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 rectangle.width of rectangle: 2 inlength of rectangle: $x+2$ inarea of square: $x^2 - 4$ 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: $A_s = x^2$

Step2: Define rectangle area

Area of rectangle: $A_r = 2(x+2)$

Step3: Compute area difference

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

Step4: Simplify the expression

$$x^2 - 2x - 4 = x^2 - (2x+4)$$
Wait, correct simplification:
$$x^2 - 2(x+2) = x^2 - 2x - 4$$
Wait, correction: The given square area in the image is incorrect. Correct square area is $x^2$. Now find $A_s - A_r$:
$$x^2 - [2(x+2)] = x^2 - 2x - 4$$
Wait, no, to show difference is 4, the correct setup should be:
Wait, the problem says square area is 4 more than rectangle. Let's re-derive:

Step1: Correct square area

Square area: $A_s = x^2$

Step2: Correct rectangle area

Rectangle area: $A_r = (x-2)(x+2)$ (this matches the intended setup for difference 4)

Step3: Expand rectangle area

$$(x-2)(x+2) = x^2 - 4$$

Step4: Calculate area difference

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

Answer:

The area of the square is $x^2$, and the area of the rectangle (with length $x+2$ and width $x-2$, correcting the given width error) is $x^2 - 4$. The difference is $x^2 - (x^2 - 4) = 4$, so the square's area is always 4 square inches larger than the rectangle's area.