If I know that the perimeter of the rectangle is 34 and the area is 60 how do I find the length and width?

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Use substitution A rectangle has a "width" (W) and a "length" (L). The perimeter (P) is the sum of all the sides, or twice the width plus twice the height, 2 * W + 2 * L = 34 and the area is the length times the width, L * W = 60 You can divide both sides of the area equation by L to get W = 60 / L Then you can substitute "60 / L" for "W" in the perimeter equation to get 2 * 60 / L + 2 * L = 34 or 120 / L + 2 * L = 34 then you can multiply both sides by L to get 120 + 2 * L^2 = 34 * L or L^2 - 17 * L + 120 = 0 which you can solve for L using the quadratic formula, with a = 1, be = -17, and c = 120. If you get two possible values for L from the quadratic formula, you should use the positive one.

Then you know that W = 60 / L so you can divide 60 by the value you got for L to find W, the width.

34 would have to break down to four numbers, two of which multiply to be 60. The good news is that there are very few numbers that multiply to become sixty. 6 x 10; 12 x 5; 30 x 2; 15 x 4.

Then you need only go thru those sets to see which ones total 34, and 12 + 12 + 5 + 5 = 34.

1 Area = (base*height)/2that should give you a starting point.

Area = (base*height)/2that should give you a starting point.

2 Set up equations for the area and perimeter. You know what the values are; plug them into the formulas. You should have two equations, and two variables (length and width).

If you remember how to solve simultaneous equations, it should be easy from there. If you don't remember, at least come up with the equations and we'll help you from there.

Set up equations for the area and perimeter. You know what the values are; plug them into the formulas. You should have two equations, and two variables (length and width).

If you remember how to solve simultaneous equations, it should be easy from there. If you don't remember, at least come up with the equations and we'll help you from there.

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