Circumference, circle and semicircle areas

Learning outcome

Calculate circular and semicircular areas and boundaries, recover a radius from given data, and distinguish stipulated pi approximations from exact geometric values.

Concepts and assumptions

A radius joins a circle’s centre to its boundary; a diameter passes through the centre and equals twice the radius. Treat stated dimensions as ideal values.

The constant pi is circumference divided by diameter. Therefore circumference = pi × diameter = 2 × pi × radius. Circumference measures boundary length, not surface coverage.

To understand circle area, imagine dividing it into many narrow sectors and alternating them. Their arrangement approaches a rectangle with height r and base half the circumference, pi × r. Its area approaches pi × r².

A semicircle is half a circular region, cut along a diameter. Its area is pi × r²/2 and its curved arc is pi × r. However, its complete perimeter is pi × r + 2r, because the straight diameter also belongs to the boundary.

Using pi = 22/7 is a stipulated approximation, not pi’s exact value. For calculations specifying it below, apply no further rounding; results are approximate geometric values. When calculator pi is specified, retain full precision until rounding each final answer to two decimal places.

Worked examples

Example 1 — A full circle. Find circumference and area for radius 7 m. Use pi = 22/7; no further rounding.

Circumference = 2 × (22/7) × 7 = 44 m. Area = (22/7) × 7² = (22/7) × 49 = 154 m². Both use the stated pi approximation.

Example 2 — Recovering radius. A circle has circumference 66 cm. Find its radius and area. Use pi = 22/7; no further rounding.

66 = 2 × (22/7) × radius. Radius = 66 × 7/44 = 10.5 cm. Area = (22/7) × 10.5² = (22/7) × 110.25 = 346.5 cm² under this convention.

Example 3 — A complete semicircular boundary. A semicircle has diameter 10 cm. Find its area and complete perimeter. Use calculator pi; round final answers to two decimal places.

Radius = 10/2 = 5 cm. Area = pi × 25/2 = 12.5 × pi ≈ 39.27 cm². Perimeter = 5 × pi + 10 ≈ 25.71 cm. The diameter contributes 10 cm; using only the arc would omit this boundary.

Common mistakes

Substituting diameter for radius; confusing circumference with area; halving circumference and forgetting the diameter; treating 22/7 as exact pi; rounding intermediate values.

Practice questions

For each question, use pi = 22/7 with no further rounding.

1. A circle has diameter 28 m. Find circumference and area.
2. A semicircle has radius 21 cm. Find area and complete perimeter.
3. A circle has area 1,386 cm² under the stated pi convention. Find radius and circumference.
4. A wire 90 cm long forms the complete boundary of a semicircle, including its diameter, without overlap. Find radius and enclosed area.

Worked solutions

All four solutions use pi = 22/7; no additional rounding is applied.

1. Radius = 28/2 = 14 m. Circumference = (22/7) × 28 = 88 m. Area = (22/7) × 196 = 616 m².
2. Area = (22/7) × 21²/2 = 693 cm². Arc = (22/7) × 21 = 66 cm; complete perimeter = 66 + 42 = 108 cm.
3. Radius² = 1,386 × 7/22 = 441, so radius = 21 cm. Circumference = 2 × (22/7) × 21 = 132 cm.
4. 90 = (22/7 + 2) × radius = (36/7) × radius. Radius = 90 × 7/36 = 17.5 cm. Area = (22/7) × 17.5²/2 = 481.25 cm².