Question 78740This question is from textbook
: The length of a rectangular sign is 3 feet longer than the width. If the sign's area is 54 square feet, find the length and the width. This question is from textbook
You can put this solution on YOUR website! The problem tells you that the Length (L) equals the Width (W) plus 3 feet. In equation
form this becomes:
.
.
The problem also tells you that the Area (A) is 54 square feet. From geometry you know that
the formula for the Area of a rectangle is that Area equals the product of the Length
times the Width. In equation form this is:
.
.
Substitute the given area to get:
.
.
This equation has two unknowns. You can't solve it unless you can eliminate one of the
unknowns. To do that you can note that L = W + 3. If you substitute W + 3 for L in the
equation you get:
.
.
And multiplying out the right side results in:
.
.
Get this into the standard quadratic form by first subtracting 54 from both sides and
then switching sides around to get:
.
.
This can be factored into:
.
.
Notice that this equation will be true if either of the 2 factors is a zero because
zero times anything is zero and therefore equal to the right side of this equation.
.
Set W+9 equal to zero and solve to find that W = -9. But what sense does a negative
width make in this case? None at all, so disregard this solution.
.
Next set W-6 equal to zero and solve to find that W = +6. That's better. The width is
6 feet, and since we know that the length is 3 feet longer than the width, the length
has to be 9 feet. Check by noting that the product of the length times the width is
9 ft times 6 ft and that equals the given area of 54 sq ft.
.
Hope this helps you to understand the problem a little better.
(