Question

A square is altered so that one dimension is increased by 5 meters and the other dimension is increased by 3 meters If the area of the resulting rectangle is 98 square meters, find the area of the original

Answer 1

Let the side length of the original square be \( s \) meters. The area of the original square is \( s^2 \).

When one dimension is increased by 5 meters and the other by 3 meters, the new dimensions become \( s + 5 \) and \( s + 3 \).

The area of the resulting rectangle is given as 98 square meters. Therefore, we can set up the equation:

\[
(s + 5)(s + 3) = 98
\]

Expanding the left side:

\[
s^2 + 3s + 5s + 15 = 98
\]

This simplifies to:

\[
s^2 + 8s + 15 = 98
\]

Next, we subtract 98 from both sides:

\[
s^2 + 8s + 15 - 98 = 0
\]

This simplifies to:

\[
s^2 + 8s - 83 = 0
\]

Now, we can use the quadratic formula to solve for \( s \):

\[
s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In this case, \( a = 1 \), \( b = 8 \), and \( c = -83 \):

\[
s = \frac{-8 \pm \sqrt{8^2 - 4 \times 1 \times -83}}{2 \times 1}
\]
\[
s = \frac{-8 \pm \sqrt{64 + 332}}{2}
\]
\[
s = \frac{-8 \pm \sqrt{396}}{2}
\]
\[
s = \frac{-8 \pm 2\sqrt{99}}{2}
\]
\[
s = -4 \pm \sqrt{99}
\]

Since we are looking for a positive dimension, we take the positive root:

\[
s = -4 + \sqrt{99}
\]

To find the area of the original square:

\[
\text{Area} = s^2 = (-4 + \sqrt{99})^2
\]
\[
= (-4)^2 + 2 \cdot (-4) \cdot \sqrt{99} + (\sqrt{99})^2
\]
\[
= 16 - 8\sqrt{99} + 99
\]
\[
= 115 - 8\sqrt{99}
\]

Thus, the area of the original square is:

\[
\boxed{115 - 8\sqrt{99}} \text{ square meters.}
\]

To approximate, \( \sqrt{99} \approx 9.95 \):

\[
115 - 8 \cdot 9.95 \approx 115 - 79.6 \approx 35.4 \text{ square meters.}
\]

So the area of the original square is approximately \( 35.4 \text{ square meters.} \)