The floor of a shed has an area of 90 square feet. The floor is in the shape of a rectangle whose length is 3 ft less than twice the width. Find the length and the width of the shed.

Question

The floor of a shed has an area of 90 square feet. The floor is in the shape of a rectangle whose length is 3 ft less than twice the width. Find the length and the width of the shed.

Answer 1

The first thing to do is draw a rectangular to have a visual. We know that Area = Length times Width. Let W represent the Width and since the Length is 3 less than 2 times the width, then L = 2w - 3. Always back into this by subtracting 3 from 2 times W. Therefore now we have 90 ft² = (W) (2W - 3). Next, we need to distribute the W which makes the first side 2W² - 3W = 90. To make this a quadratic equation, (ax² + bx + c), we move the 90 to the other side and set it all equal to 0. The equation becomes 2W² - 3W - 90 = 0. Then to solve, we must find two binomials that can be factored from this equation. The best way to find these two is to factor by grouping where we multiply a x c which is 2 x -90 = -180 and since b is -3, we find two numbers that can add to get -3. Make a T chart and use trial and error. Those two numbers are -15 & 12. We keep the first term (2W²), and the last term (-90) and split the middle term into two (-15W & 12W). As a result, our four terms are now 2W² - 15W + 12W - 90. Factor out the common terms for the first two and the second two. W(2W - 15) + 6(2W - 15). Keep the common binomial (2W - 15) and connect the other two outside factors (W + 6). We set both of these equal to 0. W + 6 = 0 which means W = -6 and since we know that the width cannot be negative, we eliminate that factor. Then we solve 2W - 15 = 0. We determine that W = 7.5 ft. Now, we go back to the original equations and plug in 7.5 ft for width and find the Length. If the L = 2W - 3, then 2(7.5) - 3 = 12 ft. Then just to make sure our numbers are correct, multiply the Length 7.5 ft times the Width, 12 ft, and we get 90 ft^2..