Use parallelogram and trapezium areas
Find and rearrange parallelogram and trapezium areas using their real perpendicular heights and parallel sides.
Exam scope and assessment notes
The first six lessons are Foundation content inherited by Higher; the last two are additional Higher sector work. Use compatible metric units, genuine perpendicular heights and the specified exposed boundary or region. Retain exact model values and π; round only as each question requests. Sectors use degrees; radian measure is excluded. Ordinary sector boundaries require a positive radius and a partial-turn sweep; a full disk has no exposed radial cuts. Compound, inverse and annular examples are chosen teaching breadth, not extra source rows. Drawings and fixed comparisons support self-checking, without marks, mastery, proof approval or physical instrument competence. Linear 4MA1 Issue 2 is checked against November 2026/June 2027 entries; regional options depend on centre eligibility and modular awards remain separate. Book 1 Shape and Space 1 and Number 4 links are navigation only, with no Book 2 tag or verified Teaching Hub body. Public proof pagination differs from the map; confirm the required edition and pages. Full-book, specialist and formal-assessment approval remain pending.
Before you begin
- Recognise parallelograms, trapezia and marked parallel sides.
- Identify a perpendicular height even in a rotated diagram.
- Rearrange a simple area equation with positive lengths.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which three quantities belong in a trapezium area formula?
What you will learn
- Pair a parallelogram base with its perpendicular height rather than its oblique side.
- Identify both parallel sides and their perpendicular separation in a trapezium.
- Recover a missing positive height or parallel side and combine contained planar areas.
Use separate labelled panels. In parallelogram ABCD, AB = 11.2 cm, the perpendicular height to AB is 6.4 cm and oblique AD = 8 cm. In trapezium PQRS, PQ ∥ SR, PQ = 16.5 cm, SR = 9.5 cm, their perpendicular separation is 7.2 cm and an oblique side is 7.5 cm. Find both areas, marking each relevant height and the trapezium’s parallel pair.
Construct and mark your own actual geometry, then enter every requested quantity, substitution and reason. Missing outlines and measures stay for your own construction and calculation.
Given panel 1
Blue: original givens. Amber: my actual metric constructions. Dashed white-filled components are my selected subtractions. Unlabelled measures stay for calculation. The board can scroll horizontally on narrow screens.
The x and y drawing directions use the same scale: 33.38 SVG drawing units per cm. Screen pixels depend on device size. A whole-view turn rotates every trace together; free clicks retain actual coordinates without snapping or correction.
parallelogram Base Cm: 11.2
parallelogram Height Cm: 6.4
parallelogram Side Cm: 8
trapezium Height Cm: 7.2
trapezium Oblique Cm: 7.5
Use separate labelled panels. In parallelogram ABCD, AB = 11.2 cm, the perpendicular height to AB is 6.4 cm and oblique AD = 8 cm. In trapezium PQRS, PQ ∥ SR, PQ = 16.5 cm, SR = 9.5 cm, their perpendicular separation is 7.2 cm and an oblique side is 7.5 cm. Find both areas, marking each relevant height and the trapezium’s parallel pair.
Compare other given figures at the same scale
Given panel 2
Blue: original givens. Amber: my actual metric constructions. Dashed white-filled components are my selected subtractions. Unlabelled measures stay for calculation. The board can scroll horizontally on narrow screens.
The x and y drawing directions use the same scale: 33.38 SVG drawing units per cm. Screen pixels depend on device size. A whole-view turn rotates every trace together; free clicks retain actual coordinates without snapping or correction.
Use separate labelled panels. In parallelogram ABCD, AB = 11.2 cm, the perpendicular height to AB is 6.4 cm and oblique AD = 8 cm. In trapezium PQRS, PQ ∥ SR, PQ = 16.5 cm, SR = 9.5 cm, their perpendicular separation is 7.2 cm and an oblique side is 7.5 cm. Find both areas, marking each relevant height and the trapezium’s parallel pair.
Mark each relevant actual base and perpendicular separation
No marking is recorded for this action.
Choose each count, region or piece kind yourself. Signed arc sweeps are measured counterclockwise from +x when positive and clockwise when negative; only one actual turn is supported. These are coordinate constructions, not a physical-instrument assessment.
My quantities, equations and reasons
Keep significant trailing zeroes in your manually entered text. Exact fields support rational and rational-plus-π notation, including a fractional coefficient such as (3/2)π. Changing any response or unit closes earlier comparison and reference results.
Required for the complete attempt. Give an exact supported rational or rational-plus-π value.
Required for the complete attempt. Give an exact supported rational or rational-plus-π value.
Required for the complete attempt.
Written equations and reasons are recorded for human review; they are not automatically proved or marked.
Required for the complete attempt.
Written equations and reasons are recorded for human review; they are not automatically proved or marked.
Required for the complete attempt.
Written equations and reasons are recorded for human review; they are not automatically proved or marked.
Required for the complete attempt.
Written equations and reasons are recorded for human review; they are not automatically proved or marked.
This tool compares fixed mathematics and actual coordinate constructions. Written reasons need human review; it awards no marks, mastery, physical-instrument competence, specialist assessment or formal assessment approval.
A sloping side is not the height
A parallelogram has area base × perpendicular height. Imagine moving a triangular end to form a rectangle of the same base and height: the occupied area is preserved even though its oblique side looks different. Name the base first, then identify the perpendicular separation of its two parallel supporting lines.
An oblique side can help describe the boundary, but it is not substituted for height. Rotating the whole figure does not change the perpendicular distance. Use the supplied right-angle or height statement rather than treating the page’s vertical direction as a mathematical condition.
A trapezium uses its parallel pair
For a trapezium, let a and b be the two parallel side lengths and h their perpendicular separation. Its area is ½(a + b)h. One explanation places a congruent copy alongside it to form a parallelogram: the doubled area has base a + b and the same height.
The two sides in the sum must be the stated parallel pair, not whichever pair looks convenient. An oblique side between them is different from their perpendicular separation. Label the parallel marks and a matching altitude before substituting, especially when a diagram has been turned.
Recover dimensions and combine areas
Write the original area equation before rearranging for an unknown. A parallelogram height is area divided by its positive base. For a trapezium, double the area and divide by the positive height to obtain the sum of the parallel sides, then remove the known parallel side if the other is requested.
Check that a recovered side or height is positive and reproduces the given area. In a composite context, calculate each region using its own base and perpendicular height. Subtract a bed or opening only when it is actually contained, and do not transfer an oblique side from one shape into another shape’s formula.
Sources for this lesson
Pause and explain
A parallelogram has base 8 cm, perpendicular height 3 cm and oblique side 5 cm. What is its area?
Worked example
Use separate labelled panels. In parallelogram ABCD, AB = 11.2 cm, the perpendicular height to AB is 6.4 cm and oblique AD = 8 cm. In trapezium PQRS, PQ ∥ SR, PQ = 16.5 cm, SR = 9.5 cm, their perpendicular separation is 7.2 cm and an oblique side is 7.5 cm. Find both areas, marking each relevant height and the trapezium’s parallel pair.
Show the worked solution
- For the parallelogram, use AB and its perpendicular height: area = 11.2 × 6.4 = 71.68 cm². AD = 8 cm is an oblique boundary side, not the height.
- The trapezium’s actual parallel pair is PQ and SR. Their sum is 16.5 + 9.5 = 26 cm; the 7.5 cm oblique side does not belong in this sum.
- Use their perpendicular separation: area = ½ × 26 × 7.2 = 93.6 cm². Both results use square centimetres, and each height is paired with the appropriate base lines.
Answer Parallelogram area = 71.68 cm²; trapezium area = 93.6 cm². The relevant heights are 6.4 cm and 7.2 cm, not the oblique side lengths.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
In one labelled panel, a parallelogram has area 84 cm², base 12 cm and oblique side 7.4 cm. Find its perpendicular height. In a separate panel, a trapezium has area 60.5 cm² and parallel sides 8.4 cm and 13.6 cm. Find their perpendicular separation. Write and check both original area equations.
Give me a hint
The oblique side is a distractor for height; divide the area by the appropriate base or average of parallel sides.
Compare my reasoning
- For the parallelogram, write 84 = 12h. Hence h = 84 ÷ 12 = 7 cm; the given 7.4 cm oblique side is a different length.
- For the trapezium, use its actual parallel pair: 60.5 = ½(8.4 + 13.6)k = 11k.
- The separation is k = 5.5 cm. Both recovered heights are positive, and substituting them into their own equations gives 84 cm² and 60.5 cm² respectively.
Parallelogram perpendicular height = 7 cm; trapezium perpendicular separation = 5.5 cm.
Look for these in your work
- I distinguish each perpendicular height from any oblique boundary.
- I identify the trapezium’s parallel pair before rearranging.
- I verify both positive recovered heights in the original equations.
A parallelogram has area 147.6 cm², base 18 cm and oblique side 10.25 cm. Find its perpendicular height. Separately, trapezium PQRS has PQ ∥ SR, area 106.5 cm², perpendicular separation 7.1 cm and SR = 11.3 cm. Find PQ. Sketch and label both independent figures.
Give me a hint
In the trapezium, first recover the sum PQ + SR, not the unknown side alone.
Compare my reasoning
- Use the parallelogram equation 147.6 = 18h, giving h = 8.2 cm. Its 10.25 cm oblique side is not used as the height.
- For the trapezium, 106.5 = ½(PQ + 11.3) × 7.1. Double and divide by 7.1 to obtain PQ + 11.3 = 30 cm.
- Therefore PQ = 18.7 cm. Both results are positive and reproduce their respective areas when substituted into the original formulae.
Parallelogram height = 8.2 cm; the other trapezium parallel side PQ = 18.7 cm.
Look for these in your work
- I draw two separate figures with the correct base/height or parallel-pair labels.
- I recover the sum of trapezium parallel sides before subtracting the known side.
- I reject an oblique-side substitution and check both original areas.
Sketch a trapezium garden with parallel boundaries 14.2 m and 21.8 m, separated perpendicularly by 9.5 m. A fully contained parallelogram bed has base 7.6 m and perpendicular height 4.5 m. Label both different heights. Find the garden area, bed area and remaining area, explaining the subtraction.
Give me a hint
Use a separate area equation for each region; the bed is a contained part of the garden.
Compare my reasoning
- The garden area uses the parallel pair and its perpendicular separation: ½(14.2 + 21.8) × 9.5 = 171 m².
- The parallelogram bed has its own base and height, giving 7.6 × 4.5 = 34.2 m². Do not substitute an outer garden dimension into this formula.
- The bed is fully contained, so subtract its area once: 171 − 34.2 = 136.8 m² of garden remains. The containment statement, rather than a drawing’s appearance, justifies this step.
Garden area = 171 m²; bed area = 34.2 m²; remaining garden area = 136.8 m².
Look for these in your work
- I identify the garden’s parallel pair and perpendicular height.
- I use the bed’s own perpendicular height in its parallelogram area.
- I justify one contained subtraction and give square-metre results.
Common mistakes
- Multiplying by an oblique side treats a sloping boundary as a perpendicular height.
- Using nonparallel trapezium sides in a + b changes the formula’s geometric inputs.
- A negative recovered height or parallel side does not describe the stated nondegenerate shape.
What can you explain now?
With the parallel side lengths fixed, what happens to a trapezium’s area if its perpendicular height doubles?
Compare with the explanation
Its area doubles because the same parallel-side sum is multiplied by twice the height.
In ½(a + b)h, the sum of parallel sides stays fixed and the factor h doubles. This statement concerns that one area formula, not a general similarity claim.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Write the two area formulae and explain the role of every length in them. Describe how to recover an unknown trapezium parallel side from its area, height and other parallel side.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.