Question

the length of a rectangle is 2cm more than twice it's widht the perimeter of a the rectangle is 28 cm find the widthof the rectangle.
please solve ​

Answer 1

Answer:

$\boxed{\sf Width \ of \ the \ rectangle = 4 \ cm}$

Given:

Length of the rectangle = 2 cm more than twice it's width

Perimeter of the rectangle = 28 cm

To find:

Width of the rectangle

Step-by-step explanation:

Let the width of the rectangle be 'w'

So,

Length of the rectangle = (2 + 2w) cm

$\sf \implies Perimeter \ of \ rectangle = 2(Length + Width) \\ \\ \sf \implies 28= 2((2 + 2w) + w)\\ \\ \sf \implies 28 = 2(2 + 2w + w)\\ \\ \sf \implies 28 = 2(2 + 3w)\\ \\ \sf \implies 28 = (2 \times 2) + (2 \times 3w)\\ \\ \sf \implies 28 = 4 + 6w \\ \\ \sf \implies 6w + 4 = 28\\ \\ \sf \implies 6w = 28 - 4\\ \\ \sf \implies 6w = 24\\ \\ \sf \implies w = \frac{24}{6} \\ \\ \sf \implies w = \frac{4 \times \cancel{6}}{ \cancel{6}} \\ \\ \sf \implies w = 4 \: cm$

Width of the rectangle = 4 cm

Answer 2

⚽Solution : -

$\implies\tt\bold{Length\:of\:rectangle=2cm}$

$\implies\tt\bold{Let\:Width\:of\:rectangle=x}$

$\implies\tt\bold{So,Length\:of\:rectangle\:will\:be=(2+2x)cm}$

⚽Formula Used :

$\implies\tt\bold{\large{\bold{\bold{\bold{\purple{\sf{Perimeter\:of\:rectangle=2(l+w)}}}}}}}$

⚽Now , Calculating :

$\implies\tt\bold{28=2(2+2x+x)}$

$\implies\tt\bold{28=2(2+2x+x)}$

$\implies\tt\bold{28=2(2+3x)}$

$\implies\tt\bold{28=(2\times{2})+(2\times{3x)}}$

$\implies\tt\bold{28=4+6x}$

$\implies\tt\bold{6x=28-4}$

$\implies\tt\bold{6x=24}$

$\implies\tt\bold{x=\cancel\dfrac{24}{6}}$

$\implies\tt\bold{\large{\bold{\bold{\bold{\green{\sf{x=4}}}}}}}$

⚽Hence , Width of Rectangle is 4 ..