Decompose triangle and rectangle areas
Use triangle and rectangle areas to build or remove planar regions, with genuine perpendicular heights and explicit overlap conditions.
Section 4.9 · Mensuration of 2D shapes
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
- Use rectangle and triangle area formulae.
- Identify perpendicular lines and an altitude foot.
- Multiply decimals and rearrange a one-step positive-length equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which length is the height paired with a chosen triangle base?
What you will learn
- Choose a base and its perpendicular height, including an outside altitude.
- Add nonoverlapping triangle and rectangle areas or subtract contained cut-outs.
- Recover a positive missing dimension and keep exact intermediate areas.
Draw a sign made from rectangle ABCE and an outward triangular top CDE along CE. AB = CE = 13.6 cm, AE = BC = 7.5 cm, and the perpendicular altitude from D to CE is 4.5 cm. Label its foot H and a right-angle mark. Find the rectangle area, triangle area and total, explaining why the areas can be added.
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: 31.41 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.
triangle Base Cm: 13.6
triangle Height Cm: 4.5
Draw a sign made from rectangle ABCE and an outward triangular top CDE along CE. AB = CE = 13.6 cm, AE = BC = 7.5 cm, and the perpendicular altitude from D to CE is 4.5 cm. Label its foot H and a right-angle mark. Find the rectangle area, triangle area and total, explaining why the areas can be added.
Propose every actual component region and its add/subtract operation
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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. 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.
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 base needs its matching perpendicular height
The area of a rectangle is its base multiplied by its perpendicular height; a square is the special case with equal dimensions. For a triangle, use half the product of a chosen base and the perpendicular distance from the opposite vertex to the base line. Attach each height to the base it belongs to.
The height need not lie inside a triangle. In an obtuse triangle its perpendicular foot can fall on an extension of the base. The formula still uses the original base segment and that perpendicular distance; neither the sloping side nor an extra extension length is substituted for them.
Build an area without overlap
A compound region can be separated into rectangles and triangles. Label the parts and choose a partition whose interiors do not overlap. Then add their areas: sharing an edge does not create shared positive area, so an attached triangle can meet a rectangle along its base without double-counting the region.
Subtraction needs a different justification. Start with a complete enclosing region and remove only the portions that are actually cut out. A hole must be contained, and separate cut-outs must be disjoint unless overlap is handled explicitly. A sketch that merely resembles a familiar shape does not establish these conditions.
Rearrange and interpret square units
If a rectangle’s area and base are known, write the area equation and divide by the positive base to find its height. For a triangle, retain the factor one-half when rearranging. Check that the recovered positive dimension reproduces the original area in compatible units.
Two multiplied length units produce square units. Keep component areas exact until their sum or difference is complete, then round only if the question asks. An area result does not by itself supply a perimeter: sloping and exposed boundary lengths require their own information and reasoning.
Sources for this lesson
Pause and explain
A shape has an 8 cm by 5 cm rectangle and an outward triangle of base 8 cm and perpendicular height 3 cm, meeting only along that base. What is its total area?
Worked example
Draw a sign made from rectangle ABCE and an outward triangular top CDE along CE. AB = CE = 13.6 cm, AE = BC = 7.5 cm, and the perpendicular altitude from D to CE is 4.5 cm. Label its foot H and a right-angle mark. Find the rectangle area, triangle area and total, explaining why the areas can be added.
Show the worked solution
- The rectangle has two perpendicular dimensions 13.6 cm and 7.5 cm. Its area is 13.6 × 7.5 = 102 cm².
- For triangle CDE, use base CE = 13.6 cm and perpendicular height DH = 4.5 cm. Its area is ½ × 13.6 × 4.5 = 30.6 cm²; a roof slope is not this height.
- The two parts meet only along CE and have no overlapping interior. Add their areas: 102 + 30.6 = 132.6 cm². The shared edge is a length, not a second area to remove.
Answer Rectangle area = 102 cm²; triangular top area = 30.6 cm²; total = 132.6 cm². The partition has no overlapping interior.
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.
Sketch an 18 cm by 12 cm floor with a 7 cm by 5 cm top-right rectangular corner removed. In the retained floor, a separate contained triangular opening KLM has base KL = 5.6 cm and perpendicular altitude 3.5 cm. The two cut-outs are disjoint. Label an appropriate altitude foot and find the remaining area, explaining each subtraction.
Give me a hint
Calculate the complete rectangle, the corner and the triangular opening separately before combining them.
Compare my reasoning
- The complete enclosing rectangle has area 18 × 12 = 216 cm². The removed rectangular corner has area 7 × 5 = 35 cm².
- The triangular opening uses its own base KL and matching perpendicular altitude. Its area is ½ × 5.6 × 3.5 = 9.8 cm².
- The stated disjoint, contained removals justify 216 − 35 − 9.8 = 171.2 cm². Neither opening overlaps the other or extends into a region already removed.
Remaining area = 171.2 cm². The 35 cm² corner and 9.8 cm² triangular opening are subtracted once each.
Look for these in your work
- I label the enclosing rectangle and both disjoint cut-outs.
- I identify the triangle’s base and perpendicular altitude and retain the factor one-half.
- I justify subtraction using containment and disjointness rather than the sketch alone.
Draw triangle ABC with AB = 17.5 cm. The perpendicular from C meets the extended base line at H, with H–A–B in that order, CH = 8.4 cm and HA = 3.2 cm. Find the triangle area. Separately, rectangle PQRS has PQ = 6.8 cm and area 45.9 cm²: find QR. Explain the height used in each calculation.
Give me a hint
The outside foot changes the position of the altitude, not the stated base AB.
Compare my reasoning
- CH is perpendicular to the line through AB, so it is the matching altitude even though H is outside the base segment. Keep AB = 17.5 cm; do not add HA.
- The triangle area is ½ × 17.5 × 8.4 = 73.5 cm². The separate 3.2 cm extension is not part of the chosen base.
- For the rectangle, write 45.9 = 6.8 × QR. Division gives QR = 6.75 cm, and the positive dimension reproduces the given area.
Triangle area = 73.5 cm²; rectangle height QR = 6.75 cm. Perpendicular distance, rather than the position of the foot or a sloping side, determines the triangle height.
Look for these in your work
- I show H–A–B order and the right angle at H.
- I retain the original base AB when the altitude falls outside it.
- I rearrange the separate rectangle equation and check the positive result.
Draw a central 2.4 m by 1.15 m rectangular banner. One outward triangular tab is attached along each short end; each tab has base 1.15 m and perpendicular projection 0.45 m. A contained 0.8 m by 0.25 m rectangular window is removed. Label the two tab heights and the window. Find the exact remaining area and a 3-significant-figure approximation.
Give me a hint
The two tab areas are added; the contained window is removed after the nonoverlapping outside parts are combined.
Compare my reasoning
- The rectangle contributes 2.4 × 1.15 = 2.76 m². Each triangular tab contributes ½ × 1.15 × 0.45 = 0.25875 m².
- The two outward tabs have no interior overlap with the rectangle or each other. Their combined added area is 0.5175 m².
- The contained rectangular window removes 0.8 × 0.25 = 0.2 m². Therefore the exact remaining area is 2.76 + 0.5175 − 0.2 = 3.0775 m².
- Only now round the requested total: 3.0775 m² ≈ 3.08 m² to 3 significant figures. Rounding each small tab first would discard information unnecessarily.
The exact remaining area is 3.0775 m², approximately 3.08 m² to 3 significant figures.
Look for these in your work
- I show both outward tabs, their perpendicular heights and the contained window.
- I add two separate triangle areas and subtract the window once.
- I keep exact intermediate areas and label the final square unit and requested precision.
Common mistakes
- Using a sloping side instead of the perpendicular height changes the triangle or parallelogram area.
- Adding overlapping areas or subtracting the same cut-out twice gives a false total.
- An outside altitude does not extend the chosen base segment.
What can you explain now?
A triangle’s altitude foot lies outside its base segment. Does that change the formula, or require adding the extension length to the base?
Compare with the explanation
No. Use ½ × the original base × its perpendicular altitude.
The height is a distance to the base line. Its location outside the segment does not change the chosen base length.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain when areas should be added and when they should be subtracted. Include the roles of nonoverlap, containment and a base’s matching perpendicular height.
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.