Question

shown attached is a rectangle, with width 2x

The area of the rectangle is 8$x^{2}$
Find the perimeter of the rectangle

Answer 1

Answer:

Perimeter of this rectangle is 12x.

Step-by-step explanation:

Using the properties given below :

• Area of rectangle = length x breadth of that rectangle
• Perimeter = 2( length + breadth )

Here,

Area of rectangle is 8x^2 and width( breadth ) is 2x.

= > length of rectangle × 2x = 8x^2

= > length of rectangle = 8x^2 / 2x

= > length of rectangle = 4x

Thus,

= > Perimeter of rectangle = 2( 2x + 4x )

= > Perimeter of rectangle = 2( 6x )

= > Perimeter of rectangle = 12x

Answer 2

Answer:

$\large{\underline{\boxed{\mathfrak\green{Perimeter \: of \: rectangle =12x \: unit }}}}$

Reference of diagram is as below:-

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.5,5.8){ B }\put(11.1,5.8){ C }\put(11.05,9.1){ D }\put(4.5,7.5){ 2x }\put(8.1,5.3){ 4x }\put(11.5,7.5){ 2x }\put(8.1,9.5){ 4x }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\end{picture}$

Given:-

• Width of rectangle = 2x unit.
• Area of rectangle = 8x^2 sq.unit.

To find:-

• Perimeter of rectangle = ?

First we need to find out length of rectangle.

$\rm We \: know, \\\\\dag \: {\boxed{\rm{Area_{rectangle} = Length \times width }}}$

$\twoheadrightarrow\sf 8x^2 = Length \times 2x \\\\\twoheadrightarrow\sf Length = 8x^2 / 2x \\\\\twoheadrightarrow\large{\underline{\boxed{\sf\blue{Length = 4x \: unit }}}}$

$\rule{100}{2}$

Now,

$\rm We \: know, \\\\\dag \: {\boxed{\rm{Perimeter_{rectangle} = 2(length + width) }}}$

$\dashrightarrow\sf Perimeter \: of \:rectangle = 2(4x + 2x) \\\\\dashrightarrow\sf Perimeter \: of \: rectangle = 2 \times 6x \\\\\dashrightarrow\large{\underline{\boxed{\sf\green{Perimeter \: of \: rectangle = 12x \: unit }}}}$

$\therefore\: {\boxed{\sf\red{Perimeter\: of \: rectangle = 12x \: unit }}}$