The area of a square wooden deck is 39 m² less than the area of a rectangular terrace. If the side length of the square deck is 3 m greater than the width of the rectangular terrace, and the length of the rectangular terrace exceeds the side length of the square deck by 7 m, what are the perimeters of the rectangular terrace and the square deck, respectively?

The correct answer is 68 m, 60 m.

Step 1: Define the variables for the dimensions.
Let $s$ be the side length of the square wooden deck in meters.
Let $w$ be the width of the rectangular terrace in meters.
Let $l$ be the length of the rectangular terrace in meters.

Step 2: Express the length and width of the terrace in terms of $s$.
From the problem statement, the side length of the square deck is 3 m greater than the width of the rectangular terrace:

$s=w+3\Rightarrow w=s-3$

Also, the length of the rectangular terrace exceeds the side length of the square deck by 7 m:

$l=s+7$

Step 3: Formulate an equation using the given areas.
The area of the square deck is $s^2$ and the area of the rectangular terrace is $l\times w$.
We are given that the area of the square deck is 39 m² less than the area of the terrace:

$s^2=(l\times w)-39$

Substitute the expressions for $l$ and $w$ into the area equation:

$s^2=(s+7)(s-3)-39$

Step 4: Solve for $s$.
Expand the right side of the equation:

$s^2=s^2-3s+7s-21-39$

$s^2=s^2+4s-60$

Subtract $s^2$ from both sides:

$0=4s-60$

$4s=60$

$s=15\text{m}$

Step 5: Determine the dimensions of both shapes.
Side length of the square deck:

$s=15\text{m}$

Width of the rectangular terrace:

$w=15-3=12\text{m}$

Length of the rectangular terrace:

$l=15+7=22\text{m}$

Step 6: Calculate the perimeters.
Perimeter of the rectangular terrace:

$P_{\text{terrace}}=2(l+w)=2(22+12)=2\left(34\right)=68\text{m}$

Perimeter of the square deck:

$P_{\text{deck}}=4s=4\times 15=60\text{m}$

Thus, the perimeters of the rectangular terrace and the square deck are 68 m and 60 m, respectively.