Question

What are the dimensions of a rectangle with an area of 48 square inches and a length 6 inches more than its width?

Answer 1

To find the dimensions of the rectangle, we can set up equations based on the information given.

Let the width be \( w \) inches. Then, the length, which is 6 inches more than the width, would be \( w + 6 \) inches.

The area of a rectangle is given by the formula:
\[
\text{Area} = \text{length} \times \text{width}
\]

In this case, we have:
\[
48 = (w + 6) \times w
\]

Expanding this equation gives:
\[
48 = w^2 + 6w
\]

Rearranging this into standard quadratic form leads to:
\[
w^2 + 6w - 48 = 0
\]

Now we can use the quadratic formula to solve for \( w \):
\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 6 \), and \( c = -48 \).

Calculating the discriminant:
\[
b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-48) = 36 + 192 = 228
\]

Now substituting back into the quadratic formula:
\[
w = \frac{-6 \pm \sqrt{228}}{2 \cdot 1}
\]
\[
w = \frac{-6 \pm \sqrt{228}}{2}
\]

We simplify \( \sqrt{228} \):
\[
\sqrt{228} = \sqrt{4 \cdot 57} = 2\sqrt{57}
\]

Therefore:
\[
w = \frac{-6 \pm 2\sqrt{57}}{2} = -3 \pm \sqrt{57}
\]

Since width cannot be negative, we take the positive solution:
\[
w = -3 + \sqrt{57}
\]

Now, we can find the length:
\[
\text{Length} = w + 6 = (-3 + \sqrt{57}) + 6 = 3 + \sqrt{57}
\]

To find numerical approximations:
\[
\sqrt{57} \approx 7.55 \implies w \approx -3 + 7.55 \approx 4.55
\]
\[
\text{Length} \approx 3 + 7.55 \approx 10.55
\]

Thus, the approximate dimensions of the rectangle are:
- Width: approximately 4.55 inches
- Length: approximately 10.55 inches

In exact form, the dimensions of the rectangle are:
- Width: \( -3 + \sqrt{57} \)
- Length: \( 3 + \sqrt{57} \)