The length of a rectangle is 3 more than its width. The perimeter of the rectangle is 58 cm. What are the rectangle's dimensions?

Question

Theodore Abad · Accepted Answer

13 cm and 16 cm

Let the width be #x# and length be #(x+3)#. now, we have, #Perimeter=2(l+w)# #58 cm=2(x+x+3)# #4x+6=58 cm# #4x=52 cm# #x=52/4 cm# #:. x=13 cm# So, width = #x# =13 cm and, length = #x+3# = 13 cm +3 =16 cm

Sadie Allen · Answer

undefined Let $ w $ represent the width of the rectangle. Since the length is 3 more than the width, the length can be represented as $ w + 3 $.

The perimeter of a rectangle is given by the formula: $ P = 2(w + l) $, where $ P $ is the perimeter, $ w $ is the width, and $ l $ is the length.

Given that the perimeter is 58 cm, we have:
$$ 58 = 2(w + w + 3) $$

Solve for $ w $ by rearranging and solving the equation:
$$ 58 = 2(2w + 3) $$
$$ 58 = 4w + 6 $$
$$ 4w = 58 - 6 $$
$$ 4w = 52 $$
$$ w = \frac{52}{4} $$
$$ w = 13 $$

So, the width of the rectangle is 13 cm.

Substitute $ w = 13 $ into the expression for the length to find the length:
$$ l = w + 3 $$
$$ l = 13 + 3 $$
$$ l = 16 $$

Therefore, the dimensions of the rectangle are: width = 13 cm and length = 16 cm.