If the width of rectangle is 5 cm more than one-half of its length the perimeter is 70 cm, what are the dimensions of the rectangle?

Question

Eva Abramson · Accepted Answer

The length is 20 cm and the width is 15 cm

First, let's call the width of our rectangle #w# and the length #l#. Based on the issue at hand: #w = 1/2l + 5# We also know the following formula to calculate a rectangle's perimeter: #p = 2*l + 2*w# So we can substitute #70# for #p# which is given in the problem and we can also substitute #1/2l + 5# for #w# and then solve for #l#: #70 = 2*l + 2*(1/2l + 5)# #70 = 2l + 1l + 10# #70 = 3l + 10# #70 - 10 = 3l + 10 - 10# #60 = 3l# #60/3 = (3l)/3# #20 = l# Now we can substitute #20# for #l# in the formula for #w# to find the width: #w = 1/2 20 + 5# #w = 10 + 5# #w = 15#

Hazel Anglin · Answer

undefined Let the length of the rectangle be $ L $ cm and its width be $ W $ cm. We are given that the width is 5 cm more than half of the length, so $ W = \frac{1}{2}L + 5 $. The formula for the perimeter of a rectangle is $ P = 2(L + W) $. Substituting the given values into this formula, we get $ 70 = 2(L + \frac{1}{2}L + 5) $. Solving for $ L $, we find $ L = 20 $. Substituting $ L = 20 $ back into the expression for $ W $, we find $ W = \frac{1}{2}(20) + 5 = 15 $. Therefore, the dimensions of the rectangle are length $ 20 $ cm and width $ 15 $ cm.

Isaiah Anthony · Answer

undefined Let's denote the length of the rectangle as $ L $ and the width as $ W $. According to the problem, we can set up the following equations based on the given information:

1. The width of the rectangle is 5 cm more than one-half of its length: $ W = \frac{1}{2}L + 5 $
2. The perimeter of the rectangle is 70 cm: $ 2L + 2W = 70 $

We can use these equations to solve for the length ($ L $) and width ($ W $) of the rectangle.