SOLUTION: The perimeter of a rectangle is 70m. The length is 8m more than twice the width. Find the demensions
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Question 280801: The perimeter of a rectangle is 70m. The length is 8m more than twice the width. Find the demensions
Answer by PRMath(133) (Show Source): You can put this solution on YOUR website!
Perimeters are the measure of a shape all the way around the shape. In the case of a rectangle then, the perimeter is made up of TWO lengths and TWO widths. If we say "L" is length and "w" is width, then our equation can be:
L + L + w + w = 70.
According to your information, the Length is 8 meters more than twice the width. Let's break that down to the following:
The Length: L
is: =
8 meters more than: 8 +
Twice the width: 2w
In other words, the equation is: L = 8 + 2w.
Now let's write that down according to what we said about the perimeter. I mean to say that we agreed with the following:
L + L + w + w = 70 Let's plug in what we know for "L"
(8 + 2w) + (8 + 2w) + w + w = 70 See where we plugged in (8 + 2w) for the L variable? Now let's combine like terms.
16 + 6w = 70 Now let's subtract 16 from both sides to begin to isolate the w.
6w = 70 - 16
6w = 54 Now let's divide both sides by 6 to further isolate the w.
w = 9
AH! Now we know that the width is 9.
If the Length is 8 meters more than 2 times the width, we can use that equation again.
L = 8 + 2w
L = 8 + 2(9) See where we plugged in 9 for the Width?
L = 8 + 18 (2 times 9 = 18)
L = 26
SO now our width is 9 and our length is 26. Does this work? Let's see:
L + L + w + w = 70
26 + 26 + 9 + 9 = 70
70 = 70 YES! It works. Yay!
I hope this helps you. :-)
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