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the length of a rectangle is 2 cm less than three times its width. its area is 21 cm. find the dimensions of rectangle. The width is ? , the length is ?

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2 Answers

Let W represent the width of the rectangle and L represent the length of the rectangle. That is, 

     width = W     and     length = L

We are given that the length (L) is 2 cm less than 3 times the width (W). From this, we arrive at the following:

     L = 3W - 2

We are also given that the area (A) of the rectangle is 21 cm2. Since the area of a rectangle is the product of the rectangle's width and the length, then we have the following:

     area = A = 21 cm2        

     A = W•L        ==>        21 = W•(3W - 2)

Distributing the W into each term inside the parentheses we get the following:

     21 = W•3W  -  W•2

     21 = 3W2 - 2W

Subtracting 21 from both sides of the equation yields the following quadratic equation:

     3W2 - 2W - 21 = 0

Solve by factoring the left hand side of the above equation:

     (3W + 7)(W - 3) = 0

By the zero product property we can set each binomial equal to 0 and solve for W in each:

     3W + 7 = 0        and        W - 3 = 0

           3W = -7                         W = 3

             W = -7/3

Since the width cannot be negative, we omit the negative answer and only consider the positive answer. That is, the width (W) of the rectangle is equal to 3 cm.

Since the length (L) of the rectangle is 2 cm less than 3 times the width, then the length of determined by using the expression we found for L:

     L = 3W - 2 = 3(3) - 2 = 9 - 2 = 7

Thus, the length of the rectangle is 7 cm.

Hey Motammem -- sometimes an "eyeball" approach works:

Begin with focus on the simplest part of the problem ... Area is 21 ... probably 3x7

Finally check 3(3) is 2 bigger than 7 ... Very best wishes :)