Primary 6 Maths: Area & Circumference of Circle

Updated

Radius, diameter, circumference and area of circles, plus semicircles, quarter circles and composite figures — the P6 circle topic that PSLE loves to combine with other shapes.

Most schools typically cover this topic in Term 2, around weeks 5–7 — though every school sets its own sequence.

Circles arrive at Primary 6 with a short, sharp scope: the parts of a circle, circumference, area, and the same two formulas applied to semicircles, quarter circles and shapes built by combining a circle with a rectangle, triangle or square. The harder exam questions rarely stop at "find the area of this circle": they ask for a shaded region made of two or three pieces, at least one of them curved.

The whole topic runs on two formulas: circumference = 2 × π × r (or π × d) and area = π × r². Each question tells the child which value of π to use: "Take π = 3.14", "Take π = 22/7", or, on the calculator paper, the calculator's π key. A radius that's a multiple of 7 usually comes with 22/7, because the 7s cancel.

The mistakes to watch for

Parts of a circle, and a corner shaded region

The radius and diameter of a circle A circle with its diameter drawn as a horizontal line through the centre and its radius drawn as a dashed vertical line from the centre to the top of the circle, next to the circumference and area formulas. d r Circumference = 2 × π × r Area = π × r² π ≈ 3.14 or 22/7 Diameter d = 2 × radius r
The diameter passes through the centre; the radius is exactly half of it. Both formulas use the radius.
A square with a quarter circle removed, shaded region highlighted A 100 by 100 square with a quarter circle of the same radius drawn from the top-left corner, curving from the top-right corner to the bottom-left corner; the region of the square beyond the curve, near the bottom-right corner, is shaded to show the area that is the square's area minus the quarter circle's area. Square side = quarter circle radius = 14 cm Shaded = square − quarter circle = 14² − ¼ × 22/7 × 14² = 196 − 154 = 42 cm²
The shaded corner is what's left of the square once the quarter circle (radius = the square's side) is taken out.

Easy: radius, diameter, circumference and area

Easy: radius from diameter

The diameter of a circular plate is 14 cm. Find its radius.

Show the worked solution

Radius = diameter ÷ 2 = 14 ÷ 2 = 7 cm.

Easy: circumference from radius (22/7)

A circle has a radius of 7 cm. Find its circumference. (Take π = 22/7.)

Show the worked solution

Circumference = 2 × π × r = 2 × 22/7 × 7 = 44 cm.

Easy: circumference from diameter (3.14)

A circle has a diameter of 10 cm. Find its circumference. (Take π = 3.14.)

Show the worked solution

Circumference = π × d = 3.14 × 10 = 31.4 cm.

Easy: area from radius (3.14)

A circle has a radius of 10 cm. Find its area. (Take π = 3.14.)

Show the worked solution

Area = π × r² = 3.14 × 10 × 10 = 314 cm².

Exam: semicircles, quarter circles and finding an unknown radius

Exam: area of a semicircle (22/7)

A semicircle has a radius of 14 cm. Find its area. (Take π = 22/7.)

Show the worked solution

Full circle's area = 22/7 × 14 × 14 = 616 cm².

Semicircle's area = 616 ÷ 2 = 308 cm².

Exam: perimeter of a semicircle (22/7)

A semicircular metal sheet has a radius of 7 cm. Find its perimeter. (Take π = 22/7.)

Show the worked solution

Full circumference = 2 × 22/7 × 7 = 44 cm. Half of that curved edge = 22 cm.

The straight edge (the diameter) is 2 × 7 = 14 cm and is part of the perimeter too.

Perimeter = 22 + 14 = 36 cm.

Exam: area of a quarter circle (22/7)

A quarter circle has a radius of 14 cm. Find its area. (Take π = 22/7.)

Show the worked solution

Full circle's area = 22/7 × 14 × 14 = 616 cm².

Quarter circle's area = 616 ÷ 4 = 154 cm².

Exam: square minus quarter circle (22/7)

The figure shows a square of side 14 cm with a quarter circle of radius 14 cm removed from one corner. Find the shaded area that remains. (Take π = 22/7.)

Show the worked solution

Area of square = 14 × 14 = 196 cm².

Area of quarter circle = 22/7 × 14 × 14 ÷ 4 = 616 ÷ 4 = 154 cm².

Shaded area = 196 − 154 = 42 cm².

Exam: find the radius from the area (3.14)

A circle has an area of 78.5 cm². Find its radius. (Take π = 3.14.)

Show the worked solution

r² = area ÷ π = 78.5 ÷ 3.14 = 25.

r = 5 cm, since 5 × 5 = 25.

PSLE challenge: composite figures with two or more circle pieces

PSLE challenge: rectangle with a semicircular end (3.14)

A running track marking is made of a rectangle 20 cm by 14 cm, with a semicircle of radius 7 cm attached along one of its 14 cm sides. Find (a) the total area and (b) the perimeter of the whole figure. (Take π = 3.14.)

Show the worked solution

(a) Rectangle's area = 20 × 14 = 280 cm². Semicircle's area = 3.14 × 7 × 7 ÷ 2 = 76.93 cm².

Total area = 280 + 76.93 = 356.93 cm².

(b) The semicircle replaces one 14 cm side, so that straight side is no longer part of the perimeter — only its curved edge is.

Straight edges left: two lengths of 20 cm and one width of 14 cm = 20 + 20 + 14 = 54 cm.

Curved edge = half the circumference = 3.14 × 7 = 21.98 cm.

Perimeter = 54 + 21.98 = 75.98 cm.

PSLE challenge: square with four quarter circles (22/7)

A square has sides of 28 cm. A quarter circle of radius 14 cm is drawn in each of the square's four corners, and the region covered by all four quarter circles is removed. Find the area that remains. (Take π = 22/7.)

Show the worked solution

Four quarter circles of the same radius join up to make one full circle of radius 14 cm.

Area of square = 28 × 28 = 784 cm². Area of the full circle = 22/7 × 14 × 14 = 616 cm².

Remaining area = 784 − 616 = 168 cm².

Check: 616 is exactly 4 × 154 (one quarter circle), confirming the four pieces really do combine into one full circle. ✓

PSLE challenge: finding a diameter from a circumference (3.14)

A circular pond has a circumference of 62.8 m. A path runs straight across the pond through its centre. Find the length of the path. (Take π = 3.14.)

Show the worked solution

The path is the diameter. Circumference = π × d, so d = circumference ÷ π = 62.8 ÷ 3.14 = 20 m.

How to practise this topic

Circle questions are formula-simple and setup-hard: the arithmetic is short once the right pieces are identified, so the practice that pays off is sketching what's being added and what's being removed before touching a calculator, and noting which value of π the question gives. Because this topic almost always shows up mixed with rectangles, squares and triangles inside a single multi-step question, it's also good practice for PSLE problem-solving more broadly.

What this topic covers

In the MOE Primary Mathematics syllabus, Primary 6 Area & Circumference of Circle includes the concepts below. (P6-M-08 is Test Paper's own topic code, not an MOE reference.)

Before this topic

These topics feed into Area & Circumference of Circle — if this one wobbles, check them first: Primary 5 Area & Perimeter · Primary 5 Decimals.

Where to go next

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