The area of a rectangle is 48 square feet. The length is 8 feet longer than the width. 1) Which picture corresponds to the problem? 2) What are the dimensions of the rectangle?
The Correct Answer and Explanation is:
Which picture corresponds to the problem?
For this, we’d expect a rectangle where one side (the length) is 8 feet longer than the other side (the width). While I can’t generate the picture, imagine a rectangular shape, and label the length LLL and the width WWW. The relationship given is:L=W+8L = W + 8L=W+8
So the length is 8 feet longer than the width, and the area of the rectangle is A=48A = 48A=48 square feet.
2) What are the dimensions of the rectangle?
We know that:Area=Length×Width\text{Area} = \text{Length} \times \text{Width}Area=Length×Width
Given that the area is 48 square feet, we can write the equation:L×W=48L \times W = 48L×W=48
Substitute L=W+8L = W + 8L=W+8 into the equation:(W+8)×W=48(W + 8) \times W = 48(W+8)×W=48
Now expand the equation:W2+8W=48W^2 + 8W = 48W2+8W=48
Rearrange the equation to form a quadratic equation:W2+8W−48=0W^2 + 8W – 48 = 0W2+8W−48=0
Now, solve for WWW using the quadratic formula:W=−b±b2−4ac2aW = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}W=2a−b±b2−4ac
Where a=1a = 1a=1, b=8b = 8b=8, and c=−48c = -48c=−48. Plug these values into the quadratic formula:W=−8±82−4(1)(−48)2(1)W = \frac{-8 \pm \sqrt{8^2 – 4(1)(-48)}}{2(1)}W=2(1)−8±82−4(1)(−48)W=−8±64+1922W = \frac{-8 \pm \sqrt{64 + 192}}{2}W=2−8±64+192W=−8±2562W = \frac{-8 \pm \sqrt{256}}{2}W=2−8±256W=−8±162W = \frac{-8 \pm 16}{2}W=2−8±16
This gives us two possible solutions:W=−8+162=4orW=−8−162=−12W = \frac{-8 + 16}{2} = 4 \quad \text{or} \quad W = \frac{-8 – 16}{2} = -12W=2−8+16=4orW=2−8−16=−12
Since width cannot be negative, we choose W=4W = 4W=4 feet.
Now, substitute W=4W = 4W=4 into the equation L=W+8L = W + 8L=W+8:L=4+8=12 feetL = 4 + 8 = 12 \, \text{feet}L=4+8=12feet
Final Answer:
- The width of the rectangle is 4 feet.
- The length of the rectangle is 12 feet.
Thus, the dimensions of the rectangle are 12 feet by 4 feet.
