The perimeter of a square and a rectangle is the same. The width of the 
rectangle is 6cm and its area is 16c𝑚2 less than the area of the square. Find the 
area of the square

Let the side length of the square be \( s \), and the length of the rectangle be \( l \) with a width of 6 cm.

### Step 1: Express the perimeter relationship
The perimeter of the square is:
\[
P_{\text{square}} = 4s
\]
The perimeter of the rectangle is:
\[
P_{\text{rectangle}} = 2(l + 6)
\]
Since the perimeters are equal:
\[
4s = 2(l + 6)
\]
Dividing both sides by 2:
\[
2s = l + 6
\]
\[
l = 2s - 6
\]

### Step 2: Express the area relationship
The area of the square is:
\[
A_{\text{square}} = s^2
\]
The area of the rectangle is:
\[
A_{\text{rectangle}} = l \times 6 = 6l
\]
We are given that the area of the rectangle is 16 cm² less than the area of the square, so:
\[
A_{\text{square}} = A_{\text{rectangle}} + 16
\]
\[
s^2 = 6l + 16
\]

### Step 3: Substitute \( l \) from Step 1 into the area equation
Substitute \( l = 2s - 6 \) into the area equation:
\[
s^2 = 6(2s - 6) + 16
\]
Simplifying:
\[
s^2 = 12s - 36 + 16
\]
\[
s^2 = 12s - 20
\]
Rearranging the equation:
\[
s^2 - 12s + 20 = 0
\]

### Step 4: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[
s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -12 \), and \( c = 20 \).

Substituting these values into the formula:
\[
s = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(20)}}{2(1)}
\]
\[
s = \frac{12 \pm \sqrt{144 - 80}}{2}
\]
\[
s = \frac{12 \pm \sqrt{64}}{2}
\]
\[
s = \frac{12 \pm 8}{2}
\]

Thus, the two possible solutions for \( s \) are:
\[
s = \frac{12 + 8}{2} = 10 \quad \text{or} \quad s = \frac{12 - 8}{2} = 2
\]

### Step 5: Determine the valid solution
If \( s = 2 \), the side length would be too small because the area of the rectangle would be negative. Therefore, the valid solution is \( s = 10 \).

### Step 6: Find the area of the square
The area of the square is:
\[
A_{\text{square}} = s^2 = 10^2 = 100 \, \text{cm}^2
\]

Thus, the area of the square is **100 cm²**.

Would you like more details on any part of this process or have any further questions?

Here are 5 related questions to explore:

1. How do you solve quadratic equations using factoring?
2. What is the formula for the perimeter of different shapes?
3. How can you derive the quadratic formula step by step?
4. Can you calculate the area of a rectangle using different dimensions?
5. How do you check if a quadratic equation has real solutions?

**Tip:** Always verify your solutions by plugging them back into the original conditions of the problem to ensure consistency.

The perimeter of a square and a rectangle is the same. The width of the rectangle is 6 cm and its area is 16 cm² less than the area of the square. Find the area of the square.

Learn how to solve a geometric word problem involving a square and a rectangle with equal perimeters. This problem uses algebraic and quadratic formulas to find the area of the square. Perfect for students in Grades 8-10.

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