Question

The width of a rectangle is 2m less than the length. the area is 48m squared. find the dimensions

Answer 1

Let the length = L
Then the width, W = L - 2

A = LW

48 = L(L - 2)

48 = L^2 - 2L

L^2 - 2L - 48 = 0

(L - 8)(L + 6) = 0

L - 8 = 0   or   L + 6 = -6

L = 8   or   L = -6

Since the length of a rectangle cannot be a negative number,
we discard the -6.

The length is 8m
The width is 2 m less than the length, so the width is 6 m.

Answer 2

The length of the rectangle is 8m and the width is 6m

This question involves the area of a rectangle and we can use the formula to solve this problem.

Data given;

• A = 48m^2
• w = (L - 2)m

Area

The area of a rectangle is given as

$A = L * W$

• l = length
• w = width

substitute the values and solve

$A = lw\\48 = L * ( L - 2)\\48 = l^2 -2l\\l^2 -2l - 48 = 0$

solve this quadratic equation to find the length

a = 1, b = -2, c = -48

$l = \frac{-b+-\sqrt{b^2 - 4ac} }{2a} \\l = \frac{-(-2)+-\sqrt{(-2)^2-4*1*(-48)} }{2*1} \\l = 8$

Taking only the positive value of l, l = 8

but we also know that

$w = l -2\\w = 8 -2\\w= 6$

from the above, the length of the rectangle is 8m and the width is 6m

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