The problem describes two rectangles, A and B. The length of rectangle A is twice the length of rectangle B, and the width of rectangle B is 2 cm more than the width of rectangle A. (a) The difference between the areas of rectangle A and rectangle B is 12 $cm^2$. Find the value of $x$. (b) Rectangles A and B are rearranged as shown in Diagram II. Calculate the perimeter of the new arrangement.

1. Problem Description

The problem describes two rectangles, A and B. The length of rectangle A is twice the length of rectangle B, and the width of rectangle B is 2 cm more than the width of rectangle A.

(a) The difference between the areas of rectangle A and rectangle B is 12 $cm^2$. Find the value of $x$.

(b) Rectangles A and B are rearranged as shown in Diagram II. Calculate the perimeter of the new arrangement.

2. Solution Steps

(a) Finding the value of $x$.

Let the length of rectangle A be $l_A$ and the width be $w_A$.

Let the length of rectangle B be $l_B$ and the width be $w_B$.

From the problem statement, we have:

$l_A = 2x$

$w_A = x$

$l_B = x$

$w_B = x+2$

Area of rectangle A, $A_A = l_A * w_A = 2x * x = 2x^2$

Area of rectangle B, $A_B = l_B * w_B = x * (x+2) = x^2 + 2x$

The difference between the areas is given as 12 $cm^2$. Therefore,

$|A_A - A_B| = 12$

$|2x^2 - (x^2 + 2x)| = 12$

$|2x^2 - x^2 - 2x| = 12$

$|x^2 - 2x| = 12$

We have two possibilities:

Case 1: $x^2 - 2x = 12$

$x^2 - 2x - 12 = 0$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1, b=-2, c=-12$.

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-12)}}{2(1)}$

$x = \frac{2 \pm \sqrt{4 + 48}}{2}$

$x = \frac{2 \pm \sqrt{52}}{2}$

$x = \frac{2 \pm 2\sqrt{13}}{2}$

$x = 1 \pm \sqrt{13}$

Since $x$ must be positive, $x = 1 + \sqrt{13} \approx 4.606$.

Case 2: $x^2 - 2x = -12$

$x^2 - 2x + 12 = 0$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1, b=-2, c=12$.

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(12)}}{2(1)}$

$x = \frac{2 \pm \sqrt{4 - 48}}{2}$

$x = \frac{2 \pm \sqrt{-44}}{2}$

Since the discriminant is negative, there are no real solutions.

Thus, $x = 1 + \sqrt{13}$.

(b) Finding the perimeter of the new arrangement.

In the new arrangement (Rajah II), the perimeter is:

$P = 2(l_A + w_B)$

$P = 2(2x + x+2)$

$P = 2(3x + 2)$

$P = 6x + 4$

Substituting $x = 1 + \sqrt{13}$:

$P = 6(1+\sqrt{13}) + 4$

$P = 6 + 6\sqrt{13} + 4$

$P = 10 + 6\sqrt{13}$

3. Final Answer

(a) $x = 1 + \sqrt{13}$

(b) $10 + 6\sqrt{13}$ cm