Question

area of square: 400 inarea of rectangle: 375 inthe squares area is 25 square inches larger than the area of the rectangle.(c) if the square had a side length of $x$ inches, show that its area will always be four square inches more than the area of the new rectanglewidth of rectangle: 2 inlength of rectangle: $x-2$ inarea of square: $x^2$ inarea of rectangle: $x^2 - 4$ inthe squares area is 4 square inches larger than the area of the rectangle

Explanation:

Step1: Define square area

Area of square with side $x$: $A_s = x^2$

Step2: Define rectangle area

Area of rectangle (length $x-2$, width 2): $A_r = 2(x-2)$

Step3: Expand rectangle area

$A_r = 2x - 4$

Step4: Calculate area difference

$A_s - A_r = x^2 - (2x - 4)$

Step5: Simplify the difference

$x^2 - 2x + 4 = (x-1)^2 + 3$? No, correct simplification:
Wait, correction: $A_r = (x-2) \times 2 = 2x - 4$
$A_s - A_r = x^2 - (2x - 4) = x^2 - 2x + 4$? No, wait the given rectangle area is $x^2 -4$, which implies length is $x+2$, width 2. Correcting:

Step1: Square area (side $x$)

$A_s = x^2$

Step2: Rectangle area (length $x+2$, width 2)

$A_r = 2(x+2) = 2x + 4$? No, given $A_r = x^2 -4$, so length is $x-2$, width $x+2$? No, $x^2-4=(x-2)(x+2)$.

Step1: Square area (side $x$)

$A_s = x^2$

Step2: Rectangle area given as $x^2-4$

$A_r = x^2 - 4$

Step3: Compute area difference

$A_s - A_r = x^2 - (x^2 - 4) = 4$

Answer:

The difference between the square's area and the rectangle's area is always 4 square inches, as shown by $x^2 - (x^2 - 4) = 4$.