Primary 5 Maths: Area & Perimeter
Area of a triangle (base × height ÷ 2), choosing the right height for obtuse triangles, and perimeter of composite figures — worked P5 examples with common mistakes.
Most schools typically cover this topic in Term 3, around weeks 5–7 — though every school sets its own sequence.
Primary 4 taught the area and perimeter of rectangles and squares. Primary 5 adds a new shape and a new idea: the area of a triangle, and the height that has to be perpendicular to the base — which is not always one of the triangle's drawn sides. Once triangles are combined with rectangles, the same figures also drive most of the composite-shape perimeter questions in this topic. Length × breadth also comes back later in P5 with a third measurement added, in volume of cubes and cuboids.
The core formula is short: Area of a triangle = base × height ÷ 2. The base can be any one of the triangle's three sides; the height is always the perpendicular distance from that base to the opposite corner — and for an obtuse triangle, that perpendicular line can land outside the triangle altogether, on an extension of the base. This is exactly the scope set out for Primary 5 in MOE's 2021 Primary Mathematics Syllabus.
P5 Area & Perimeter: the mistakes to watch for
- Using a slanted side as if it were the height. The height must meet the base at a right angle. A triangle drawn with one slanted side and one vertical side often tempts children into multiplying the two given sides directly, even when neither is perpendicular to the chosen base.
- Assuming the height must fall inside the triangle. For an obtuse triangle, the perpendicular from the apex to the base often lands beyond the base's endpoint, on the base extended. The formula still works exactly the same way — base × height ÷ 2 — using that same perpendicular length.
- Forgetting the ÷ 2. Multiplying base × height and stopping there gives the area of the rectangle that contains the triangle, which is always double the triangle's actual area.
- Assuming a smaller perimeter after a corner is cut away. Cutting a rectangular notch out of a corner removes two edges but adds two new ones of the same total length — the perimeter is often unchanged. Add up the actual sides; don't guess from the picture.
P5 Area: base and height of a triangle
P5 Area of a triangle: warm-ups with base, height and area
Easy: area from base and height
A triangle has a base of 12 cm and a height of 7 cm. Find its area.
Show the worked solution
Area = base × height ÷ 2 = 12 × 7 ÷ 2 = 42 cm².
Easy: finding the base from the area
A triangle has an area of 54 cm² and a height of 9 cm. Find its base.
Show the worked solution
Base × height = area × 2 = 54 × 2 = 108.
Base = 108 ÷ 9 = 12 cm.
Easy: finding the height from the area
A triangle has an area of 40 cm² and a base of 16 cm. Find its height.
Show the worked solution
Base × height = area × 2 = 40 × 2 = 80.
Height = 80 ÷ 16 = 5 cm.
Easy: perimeter of an L-shaped figure
An L-shaped piece of card has six straight sides. Going around it, the sides measure 10 cm, 5 cm, 4 cm, 3 cm, 6 cm and 8 cm. Find its perimeter.
Show the worked solution
Perimeter = sum of all sides = 10 + 5 + 4 + 3 + 6 + 8 = 36 cm.
P5 Area & Perimeter: exam-style triangles and composite figures
Exam: area of an obtuse triangle
An obtuse triangle has a base of 10 cm. Its height, measured at right angles to the base, is 6 cm — for this triangle, the height line falls outside the triangle and meets the base extended. Find the area of the triangle.
Show the worked solution
The formula doesn't change: Area = base × height ÷ 2 = 10 × 6 ÷ 2 = 30 cm².
Exam: area of a rectangle with a triangular roof
A figure is made of a rectangle 10 cm by 6 cm, with a triangle of base 10 cm and height 4 cm attached on top of the rectangle, sharing the 10 cm side. Find the total area of the figure.
Show the worked solution
Area of rectangle = 10 × 6 = 60 cm².
Area of triangle = 10 × 4 ÷ 2 = 20 cm².
Total area = 60 + 20 = 80 cm².
Exam: perimeter after a corner is cut away
A rectangular sheet of metal measures 15 cm by 9 cm. A rectangular notch measuring 4 cm by 3 cm is cut out of one corner. Find the perimeter of the remaining figure.
Show the worked solution
Cutting a rectangular notch from a corner removes two edges of the original rectangle but replaces them with two new edges of exactly the same total length — one going in, one going across.
Perimeter of the figure = perimeter of the original rectangle = 2 × (15 + 9) = 48 cm.
Check: whatever length is removed from the top and side is exactly matched by the two new inner edges of the notch. ✓
Exam: finding a missing height from a composite area
A figure is made of a rectangle 8 cm by 5 cm, with a triangle attached on top sharing the 8 cm side as its base. The total area of the figure is 52 cm². Find the height of the triangle.
Show the worked solution
Area of rectangle = 8 × 5 = 40 cm².
Area of triangle = 52 − 40 = 12 cm².
Height of triangle = 12 × 2 ÷ 8 = 3 cm.
PSLE challenge: triangles combined with area and perimeter
PSLE challenge: a trapezium cut from a rectangle
A rectangular plot of land measures 18 m by 10 m. Two triangular corners are cut off to leave a trapezium-shaped flower bed: one triangle has a base of 6 m along the top edge and the full 10 m height of the plot; the other has a base of 4 m along the top edge and the same 10 m height. Find the area of the flower bed that remains.
Show the worked solution
Area of rectangle = 18 × 10 = 180 m².
Area of first triangle = 6 × 10 ÷ 2 = 30 m². Area of second triangle = 4 × 10 ÷ 2 = 20 m².
Remaining area = 180 − 30 − 20 = 130 m².
PSLE challenge: base and height in a fixed relationship
A triangle's height is twice its base. Its area is 36 cm². Find its base and its height.
Show the worked solution
Let the base be b cm, so the height is 2 × b cm.
Area = base × height ÷ 2 = b × 2b ÷ 2 = b × b.
So b × b = 36, which means b = 6 cm (since 6 × 6 = 36), and the height = 2 × 6 = 12 cm.
Check: 6 × 12 ÷ 2 = 36 cm². ✓
PSLE challenge: area and fencing for a garden with a pond
A square garden has sides of 20 m. A rectangular pond measuring 6 m by 4 m is built entirely inside the garden, away from its edges. Find (a) the area of the garden not covered by the pond, and (b) the total length of fencing needed to go around both the garden's outer edge and the pond's edge.
Show the worked solution
(a) Area of garden = 20 × 20 = 400 m². Area of pond = 6 × 4 = 24 m².
Area not covered = 400 − 24 = 376 m².
(b) Perimeter of garden = 4 × 20 = 80 m. Perimeter of pond = 2 × (6 + 4) = 20 m.
Total fencing = 80 + 20 = 100 m.
How to practise this topic
Sketch first, calculate second: mark which side is being used as the base, drop the height in as a dashed perpendicular line — even when it lands outside the triangle — and only then multiply. For composite figures, decide whether the total is a sum or a difference of simpler shapes before reaching for a calculator. This topic leans heavily on multi-step reasoning, so once base × height ÷ 2 is automatic, P5 Word Problems practice is the natural next step, and the same formula returns at P6 once circles are added to the mix in P6 Area & Circumference of a Circle.
What this topic covers
In the MOE Primary Mathematics syllabus, Primary 5 Area & Perimeter includes the concepts below. (P5-M-09 is Test Paper's own topic code, not an MOE reference.)
- Area of a triangle
- Identifying base and height of a triangle
- Area of composite figures
- Perimeter of composite figures
Before this topic
These topics feed into Area & Perimeter — if this one wobbles, check them first: Primary 4 Area & Perimeter · Primary 5 Geometry (Angles & Shapes).
Where to go next
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