The width of a rectangle is 3 inches less than its length. The area of the rectangle is 340 square inches. What are the length and width of the rectangle?

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The width of a rectangle is 3 inches less than its length. The area of the rectangle is 340 square inches. What are the length and width of the rectangle?

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Answer 1 · 2016-04-19T01:00:38

Length and width are 20 and 17 inches, respectively.

Explanation:

First of all, let us consider $x$ $x$ the length of the rectangle, and $y$ $y$ its width. According to the initial statement:

$y = x - 3$ $y = x-3$

Now, we know that the area of the rectangle is given by:

$A = x \cdot y = x \cdot (x - 3) = x^{2} - 3 x$ $A = x cdot y = x cdot (x-3) = x^2-3x$

and it is equal to:

$A = x^{2} - 3 x = 340$ $A = x^2-3x=340$

So we get the quadratic equation:

$x^{2} - 3 x - 340 = 0$ $x^2-3x-340=0$

Let us solve it:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = {-b pm sqrt{b^2-4ac}}/{2a}$

where $a, b, c$ $a, b, c$ come from $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ . By substituting:

$x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 \cdot 1 \cdot (- 340)}}{2 \cdot 1} =$ $x = {-(-3) pm sqrt{(-3)^2-4 cdot 1 cdot (-340)}}/{2 cdot 1} =$
$= \frac{3 \pm \sqrt{1369}}{2} = \frac{3 \pm 37}{2}$ $={3 pm sqrt{1369}}/{2} = {3 pm 37}/2$

We get two solutions:

$x_{1} = \frac{3 + 37}{2} = 20$ $x_1 = {3 + 37}/2 = 20$
$x_{2} = \frac{3 - 37}{2} = - 17$ $x_2 = {3-37}/2 = -17$

As we are talking about inches, we must take the positive one.

So:

$Length = x = 20 inches$ $"Length" = x = 20 " inches"$
$Width = y = x - 3 = 17 inches$ $"Width" = y = x-3 = 17 " inches"$

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The width of a rectangle is 3 inches less than its length. The area of the rectangle is 340 square inches. What are the length and width of the rectangle?

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Explanation:

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