BTT Mathematics / Primary Mathematics Learning Hub / Guide 18
Triangle area depends on a base and its perpendicular height. The height is not automatically one of the triangle’s sloping sides. It is the shortest perpendicular distance from the chosen base line to the opposite vertex.
A triangle with base 10 cm and perpendicular height 6 cm has area one half of a 10 cm by 6 cm rectangle or parallelogram: ½×10×6=30 cm². The same triangle can be turned or tilted without changing its area.
This guide develops triangle area, missing dimensions, parallelograms, composite rectilinear and triangular figures, and area-preserving rearrangement. Later composite examples are extensions; select sections according to current school scope.
Base and height · Triangle area · Missing dimensions · Parallelograms · Composite figures · 24 questions · Worked answers
1. Base and height must be perpendicular
Any side of a triangle can be chosen as a base. Once a base is chosen, the corresponding height is measured perpendicular to that base from the opposite vertex.
A sloping side may not be the height
Suppose a triangle has a horizontal base of 12 cm, a sloping side of 10 cm and a perpendicular vertical height of 8 cm. For area using the 12 cm base, the correct height is 8 cm, not the 10 cm sloping side.
Using ½×12×10 would calculate a quantity from two lengths that are not perpendicular and would overstate the area.
The height can lie outside an obtuse triangle
For an obtuse triangle, the perpendicular from a vertex may meet an extension of the opposite side rather than the side segment itself. The base line still determines the height.
This is an extension in visual reasoning. The area formula still uses a perpendicular distance.
Orientation does not matter
Rotate a triangle on the page. Its base and height labels may look different, but if the same geometric lengths are used the area is unchanged.
A learner who searches only for a horizontal base may know an image rather than the relationship. Ask which side is being chosen as base and where a perpendicular height can be drawn.
2. Triangle area is half a matching parallelogram
Two congruent copies of a triangle can be rearranged to make a parallelogram with the same base and perpendicular height. Therefore one triangle occupies half the parallelogram’s area.
Area of triangle = ½ × base × perpendicular height.
Worked example A: Direct area
Base=14 cm and height=9 cm.
Area=½×14×9=7×9=63 cm².
Worked example B: Odd base
Base=15 cm and height=8 cm.
Area=½×15×8=15×4=60 cm². Halving the even factor first keeps the arithmetic simple.
Worked example C: Same base, same height
Two triangles stand on the same 12 cm base and both have perpendicular height 7 cm, but their top vertices lie at different horizontal positions. Both have area ½×12×7=42 cm².
The shape of the triangle can change while base and height remain fixed. Area depends on those perpendicular dimensions, not on how far left or right the top vertex sits.
Worked example D: Right triangle
A right triangle has perpendicular legs 6 cm and 9 cm. Either leg can serve as base with the other as height.
Area=½×6×9=27 cm².
3. Reverse area problems reconstruct a missing dimension
Worked example E: Find height
A triangle has area 48 cm² and base 12 cm. From A=½bh:
48=½×12×h=6h, so h=8 cm.
A Primary route doubles the area first: twice the triangle area is ninety-six, which equals base×height. Then 96÷12=8.
Worked example F: Find base
A triangle has area 35 cm² and height 5 cm. Twice the area is seventy. Base=70÷5=14 cm.
Check the found dimension
Substitute the result into the original area formula. For base fourteen and height five, ½×14×5=35 cm².
The final dimension should also be plausible compared with the diagram and other given lengths, though appearance alone is not proof.
Do not divide by two twice
A common error is to double the area and then still multiply by one half during the reverse step. Once 2A=bh has been formed, solve that product directly.
4. Parallelogram area uses base times perpendicular height
A parallelogram can be cut and rearranged into a rectangle without changing area. Its area is base×perpendicular height.
Worked example G: Direct parallelogram area
Base=11 cm, perpendicular height=6 cm. Area=66 cm².
The sloping side length is irrelevant to area unless it is being used as the base with its own corresponding perpendicular height.
Triangle as half a parallelogram
A triangle and parallelogram share the same base 10 cm and height 8 cm. The parallelogram area is 80 cm²; the triangle area is 40 cm².
This one-to-two relationship is a useful check.
Area-preserving shear
If the top of a parallelogram slides sideways while base and perpendicular height remain fixed, its area remains the same. The sloping side changes but the covered surface does not.
This helps learners stop associating “more slanted” with “larger area.”
5. Composite area can be added, subtracted or rearranged
Worked example H: Rectangle split by a diagonal
A 12 cm by 8 cm rectangle is divided by a diagonal into two congruent triangles. Rectangle area=96 cm², so each triangle has area 48 cm².
This also matches ½×12×8.
Worked example I: Rectangle plus triangle
A house-shaped figure consists of a 10 cm by 6 cm rectangle and a triangular roof with base 10 cm and height 4 cm.
Rectangle area=60 cm². Roof area=½×10×4=20 cm². Total=80 cm².
Worked example J: Large rectangle minus triangular corner
A 14 cm by 10 cm rectangle has a right-triangular corner removed with perpendicular sides 4 cm and 6 cm.
Rectangle area=140 cm². Triangle area=½×4×6=12 cm². Remaining area=128 cm².
Worked example K: Two triangles in one rectangle
A point lies on the top edge of a 16 cm by 9 cm rectangle. Join the point to both bottom corners. The two triangles at the bottom together have bases whose total is 16 cm and the same height 9 cm.
Combined area=½×16×9=72 cm², regardless of where the top point lies along that top edge.
This is a strong example of invariance: the separate triangle areas change, but their total does not.
Worked example L: Trapezium by decomposition
A trapezium has parallel horizontal sides 8 cm and 14 cm, with perpendicular height 5 cm. Split it into a rectangle 8 cm by 5 cm and a right triangle with base 6 cm and height 5 cm.
Area=40+½×6×5=40+15=55 cm².
This decomposition avoids needing to memorise a separate trapezium formula, though the result agrees with ½×(8+14)×5.
Perimeter and area remain separate
When a triangle is attached to a rectangle, the shared edge is inside the combined figure. It contributes to component calculations but not to external perimeter.
For area, internal shared edges do not subtract surface; the two non-overlapping regions are simply added.
6. Practice: 24 questions
Use perpendicular height, not a visually convenient side. Keep worked answers covered.
Questions 1–8: Triangle foundations
1. Find the area of a triangle with base 10 cm and height 6 cm.
2. Find the area of a triangle with base 14 cm and height 9 cm.
3. Find the area of a right triangle with perpendicular sides 7 cm and 8 cm.
4. A triangle has base 15 cm and perpendicular height 4 cm. Find its area.
5. Two triangles have the same base 12 cm and height 7 cm. Must their areas be equal?
6. A triangle has area 48 cm² and base 12 cm. Find height.
7. A triangle has area 35 cm² and height 5 cm. Find base.
8. A triangle has base 20 cm and area 90 cm². Find height.
Questions 9–16: Parallelograms and decomposition
9. Find the area of a parallelogram with base 11 cm and height 6 cm.
10. A parallelogram has area 84 cm² and base 12 cm. Find height.
11. A triangle and parallelogram share base 10 cm and height 8 cm. Find both areas.
12. A 12 cm by 8 cm rectangle is divided by a diagonal. Find each triangular half.
13. A 10 cm by 6 cm rectangle has a triangular roof of base 10 cm and height 4 cm. Find total area.
14. A 14 cm by 10 cm rectangle has a right triangle of base 4 cm and height 6 cm removed. Find remaining area.
15. A trapezium has parallel sides 8 cm and 14 cm with height 5 cm. Find its area by decomposition.
16. A parallelogram has base 15 cm, sloping side 9 cm and perpendicular height 7 cm. Find area.
Questions 17–24: Composite reasoning
17. A square of side 12 cm is split by a diagonal. Find the area of each triangle.
18. A rectangle 16 cm by 9 cm has a point anywhere on the opposite 16 cm edge. Two triangles are formed by joining that point to the bottom corners. Find their combined area.
19. A 20 cm by 12 cm rectangle has two non-overlapping triangular corners removed, each with base 4 cm and height 6 cm. Find remaining area.
20. A triangle has area 72 cm² and height 9 cm. Find base.
21. A parallelogram has area 96 cm² and height 8 cm. Find base.
22. A triangle has base 18 cm and height 12 cm. If the height is doubled and the base unchanged, by what factor does area change?
23. If both base and height of a triangle are doubled, by what factor does area change?
24. Explain why a sloping side cannot automatically be used as the height.
7. Worked answers
Answers 1–8
1. 30 cm². ½×10×6.
2. 63 cm². ½×14×9.
3. 28 cm². ½×7×8.
4. 30 cm². ½×15×4.
5. Yes. Both use the same base and perpendicular height, so both have area 42 cm².
6. 8 cm. Double area to 96, then 96÷12.
7. 14 cm. Double area to 70, then 70÷5.
8. 9 cm. Double area to 180, then divide by base twenty.
Answers 9–16
9. 66 cm². 11×6.
10. 7 cm. 84÷12.
11. Triangle 40 cm²; parallelogram 80 cm².
12. 48 cm² each. Rectangle area is ninety-six and the diagonal halves it.
13. 80 cm². Rectangle sixty plus triangle twenty.
14. 128 cm². Rectangle 140 minus triangle twelve.
15. 55 cm². Rectangle 8×5=40 plus right triangle ½×6×5=15.
16. 105 cm². Use perpendicular height seven: 15×7. The sloping side nine is not the corresponding height.
Answers 17–24
17. 72 cm² each. Square area is 144 cm²; the diagonal makes equal halves.
18. 72 cm². Combined bases total sixteen and common height is nine: ½×16×9.
19. 216 cm². Rectangle area 240. Each removed triangle is 12 cm², so remove twenty-four.
20. 16 cm. 2A=144; 144÷9=16.
21. 12 cm. 96÷8.
22. ×2. Area is directly proportional to height when base stays fixed.
23. ×4. Both factors double: 2×2=4.
24. Height must be perpendicular to the chosen base. A sloping side generally measures a different direction and does not give the perpendicular separation needed for area.
8. Diagnose height before arithmetic
If triangle area work fails, ask the learner to draw a right-angle marker where the height meets the base line. If no right angle can be justified, the selected segment may not be the corresponding height.
Use doubling to reverse the formula
Instead of manipulating symbols, double the triangle area to reconstruct the matching parallelogram area. Then base×height becomes a familiar multiplication relationship.
Use two decompositions to check a composite area
Solve a figure by addition, then see whether it can also be solved by subtraction from a larger rectangle. Agreement between structurally different methods is strong evidence.
Connect area to scaling
Doubling one perpendicular dimension doubles area. Doubling both multiplies area by four. These scale rules provide fast plausibility checks.
Continue through the Primary Mathematics series
For rectangles, perimeter and volume, use Measurement, Units, Perimeter, Area and Volume. For angle and shape properties, use Geometry, Angles, Symmetry and Coordinates. For circular composite figures, use Circles and Composite Figures.
Return to the BTT Primary Mathematics Learning Hub.
Original learning guide. Examples and composite figures are educational illustrations rather than official examination questions.