Math // Measurement

Area of Triangles & Parallelograms

Calculate the area of parallelograms and triangles using formulae (A = b × h for parallelograms, A = ½ × b × h for triangles)

Ages
10 to 11
Type
Method
Builds on
4 topics
Leads to
1 topic
Standards
1 code

Quick check

If the learner is cutting triangular pieces of cloth where the base is 8 cm and the height is 5 cm, can they use a formula to work out the area of each piece?

Ask this to see whether the topic is secure.

Topic // Mastery

What mastery looks like

What a learner can do once this topic is secure.
  • Calculate the area of a parallelogram with base 8 cm and height 5 cm
  • Calculate the area of a triangle with base 12 cm and height 7 cm
  • Explain why the area of a triangle is half the area of a related parallelogram

Topic // Map

Where this fits

What to learn first and what this topic opens up, with the reason for each link. Required links come first.
  • RequiredMust come first
  • HelpfulHelps, but not essential

Before this

4 topics

You are here

Area of Triangles & Parallelograms

Ages 10 to 11 // Method

Leads to

1 topic
  • Area and volume formulaeRequired

    Geometry // Ages 11 to 14

    KS3 area/volume formulae extend KS2 area of parallelograms and triangles

Topic // Research-informed

Teaching notes

From the Marble Skill Taxonomy. These notes have not been validated in classrooms, so treat them as a starting point for your own judgement.

The approach

Year 6 learners derive and apply area formulae for parallelograms (A = base × perpendicular height) and triangles (A = ½ × base × perpendicular height) through dissection rather than being told the rules. A parallelogram is cut and rearranged into an equal-area rectangle; two congruent triangles are joined into a parallelogram so a triangle is seen to be exactly half (justifying the ÷2). The load-bearing ideas are: the height must be the PERPENDICULAR distance to the chosen base (not a slant side), any side can be the base with its own matching perpendicular height, and the halving step follows from the dissection rather than being an arbitrary rule. Decoy slant lengths teach that a 'given' number is not always a 'needed' number.

Common misconceptions

  • Uses the slant side instead of the perpendicular height, partly from believing every number given must be used.
  • Forgets to halve for a triangle (computes base × height and stops).
  • Thinks the base must be the horizontal or bottom side, so cannot identify base and height in rotated or obtuse triangles.

Useful representations

  • Parallelogram cut-and-rearranged into a rectangle
  • Two congruent triangles joined into a parallelogram/rectangle
  • Right-angle mark showing base–height perpendicularity
  • Rotated shapes with base and perpendicular height labelled
  • Decoy-slant diagrams (extra unnecessary length shown)

Teaching practices

  • Derive each formula by dissection: rearrange a parallelogram into a rectangle; join two copies of a triangle into a parallelogram so the triangle is half of it.
  • Insist the height is perpendicular to the chosen base; have children mark the right angle and show non-examples of lines that are NOT the perpendicular height.
  • Include shapes in varied orientations so children see the base need not be horizontal and any side can be the base with its own perpendicular height.
  • Deliberately give diagrams with a decoy slant length (e.g. base 7, height 6, slant 8) so children practise selecting the perpendicular height and ignoring unnecessary numbers.
  • Tie the ÷2 to the two-triangles-make-a-parallelogram image so halving is meaningful and not forgotten.

Activities to try

  1. 1

    Dissection workbench: rearrange to reveal the formulae

    Why it works: The area formulae are consequences of rearranging shapes: a parallelogram is a rectangle in disguise, and a triangle is half a parallelogram, which is where the ÷2 comes from. The height is always perpendicular to whichever side is chosen as the base, and not every given number is needed.

    1. Take a parallelogram whose perpendicular height from a top corner lands on the base. Cut along that height to remove a right-angled triangle, then slide it to the other end so it forms a rectangle of the SAME area. Confirm that base × perpendicular height is the area, because the rectangle's length and width are the parallelogram's base and perpendicular height.
    2. Take a triangle, duplicate and rotate the copy, and fit the two copies together into a parallelogram; observe that the single triangle is exactly HALF, so its area is base × perpendicular height ÷ 2.
    3. Place or drag the height line so it meets the base at 90°, marking the right angle; check a line that is NOT perpendicular and explain why it is not the height.
    4. Rotate the shape so the base is no longer horizontal, then identify which side is now the base and where its perpendicular height falls.
    5. Work with a diagram that includes a decoy slant length (e.g. base 7, perpendicular height 6, slant 8) and deliberately ignore the slant when computing the area.
  2. 2

    Paper-and-scissors dissection (hands-on alternative)

    Why it works: Physically cutting and rearranging makes the equal-area and 'half of a parallelogram' relationships concrete, and turning the paper shape shows the base does not have to be horizontal.

    1. Cut out a paper parallelogram, draw its perpendicular height, snip along it to remove a right-angled triangle and slide it across to physically make a rectangle; measure the rectangle's length and width and compare them with the parallelogram's base and perpendicular height.
    2. Cut two identical paper triangles and fit them together to make a parallelogram or rectangle; state in words why one triangle is half.
    3. Physically turn a triangle so a different side is the base, then draw the perpendicular height from the opposite vertex to that base and mark the right angle.
  3. 3

    'Who is right?' and calculation practice with decoy lengths

    Why it works: Fluency in selecting the perpendicular height, ignoring unnecessary numbers, and applying the ÷2 only to triangles, while being able to justify each choice.

    1. Compute the area of a triangle given a base, a marked perpendicular height and a decoy slant length, choosing the correct length and halving (e.g. base 7, perpendicular height 6, slant 8 → 7 × 6 ÷ 2 = 21 cm²).
    2. Compute the area of a triangle with base 12 cm and perpendicular height 7 cm, halving the product: 12 × 7 ÷ 2 = 42 cm².
    3. Decide in a 'who is right?' task which of two learners picked the correct length for the height, and justify the choice using the right-angle mark.
    4. Complete a worked step that forces the ÷2 explicitly, and separately compute a parallelogram area without halving (e.g. base 8, perpendicular height 5 → 40 cm²) to show that ÷2 is NOT applied to parallelograms.

Topic // Standards

Curriculum standards

Codes for the curriculum statements this topic maps to. Each publisher owns its standard text.
The national curriculum in England: Key stages 1 and 2 framework document
Ma6/3.1f

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