Form an expression for the perimeter of the rectangle, remember that opposite sides are equal and label each of the other two sides.

2-10-forming-and-solving-equations-hard-q5

Use this to form an expression for the perimeter of the rectangle.

$\begin{array}{rcl} P_{r e c t} & = & 4 x + (3 x + 4) + 4 x + (3 x + 4) \end{array}$

Form an expression for the perimeter of the trapezium by adding the expressions given for each of the sides.

$\begin{array}{rcl} P_{t r a p} & = & 5 x + (x - 3) + 5 x + (7 x - 3) \end{array}$

Either expression formed [1]

Form an equation by setting the expressions for the perimeter of the rectangle and the perimeter of the trapezium equal to each other.

$\begin{array}{rcl} 4 x + (3 x + 4) + 4 x + (3 x + 4) & = & 5 x + (x - 3) + 5 x + (7 x - 3) \end{array}$

[1]

Simplify both sides by collecting the like terms together.

$\begin{array}{rcl} 4 x + 3 x + 4 + 4 x + 3 x + 4 & = & 5 x + x - 3 + 5 x + 7 x - 3 \\ 4 x + 3 x + 4 x + 3 x + 4 + 4 & = & 5 x + x + 5 x + 7 x - 3 - 3 \\ 14 x + 8 & = & 18 x - 6 \end{array}$

Solve the equation to find $x$ , begin by subtracting the term with the smallest coefficient of $x$ .

$\begin{array}{rcl} 14 x + 8 & = & 18 x - 6 \\ - 14 x & = & - 14 x \\ 8 & = & 4 x - 6 \\ + 6 & = & + 6 \\ 14 & = & 4 x \\ \div 4 & = & \div 4 \\ \frac{14}{4} & = & x \end{array}$

[1]

Substitute $x = 3.5$ into the expression for the length and width of the rectangle.

$w = 4 x = 4 (3.5) = 14 l = 3 x + 4 = 3 (3.5) + 4 = 14.5$

So w = 14 cm and l = 14.5 cm.
Use this to find the area of the rectangle.

$\begin{array}{rcl} {Area}_{(r e c t a n g l e)} & = & 14 \times 14.5 = 203 {cm}^{2} \end{array}$

[1]

$203 {cm}^{2}$ [1]

Forming & Solving Equations (Cambridge O Level Maths)

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