{3}+y+\frac{1}{2}=\frac{2x}{3}+2y+\frac{5}{2}\Rightarrow \frac{5x}{3}-\frac{2x}{3}+y-2y=\frac{5}{2}-\frac{1}{2}\Rightarrow x-y=2\left(2\right)$

Multiply $\left(2\right)$ by $2$ , $2x-2y=4\left(3\right)$

Subtract $\left(3\right)$ from $\left(1\right)$, we get

$2x-3y=3\underset{–}{-2x+2y=-4}-y=-1\Rightarrow y=1$

From $\left(2\right)$ , $x-1=2\Rightarrow x=3$

$\therefore a=2x+\frac{y}{2}=2\times 3+\frac{1}{2}=\frac{13}{2}b=\frac{5x}{3}+y+\frac{1}{2}=5+1+\frac{1}{2}=\frac{10+2+1}{2}=\frac{13}{2}c=\frac{2x}{3}+2y+\frac{5}{2}=2+2+\frac{5}{2}=\frac{4+4+5}{2}=\frac{13}{2}$

Perimeter $=3\times \frac{13}{2}=\frac{39}{2}$