The length of a rectangle is 2 feet more than it's width. How do you find the dimensions of the rectangle if its area is 63 square feet?

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Answer 1 · 2016-12-25T02:14:34

$7$ $7$ by $9$ $9$ feet.

We let the length be $x + 2$ $x + 2$ and the width be $x$ $x$ .

The area of a rectangle is given by $A = l \cdot w$ $A = l * w$ .

$A = l \cdot w$ $A = l * w$

$63 = x (x + 2)$ $63 = x(x + 2)$

$63 = x^{2} + 2 x$ $63 = x^2 + 2x$

$0 = x^{2} + 2 x - 63$ $0 = x^2 + 2x - 63$

$0 = (x + 9) (x - 7)$ $0 = (x + 9)(x - 7)$

$x = - 9 and 7$ $x = -9 and 7$

A negative answer is impossible here, so the width is $7$ $7$ feet and the length is $9$ $9$ feet.

Hopefully this helps!