Question

The length of a rectangle is 12 m more than twice the width. The area of the rectangle is 320 spuare meter. Write an equation that can be used to find the lenght and width of the rectangle. Also find the dimension of the rectangle

Answer 1

Answer:

given length of rectangle is 12 m more than twice the width

area of the rectangle is 320 m

let the width of rectangle be x

then the length of rectangle =2x+12 +(given)

we know that area of rectangle=lxb

x(2x+12)=320

Answer 2

Given :

• The length of a rectangle is 12 m more than twice the width.
• The area of the rectangle is 320 m².

To Find :

• Dimensions of the Rectangle .

Solution :

$\longmapsto\tt{Let\:width/Breadth\:be=x}$

As Given that The length of a rectangle is 12 m more than twice the width. So ,

$\longmapsto\tt{Length=2x+12}$

Using Formula :

$\longmapsto\tt\boxed{Area\:of\:Rectangle=l\times{b}}$

Putting Values :

$\longmapsto\tt{320=(2x+12)\times{x}}$

$\longmapsto\tt{320={2x}^{2}+12x}$

$\longmapsto\tt\bf{{2x}^{2}+12x-320=0}$

By Splitting Middle Term :

$\longmapsto\tt{{2x}^{2}+32x-20x-320=0}$

$\longmapsto\tt{2x(x+16)-20(x+16)=0}$

$\longmapsto\tt{(2x-20)\:(x+16)=0}$

• x = 10
• x = -16

Dimensions can't be negative . So , The value of x is 10 .

Therefore :

$\longmapsto\tt{Breadth\:of\:Rectangle=x}$

$\longmapsto\tt\bf{10\:m}$

$\longmapsto\tt{Length\:of\:Rectangle=2(10)+12}$

$\longmapsto\tt\bf{32\:m}$