A rectangle is 12 longer than it is wide. If the perimeter is 136 what are the dimensions

Question

asked Feb 13, 2024 123k views

2 Answers

Karl Barker · Answer 1 · 2024-02-17T19:19:50+0000

Answer:

28, 40

Explanation:

Let x be the width of the rectangle.

Then the length is x + 12.

The perimeter is 2 times the sum of the width and length.

So, we have:

$\sf 2(x + x + 12) = 136$

Solve like terms:

$\sf 2(2x + 12) = 136$

Divide both side by 2.

$\sf (2x + 12) = (136)/(2)$

$\sf (2x + 12) =68$

Subtract 12 on both sides

$\sf 2x+12-12 =68-12$

$\sf 2x=56$

Dividing both side by 2.

$\sf x=(56)/(2)$

$\sf x=28$

Therefore, the width is 28 and the length is 28 + 12 = 40.

So the dimensions of a rectangle is 28, 40