the length of a rectangle is 3yd longer than its width. if the perimeter of the rectangle is 46 yd, find its area

Answer 1

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Problems like this are solved with systems of equations. All you need to get started are the basic formulas and a little understanding of "translating" words to math.

Let L = the length of the rectangle and let W = the width.

If the length is 3 yds longer than the width, then we could say that L = W+3. OK so far? We'll call that equation (1)

We are also given that the perimeter is 46 yds.

The formula for perimeter is P = 2L + 2W, so in this case 2L + 2W = 46. That's equation (2)

Equation (1) gives us an expression for L, so let's substitute that expression into (2)

2L + 2W = 46

2(W+3) + 2W = 46 (Substituting W+3 for L)

2W + 6 + 2W = 46 (Distributive prop)

4W+6 = 46 (combining like terms)

4W=40 (Subtract 6 from both sides)

W=10 (Dividing by 4)

Now, we substitute this value of W into equation (1)

L=W+3

L=10+3

L=13

Now we know both length and width. Since Area = L x W,

the area of the rectangle is 10 x 13 = 130 yds²

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