Lesson 68 of 100 | Quantitative Aptitude / Plane Mensuration / समतल क्षेत्रमिति
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.
- A circle has diameter 28 m. Find circumference and area.
- A semicircle has radius 21 cm. Find area and complete perimeter.
- A circle has area 1,386 cm² under the stated pi convention. Find radius and circumference.
- 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.
- Radius = 28/2 = 14 m. Circumference = (22/7) × 28 = 88 m. Area = (22/7) × 196 = 616 m².
- Area = (22/7) × 21²/2 = 693 cm². Arc = (22/7) × 21 = 66 cm; complete perimeter = 66 + 42 = 108 cm.
- Radius² = 1,386 × 7/22 = 441, so radius = 21 cm. Circumference = 2 × (22/7) × 21 = 132 cm.
- 90 = (22/7 + 2) × radius = (36/7) × radius. Radius = 90 × 7/36 = 17.5 cm. Area = (22/7) × 17.5²/2 = 481.25 cm².