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Area of triangles and parallelograms - Mathematics Year 6

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Introduction

Why is a triangle's area half a base-height product, while a parallelogram uses the whole product? Derive both rules, choose the correct height even after a shape turns, and solve a composite-area problem with an independent check.

Success criteria. Calculate six specified areas, identify heights in two actual source diagrams, explain both derivations and communicate a complete contextual solution with squared units.

Every numerical dimension below is a supplied hypothetical value. The source diagrams illustrate relationships; their screen pixels do not provide measurements. You may describe rearrangements with words or labelled sketches instead of physically cutting paper.

Derive two area rules from rectangles

Area measures coverage. A square centimetre is a square one centimetre on each side. Counting equal square units gives area. A rectangle eight centimetres long and five centimetres wide contains forty such units. The rule length times width follows the rows-and-columns structure.

Parallelogram rearrangement. A parallelogram has opposite sides parallel. Choose one side as a base and use the perpendicular distance between that base line and the opposite parallel side. Imagine removing a triangular end piece and translating it to the other end. No piece is stretched, overlapped or discarded. The resulting rectangle has the same base and perpendicular height.

In the original Commons image, B labels the base and H the vertical height. A labels a horizontal offset and theta an angle. The source does not draw a movement arrow; the cut-and-slide process is a reasoning action we describe. Its surrounding rectangle also provides a check: width B plus A, times H, minus the two flanking triangles, leaves B times H.

For the supplied base 8 cm and perpendicular height 5 cm:

$8 \times 5 = 40$

Area is 40 cm². A sloping side of another length would affect the perimeter, but it is not automatically the perpendicular height.

An acute triangle with base b, a dashed altitude h and a small right-angle mark at its foot. — The dashed h meets base b at a right angle. It is a height, although it is not a boundary side.

Why does a triangle need one half? Take two congruent copies of any triangle. Rotate one copy through a half turn and join suitable matching edges so the two copies form a parallelogram with the same chosen base and perpendicular height. Both copies have the same area, so one triangle covers half of that parallelogram.

For the supplied triangle base 10 cm and perpendicular height 6 cm:

$10 \times 6 \div 2 = 30$

Area is 30 cm². Doubling it gives 60 cm², the corresponding base-height parallelogram. The division by two reflects the two equal triangle copies; it is not an arbitrary instruction added after multiplying.

A right-triangle route. A diagonal divides a rectangle into two congruent right triangles. Each therefore has half the rectangle's area. For a general triangle, the two-copy parallelogram argument extends the same idea without assuming every triangle is right-angled.

Corbettmaths' original Area of a Triangle - Primary shows the 5 cm by 4 cm example giving 10 cm². Compare that half-product explanation with our independently supplied ten-by-six task. The video does not supply the dimensions of the separate Commons diagrams.

Explain, then calculate. Name the shape, choose a base-height pair, establish the right angle, substitute the stated lengths, and give a squared unit. A calculation with two convenient side lengths is insufficient if they do not form the required base and perpendicular height.

Find the correct height when orientation changes

A height belongs to a chosen base. The word height does not mean 'the side that happens to point upwards on this page'. Turn the page or rotate the shape: its area does not change, but the same base can now be vertical or sloping.

Height identification in two source diagrams. In the acute triangle, b is the horizontal base and h is the dashed perpendicular from the opposite vertex. The small right-angle mark is the decisive relationship. In the parallelogram, B is the base and H is its perpendicular height; A is a horizontal offset. Identify h and H independently and explain why neither a general sloping side nor A replaces them.

A rotated coordinate model. Consider triangle vertices A = (0, 0), B = (0, 6), C = (4, 3), with coordinates in centimetres. AB is now a vertical base of length six. The perpendicular height from C to the line AB runs horizontally to D = (0, 3), length four.

B (0,6)
|\
| \
D--C (4,3)
| /
|/
A (0,0)

This is a conceptual labelled sketch, not a scale drawing. AB is six centimetres and CD four, giving:

$6 \times 4 \div 2 = 12$

The area is 12 cm². A second method splits the triangle at D into two right triangles. Each has perpendicular dimensions three and four, so each covers six square centimetres. Their areas total twelve. The horizontal CD, rather than the vertical appearance of the page, supplies the height for base AB.

An exterior height. For a different triangle use A = (0, 0), B = (6, 0) and C = (−2, 4). The chosen base AB is six centimetres long. The perpendicular from C meets the EXTENSION of AB at H = (−2, 0), outside the triangle. Its length is four. The triangular area is still twelve square centimetres:

$6 \times 4 \div 2 = 12$

An independent subtraction uses the larger right triangle HBC, with base eight and height four, minus the smaller right triangle HAC, with base two and height four:

$8 \times 4 \div 2 - 2 \times 4 \div 2 = 12$

The need to extend a base does not invalidate its perpendicular height. Sketch those labelled points to see which region is added and subtracted.

Changing the base needs a matching height. You may choose another triangle side as base, but must then use the perpendicular distance from its opposite vertex to THAT base line. Pairing one base with the height for a different base generally gives a wrong area.

Recover an unknown height. A triangle has area 18 cm² and chosen base 9 cm. Double the area to find the matching parallelogram area, then divide by nine:

$2 \times 18 \div 9 = 4$

Its perpendicular height is 4 cm. Substituting back, nine times four divided by two recovers eighteen. For a parallelogram of area 18 cm² and base 9 cm, the height would instead be 2 cm. The shape type determines the inverse relation.

Locate the first error. A learner sees base six, perpendicular height four and sloping side five centimetres, then calculates thirty square centimetres for a parallelogram. Multiplication is performed correctly; the first error is choosing five instead of the perpendicular four. Correct area is twenty-four square centimetres. Explain the geometric choice rather than merely replacing the final answer.

Calculate six areas and solve a composite problem

Independent area set. The heights below are explicitly perpendicular to their stated bases. Complete all six before using the Learning Check feedback.

Task	Shape	Base in cm	Perpendicular height in cm
T1	Triangle	10	6
T2	Triangle	7	4
T3	Triangle	2.4	5
P1	Parallelogram	8	5
P2	Parallelogram	6	3.5
P3	Parallelogram	9	4

Choose the formula from the shape, not from whether the numbers are whole or decimal. T3 has a decimal base but still uses the same half-product rule. P2 needs no halving merely because its height is 3.5.

Check by reconstruction. Double each triangle area to recover base times height. For each parallelogram, divide area by base to recover height. Those checks test the rule's structure rather than simply pressing the same keys twice. Compare T1 with the chapter-one derivation and P1 with the eight-by-five example.

A composite sign. A fictional sign consists of an 8 cm by 4 cm rectangle with a triangular top whose base is the rectangle's entire 8 cm top edge and whose perpendicular height is 3 cm. The parts touch along an edge and do not overlap.

$8 \times 4 + 8 \times 3 \div 2 = 44$

Total area is 44 cm². A bounding 8 cm by 7 cm rectangle covers 56 cm². The two top corner triangles together cover 12 cm². Subtracting gives the same sign area:

$8 \times 7 - 12 = 44$

The shared edge is not an extra area. If the question asked for perimeter, we would need suitable sloping-side lengths and a different boundary calculation; area alone does not determine that perimeter.

A cut-out panel. Another fictional panel starts as a 10 cm by 6 cm rectangle. Remove a triangle entirely within the panel whose base is 4 cm and perpendicular height 3 cm. The removed area is six square centimetres:

$10 \times 6 - 4 \times 3 \div 2 = 54$

The remaining panel area is 54 cm². State that the triangular region is removed once and lies inside the original rectangle. Without that positional information, subtracting areas could describe a different object.

A useful reasonableness bound. The panel must have less area than the original sixty-square-centimetre rectangle and more than zero. Fifty-four satisfies that bound, but the bound alone does not prove accuracy. An answer of fifty would also satisfy it; retain the exact decomposition and inverse check.

Units before arithmetic. For a triangle with base 80 mm and perpendicular height 5 cm, first convert eighty millimetres to eight centimetres. Its area is twenty square centimetres. In millimetres, use base eighty and height fifty to obtain two thousand square millimetres. These agree because one square centimetre contains one hundred square millimetres, not ten.

Your complete solution. Identify the region, choose and justify a perpendicular base-height pair or non-overlapping decomposition, convert compatible units, calculate with the appropriate formula and perform a second check. If dimensions are not given or a perpendicular relationship is uncertain, name the missing information instead of measuring an unscaled screen picture.

Assessment route. Submit the six areas, h/H identification in the two actual source diagrams, a rotated-height explanation and either composite problem with two methods. Include the two-copy derivation explaining why triangles use one half. This is a focused extension of earlier area work, rather than evidence of every annual mathematics objective.

Interactive Tasks

Quiz: Test Your Knowledge

Why is a triangle area half of base times perpendicular height? (Two congruent copies form a matching parallelogram) (!Every triangle has two right angles) (!Only half its boundary is counted) (!The base must always be the shortest side)

Which source label is the parallelogram perpendicular height? (H) (!A) (!The angle symbol theta) (!The sloping boundary without a right-angle relation)

In the actual acute-triangle source, what establishes the height? (The dashed h and its right-angle relationship with b) (!Its position at the top of the screen) (!The length of the image filename) (!A numerical dimension inferred from pixels)

What is T2 with base seven and perpendicular height four centimetres? (Fourteen square centimetres) (!Twenty-eight square centimetres) (!Eleven square centimetres) (!Fourteen centimetres)

What is P2 with base six and perpendicular height three point five centimetres? (Twenty-one square centimetres) (!Ten point five square centimetres) (!Nine point five square centimetres) (!Twenty-one cubic centimetres)

For vertical base AB in the rotated model, what is the perpendicular height? (Horizontal CD of four centimetres) (!The six-centimetre vertical base itself) (!The display height of the code box) (!An unknown pixel count)

Where may an obtuse triangle altitude meet its chosen base line? (On an extension outside the triangle) (!Only inside the triangle) (!Only at the midpoint of the base segment) (!Nowhere because obtuse triangles have no area)

What is the composite sign area in the worked example? (Forty-four square centimetres) (!Fifty-six square centimetres) (!Twelve square centimetres) (!Thirty-two centimetres)

Which calculation finds a triangle height from area eighteen and base nine? (Double eighteen then divide by nine) (!Divide eighteen by nine without doubling) (!Multiply eighteen by nine) (!Subtract nine from eighteen)

How do square centimetres convert to square millimetres? (Multiply by one hundred) (!Multiply by ten) (!Divide by ten) (!Keep the number unchanged regardless of unit)

Memory Game

Area	Flat coverage measured in square units
Perpendicular height	The right-angle distance to the chosen base line
Base	The chosen side used with its matching height
Congruent copies	Shapes of the same size and shape
Composite region	A region split into simpler non-overlapping parts
Altitude	A perpendicular from a vertex to the opposite base line

Drag and Drop

Triangle rule	Multiply base and perpendicular height then halve
Parallelogram rule	Multiply base and perpendicular height without halving
Height check	Use a right-angle relationship with the selected base
Composite subtraction	Remove an included region once from the whole
Inverse triangle check	Double area then divide by the selected base

Crossword Puzzle

AREA	Coverage of a flat region
HEIGHT	A perpendicular distance for a chosen base
BASE	The selected side paired with a height
TRIANGLE	A polygon with three sides
RECTANGLE	The four-right-angle shape used to explain area
ALTITUDE	A perpendicular line segment from a vertex to its opposite side line

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Six independent areas: Complete T1–T3 and P1–P3 from the actual task table. Show shape, numerical substitution and squared units.
Two source heights: Identify h in the acute triangle and H in the parallelogram. Use right-angle evidence and distinguish the source offset A.
Explain the triangle half: Describe or sketch the two-copy parallelogram derivation. Explain why both triangles have equal area.
Recover two heights: Compare a triangle and parallelogram each with area eighteen square centimetres and base nine centimetres. Recover each height and substitute back.

Standard

Rotated height: Use A, B, C and D in chapter two. Find the vertical base, horizontal height and area; check by two right triangles.
Exterior altitude: Plot the second triangle and H on the extended base line. Calculate area directly and by subtracting two right-triangle areas.
Composite sign: Calculate the sign area by addition and bounding-rectangle subtraction. Explain the role of the shared edge.
Cut-out panel: Calculate the panel remaining after the triangular cut-out. Give a bound and a more decisive reconstruction check.

Advanced

Units and squared factors: Solve the eighty-millimetre by five-centimetre triangle in two consistent unit systems. Explain the square-unit conversion factor.
Wrong-height diagnosis: Locate the first error when a parallelogram base six, perpendicular height four and sloping side five gives area thirty. State the corrected result.
Same area new dimensions: Find a different triangle base-height pair with area thirty square centimetres. Check the product and explain why the sloping sides need not match.
Unfamiliar composite dossier: Create or solve a new non-overlapping composite region. State dimensions, perpendicular relationships, units, two solution methods and any missing information.

Learning Check

Six area results: Check all six supplied tasks.
Source diagrams and derivations: Name the heights and explain both area rules.
Two unfamiliar orientations: Check the rotated and exterior-height examples.
Composite solutions: Check the sign and cut-out panel by complete decompositions.
Inverse heights and mixed units: Check the reverse and unit-conversion tasks.
Reasoned complete solution: Evaluate an original area explanation and misconception correction.

Worked checks and assessment guidance

T1 = 30 cm²; T2 = 14 cm²; T3 = 6 cm²; P1 = 40 cm²; P2 = 21 cm²; P3 = 36 cm². Triangles use half the base-height product; parallelograms use the full product. Doubling triangle answers or dividing parallelogram area by base provides a structural check.
The acute triangle uses dashed h perpendicular to b; the parallelogram uses H perpendicular to B, not offset A. A cut-and-slide parallelogram rearrangement preserves area and makes a rectangle. Two congruent triangle copies make a matching parallelogram, so one triangle occupies half its area.
Rotated A(0,0), B(0,6), C(4,3): base AB six and horizontal CD four give 12 cm². Two right triangles of area six each confirm it. Exterior example A(0,0), B(6,0), C(−2,4): base six and height four give 12 cm²; HBC area sixteen minus HAC area four gives twelve.
Sign: rectangle thirty-two plus triangle twelve gives 44 cm². Bounding rectangle fifty-six minus corner regions twelve also gives forty-four. Panel: rectangle sixty minus triangular cut-out six gives 54 cm²; adding the removed six reconstructs sixty. Shared edges do not create overlapping area.
Area eighteen with base nine gives triangle height four centimetres and parallelogram height two centimetres. For base80 mm and perpendicular height5 cm, use8×5÷2=20 cm² or80×50÷2=2000 mm². One square centimetre equals one hundred square millimetres.
Require specified region, justified perpendicular base-height relationship, compatible units, correct formula/decomposition and an independent check. The base-six parallelogram with perpendicular height four has area24 cm²; choosing sloping side five was the first wrong step. An unscaled screen picture cannot supply absent physical measurements.

Learning evidence

Keep six checked area calculations, height identifications in both actual source diagrams, both derivations and one unfamiliar composite or rotated solution with an independent check. Written or spoken reasoning may replace cutting or fine drawing. This records mathematical understanding rather than observed practical construction competence or complete annual coverage.

OER and Wikipedia

Checked 11 November 2026. Department for Education: mathematics programmes of study, Year6 measurement, informs derivation and use of triangle/parallelogram areas and appropriate units. Examples, six-task table, rotated coordinate models and composite problems are original teaching material.

Source images are unchanged. Acute-triangle b/h and right-angle mark, and parallelogram B/H/A meanings, were inspected. Numerical lengths are assigned examples rather than measurements from source pixels.

Original embedded Area of a Triangle - Primary, corbettmaths, supports the half-area model. It is embedded at its publisher, not copied or labelled as Commons-licensed.

Triangle (acute) bh.PNG, H Padleckas at English Wikibooks, Public domain. Unchanged.

Parallelogram area.svg, en:User:Cburnett, CC BY-SA 3.0. Unchanged.