A piece of wire, 44 cm in length, is cut into two parts. One part is used to make a rectangle and the other a square. The length of the rectangle is 200% longer than its width. The width of the rectangle is x centimetres.

1. Express the rectangle length using the given percentage increase from the width.

Width of rectangle $=x$ cm.

Since the length is $200\%$ longer than the width,
$\text{length}=x+200\%\text{ of }x=x+2x=3x.$

1. Find the perimeter of the rectangle in terms of $x$.

From part (a)(i), the length of the rectangle is $3x$ cm and its width is $x$ cm.

Perimeter of rectangle
$=2(3x+x)=8x\text{ cm}.$

2. Use the remaining wire to find and simplify the side length of the square.

Wire used for the square
$=44-8x\text{ cm}.$

Since a square has 4 equal sides, length of the square
$=\frac{44-8x}{4}=11-2x\text{ cm}.$

1. Write the areas of the rectangle and square in terms of $x$.

Rectangle area:
$x(3x)=3x^2$
Square area:
$(11-2x)^2$

2. Form the equation from the statement that the rectangle area is $1\text{ cm}^2$ smaller than the square area.

Since the area of the rectangle is $1\text{ cm}^2$ smaller than the area of the square,
$3x^2=(11-2x)^2-1$

3. Expand and simplify the equation to the required quadratic form.

$3x^2=(11-2x)^2-1$
$3x^2=121-44x+4x^2-1$
$3x^2=120-44x+4x^2$
$0=x^2-44x+120$
$x^2-44x+120=0$

1. Identify the coefficients and substitute into the quadratic formula.

For $x^2-44x+120=0$,
$a=1,\quad b=-44,\quad c=120.$
Using
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$
$x=\frac{-(-44)\pm\sqrt{(-44)^2-4(1)(120)}}{2(1)}.$

2. Simplify the discriminant and expression for the roots.

$x=\frac{44\pm\sqrt{1936-480}}{2}$
$x=\frac{44\pm\sqrt{1456}}{2}.$

3. Evaluate both roots and round each to 5 significant figures.

$x=\frac{44+\sqrt{1456}}{2}=41.092756\ldots$
$x=\frac{44-\sqrt{1456}}{2}=2.907244\ldots$
Correct to 5 significant figures,
$x=41.093\quad \text{or}\quad x=2.9073.$

1. State the necessary condition for the square side length to be physically possible.

From part (a)(ii), the side length of the square is $11-2x$ cm.

Since it is a length,

$11-2x>0.$

2. Use the condition to reject the inadmissible solution.

$11-2x>0$
$11>2x$
$x<5.5$

The solutions from part (b)(ii) are $x=2.9073$ or $41.093$.

Since $41.093>5.5$, it would make the side length of the square negative. Hence $x=41.093$ must be rejected.

1. Use the valid width to find the rectangle perimeter.

From part (b)(iii), $x=2.9073$.

Length of rectangle $=3x$ and width $=x$.

$\text{Perimeter}=2(3x+x)=8x$

$=8(2.9073)=23.2584\approx 23.258$