Find the value of x so that the figures have the same perimeter.

F 1.5 G 2 H 3.2 J 4

The perimeter of a rectangle is given as $2\left(l+w\right)$, where $l$is the length of the rectangle and $w$is the width of the rectangle.

From the figure, the length of the first rectangle is $6$units and the width of the first rectangle is $x$units. Then, the perimeter of the first rectangle is $2\left(6+x\right)$units.

Similarly, from the given figure, the length of the second rectangle is $x$units and the width of the second rectangle is $2x+2$units. Then, the perimeter of the second rectangle is $2\left(x+2x+2\right)$units.

It is given that the figures have the same perimeter. Then, the perimeters of the first and second rectangle are equal. So, “$2\left(6+x\right)=2\left(x+2x+2\right)$”.

This is the required equation.

Solving,

$2\left(6+x\right)=2\left(x+2x+2\right)$ (Given equation)

$\frac{2\left(6+x\right)}{2}=\frac{2\left(x+2x+2\right)}{2}$ (Divide both sides by 2)

$6+x=3x+2$ (Simplify)

$6+x-3x=3x+2-3x$ (Subtract $3x$from both sides)

$6-2x=2$ (Simplify)

$6-2x-6=2-6$ (Subtract 6 from both sides)

$-2x=-4$ (Simplify)

$\frac{-2x}{-2}=\frac{-4}{-2}$ (Divide both sides by $-2$)

$x=2$ (Simplify)

So, $x=2$is the solution of the given equation.

It was assumed that the perimeters of both figures are equal. This assumption led to the obtained solution $x=2$.

Thus, the value of x so that the figures have the same perimeter is 2.