the 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 ?
thanks
the 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 ?
thanks
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 cm^{2}. 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 cm^{2}
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 = 3W^{2} - 2W
Subtracting 21 from both sides of the equation yields the following quadratic equation:
3W^{2} - 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 :)