The perimeter of a rectangle is 80 cm. The length of the rectangle is 4cm more than twice its width. Find the dimensions of the rectangle.

Question

I’m confused how to do this.

Answer 1

w

2w + 4

The width of the rectangle is unknown, so we will let the width of the rectangle = "w"

The length of the rectangle is 4cm more than twice the width (w), so the length can be written as 2 times the width plus 4 OR 2w + 4

We know that the perimeter of this rectangle is 80 cm, so when we add the lengths of all 4 sides, we will get a total of 80.

w + w + (2w + 4) + (2w + 4) = 80

2w + 2(2w + 4) = 80

2w + 4w + 8 = 80

6w + 8 = 80

-8 -8

6w = 72

÷6 ÷6

w = 12 cm

So, the width is 12 cm and the length is:

2(12) + 4

24 + 4 = 28 cm

12 cm

28 cm

Answer 2

Don't overthink. It's really that easy, I promise.

P = 2l + 2w

P = 2(l + w)

P/2 = l + w

l = 2w + 4, right?

80/2 = 2w + 4 + w

40 = 3w + 4

3w = 36

w = 12 cm

l = 2(12) + 4

l = 24 + 4

l = 28 cm

Answer 3

1. Draw the rectangle and label the sides with L and W. (L = length, W = width)
2. Set up the perimeter equation.
3. L + L + W + W = 80
4. 2L + 2W = 80
5. Translate information to an Algebra equation
6. The length of the rectangle is 4 more than twice the width
7. L = 4 + 2W
8. Substitute into perimeter equation
9. 2L + 2W = 80
10. 2(4 + 2W) + 2W = 80 (substitute for L)
11. 8+ 4W + 2W = 80 (distribute)
12. 8 + 6W = 80 (combine like terms)
13. 6W = 72 (isolate the variable)
14. W = 12 cm
15. Solve for L
16. L = 4 + 2W
17. L = 4 + 2(12)
18. L = 28 cm
19. Double check information
20. Is 28 four more than twice 12?
21. Is 2(28) + 2(12) = 80?