A rectangle has a width one less the length. If the perimeter is 30 units, find the width and the length.

Answer 1

Nichole,

You asked -

A rectangle has a width one less the length. If the perimeter is 30 units, find the width and the length.

Let us first assign variable definitions. p = perimeter of the rectangle, l = length of the rectangle and w = width of the rectangle. Now we know from geometry that the perimeter of a rectangle is 2 times the width plus 2 times the length. So translating that into a formula with our variables is

p = 2w + 2l

from the problem statement, we know that p = 30 so let's substitute that into our formula

30 = 2w + 2l

from the problem statement, the width and length are related: width one less the length

let's write that as an equation

w = l -1

Now let's substitute this equation into the formula for the perimeter from above

30 = 2*(l-1) + 2l; now we have one equation and one variable so we can solve with our algebra tools

30 = 2l-2 + 2l; used the distributive property on the first term of the right side of the equation

30 = 4l-2; collected like terms of the variable "l"

30+2 = 4l-2+2; isolate the variable term by adding 2 to both sides

32 = 4l; simplify

32/4 = 4l/4; isolate l by dividing by 4 into both sides

8 = l; the length is 8 units, now to find the width use the equation from above w = l-1 = 8-1 = 7 units

Let's check our answer in the formula

30 = 2w + 2l =2*8 + 2*7 = 16 + 14 = 30; both side are equal so we found the correct answer.

Let me know if you have any questions!

Yours truly,

Ray