IGCSEMathsEdexcelMaths BRevision NotesAlgebraForming & Solving EquationsForming Equations from Shapes

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

Exam code: 4MB1

Mark Curtis

Written by: Mark Curtis

Reviewed by: Jamie Wood

Updated on

Edexcel Maths BEdexcel Maths A: FoundationEdexcel Maths A (Modular): Higher Unit 2Edexcel Maths A: HigherCambridge (CIE) International Maths: ExtendedCambridge (CIE) International Maths: CoreCambridge (CIE) Maths: ExtendedCambridge (CIE) Maths: Core

Forming equations from shapes

How do I form equations from shapes?

  • You need to use all the information given on the diagram and any specific properties of that shape

  • Common 2D shapes that you should know properties for are

    • Triangles: equilateral, isosceles, scalene, right-angled

    • Quadrilaterals: square, rectangle, kite, rhombus, parallelogram, trapezium

  • You may be asked about perimeter, area or angles

  • You may be asked about polygons

    • Regular vs irregular polygons

    • Interior vs exterior angles

      • The sum of interior angles is 180(n-2) for an n-sided polygon

  • You may be asked about angles in parallel lines

    • Alternative, corresponding and co-interior

  • You may be asked about 3D shapes involving surface area and volume

    • Prisms have constant cross sections 

      • Volume is cross-section area multiplied by length

Is there anything else that can help?

  • Sketch a diagram if none is given

  • Split up uncommon shapes into the sum or difference of common shapes

  • Look out for important extra information

    • For example, a trapezium "with a line of symmetry"

  • With irregular shapes, assume all angles and lengths are different (unless told otherwise)

  • Put brackets around algebraic expressions when substituting them into geometric properties

Forming and solving an equation from an irregular polygon

Examiner Tips and Tricks

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.

Worked Example

A rectangle has a length of 3 x plus 1 cm and a width of 2 x minus 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 left parenthesis 3 x plus 1 right parenthesis plus 2 left parenthesis 2 x – 5 right parenthesis

Expand the brackets

space 6 x plus 2 plus 4 x minus 10

Simplify by collecting like terms

10 x minus 8

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

10 x – 8 equals 22

Solve this equation by adding 8 then dividing by 10

table row cell 10 x end cell equals cell 22 plus 8 end cell row cell 10 x end cell equals 30 row x equals cell 30 over 10 end cell row x equals 3 end table

x equals 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

Worked Example

Diagram of a prism ABCDEFGH with trapezium-shaped cross-section. ABCD and EFGJH are the two trapezium faces. AB= 4x cm, AD= 2x cm, DE = (x + 3) cm, and EF = 3x cm.

The figure above shows right prism A B C D E F G H whose cross section has the form of a trapezium. Trapeziums ABCD and EFGH are the two end faces of the prism.

A B equals G H equals 4 x space cm
A D equals E H equals 2 x space cm
C D equals E F equals 3 x space cm
A H equals B G equals C F equals D E equals open parentheses x plus 3 close parentheses space cm

The volume of the prism is 784 cm3.

Show that x cubed plus 3 x squared minus 112 equals 0.

Answer:

The volume of a prism is given by V equals area space of space cross space section cross times length

  • Here the length is open parentheses x plus 3 close parentheses

To find the area of the cross section, use the area of a trapezium formula, A equals 1 half open parentheses a plus b close parentheses h

  • Here a equals 4 x, b equals 3 x and h equals 2 x

table row A equals cell 1 half open parentheses 4 x plus 3 x close parentheses open parentheses 2 x close parentheses end cell row blank equals cell x open parentheses 7 x close parentheses end cell row blank equals cell 7 x squared end cell end table

Put that into the prism volume formula

table row V equals cell 7 x squared cross times open parentheses x plus 3 close parentheses end cell row blank equals cell 7 x cubed plus 21 x squared end cell end table

You know the volume is equal to 784, so

table row cell 7 x cubed plus 21 x squared end cell equals 784 end table

Divide both sides of the equation by 7

table row cell fraction numerator 7 x cubed over denominator 7 end fraction plus fraction numerator 21 x squared over denominator 7 end fraction end cell equals cell 784 over 7 end cell row cell x cubed plus 3 x squared end cell equals 112 end table

Subtract 112 from both sides to get the form required

x cubed plus 3 x squared minus 112 equals 0

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    Author
    Reviewer
    Mark Curtis

    Author: Mark Curtis

    Expertise: Maths Content Creator

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

    Jamie Wood

    Reviewer: Jamie Wood

    Expertise: Maths Content Creator

    Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.