Forming Equations from Shapes (Edexcel IGCSE Maths B): Revision Note

Read the question carefully - does it want an angle? perimeter? total area? curved surface area? etc.

For surface area and volume questions, check the list of formulas given in the exam.

A rectangle has a length of $3x+1$ cm and a width of $2x-5$ cm.

Its perimeter is equal to 22 cm.

(a) Use the above information to find the value of x.

Answer:

The perimeter of a rectangle is 2 × length + 2 × width

$2(3x+1)+2(2x–5)$

Expand the brackets

$6x+2+4x-10$

Simplify by collecting like terms

$10x-8$

This perimeter is 22, so set this expression equal to 22

$10x–8=22$

Solve this equation by adding 8 then dividing by 10

$\begin{array}{rcl}10x& =& 22+8\\ 10x& =& 30\\ x& =& \frac{30}{10}\\ x& =& 3\end{array}$

$x=3$

(b) Find the area of the rectangle.

Answer:

The area of a rectangle is its length multiplied by is width
Substitute the value of x  from part (a) into the length and width given in the question

length is 3 × 3 + 1 = 10

width is 2 × 3 - 5 = 1

Find the area (multiply length by width)

10 × 1

Include the correct units for area

Area = 10 cm²

The figure above shows right prism $ABCDEFGH$ whose cross section has the form of a trapezium. Trapeziums ABCD and EFGH are the two end faces of the prism.

$AB=GH=4x\mathrm{cm}$
$AD=EH=2x\mathrm{cm}$
$CD=EF=3x\mathrm{cm}$
$AH=BG=CF=DE=\left(x+3\right)\mathrm{cm}$

The volume of the prism is 784 cm³.

Show that $x^3+3x^2-112=0$.

Answer:

The volume of a prism is given by $V=\mathrm{area}\mathrm{of}\mathrm{cross}\mathrm{section}\times \mathrm{length}$

• Here the length is $\left(x+3\right)$

To find the area of the cross section, use the area of a trapezium formula, $A=\frac{1}{2}\left(a+b\right)h$

• Here $a=4x$, $b=3x$ and $h=2x$

$\begin{array}{rcl}A& =& \frac{1}{2}\left(4x+3x\right)\left(2x\right)\\ & =& x\left(7x\right)\\ & =& 7x^2\end{array}$

Put that into the prism volume formula

$\begin{array}{rcl}V& =& 7x^2\times \left(x+3\right)\\ & =& 7x^3+21x^2\end{array}$

You know the volume is equal to 784, so

$\begin{array}{rcl}7x^3+21x^2& =& 784\end{array}$

Divide both sides of the equation by 7

$\begin{array}{rcl}\frac{7x^3}{7}+\frac{21x^2}{7}& =& \frac{784}{7}\\ x^3+3x^2& =& 112\end{array}$

Subtract 112 from both sides to get the form required

$x^3+3x^2-112=0$