Question

The rectangle shown has a perimeter of 56 cm and the given area it's length is 7 more than twice it's with right inside of a system of equations to find that dimensions

Answer 1

Answer:

The dimensions are length 7 cm and width 21 cm.

Step-by-step explanation:

Given:

The rectangle shown has a perimeter of 56 cm.

It's length is 7 more than twice its width.

Now, to find the dimensions.

Let the width be $x.$

And the length = $7+2x.$

Perimeter = 56 cm.

Now, putting formula to get the dimensions:

$Perimeter=2(length+width)$

$56=2((7+2x)+x)$

$56=2(7+2x+x)$

$56=2(7+3x)$

$56=14+6x$

Subtracting both sides by 14 we get:

$42=6x$

Dividing both sides by 6 we get:

$7=x\\x=7\ cm.$

So, the width = 7 cm.

And the length:

  7+2$x$

= 7 + 2\times 7

= 7 + 14

= 21 cm.

Therefore, the dimensions are length 7 cm and width 21 cm.