Circle: Circumference, Area, Wheels, Semicircles and Rings
Circle: Circumference, Area, Wheels, Semicircles and Rings
Circle questions use only two formulas, C = 2πr and A = πr², but NTPC turns them into wheel revolutions, semicircle perimeters, rings and circles inside or around squares. Use π = 22/7 unless told otherwise; radii that are multiples of 7 are a strong hint.
1. Formula Box
| Figure | Area | Perimeter |
|---|---|---|
| Circle (radius r) | πr² | 2πr |
| Semicircle | πr²/2 | πr + 2r (the diameter is part of the boundary) |
| Quadrant | πr²/4 | πr/2 + 2r |
| Ring (outer R, inner r) | π(R² − r²) = π(R + r)(R − r) | |
| Circle inscribed in a square of side a | π(a/2)² | |
| Circle around a square of side a | π(a/√2)² = πa²/2 | Twice the inscribed circle's area |
Distance covered in one revolution of a wheel = circumference
Number of revolutions = distance ÷ circumference
| r | 7 | 14 | 21 | 28 | 35 |
|---|---|---|---|---|---|
| Circumference (π = 22/7) | 44 | 88 | 132 | 176 | 220 |
| Area | 154 | 616 | 1,386 | 2,464 | 3,850 |
✗ Perimeter of a semicircle of radius 14 = 44 | ✓ Add the diameter: 44 + 28 = 72
✗ Doubling the radius doubles the area | ✓ Area ∝ r², so it becomes 4 times
हिंदी नोट: वृत्त की परिधि 2πr और क्षेत्रफल πr² होता है। अर्धवृत्त की परिमाप में व्यास भी जुड़ता है: πr + 2r। पहिए के एक चक्कर में तय दूरी = उसकी परिधि।
Exam Pointer: No circle-only question carried a verified NTPC paper label in our check (pages found were SSC, DSSSB and other exams), so these patterns are tagged syllabus-based. They still underpin the verified cylinder, cone and sphere questions of the 3D chapter, so learn them fully.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Circumference and area from each other
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. The circumference of a circle is 88 cm. Find its area.
2 × 22/7 × r = 88 gives r = 14 cm. Area = 22/7 × 196 = 616 cm².
Answer: 616 cm²
EXAMATLAS LEVEL
Q. The area of a circle is 1,386 cm². Find its circumference, and the area of a square with the same perimeter.
πr² = 1,386 gives r² = 441 and r = 21 cm. Circumference = 132 cm. Square side = 33 cm, area = 1,089 cm² (less than the circle's area).
Answer: 132 cm; 1,089 cm²
Pattern 2: Wheel revolutions
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. A wheel has diameter 70 cm. How many revolutions does it make in covering 2.2 km?
Circumference = 22/7 × 70 = 220 cm. Revolutions = 2,20,000/220 = 1,000.
Answer: 1,000
EXAMATLAS LEVEL
Q. The front wheel of a cart has radius 35 cm and the rear wheel 56 cm. How many more revolutions does the front wheel make than the rear wheel over 1.76 km?
Circumferences: 220 cm and 352 cm. Over 1,76,000 cm: 800 and 500 revolutions. Difference = 300.
Answer: 300
Pattern 3: Semicircles and quadrants
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the perimeter and area of a semicircle of radius 14 cm.
Perimeter = 22/7 × 14 + 28 = 44 + 28 = 72 cm. Area = 616/2 = 308 cm².
Answer: 72 cm; 308 cm²
EXAMATLAS LEVEL
Q. Find the area and perimeter of a quadrant of a circle of radius 21 cm.
Area = 1,386/4 = 346.5 cm². Perimeter = (1/4) × 132 + 2 × 21 = 33 + 42 = 75 cm.
Answer: 346.5 cm²; 75 cm
Pattern 4: Rings and circles with squares
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the area of a ring between circles of radii 21 cm and 14 cm.
π(R + r)(R − r) = 22/7 × 35 × 7 = 770 cm².
Answer: 770 cm²
EXAMATLAS LEVEL
Q. A circle is inscribed in a square of side 14 cm and another circle passes through its four corners. Find the ratio of their areas.
Inscribed radius = 7, area 154 cm². Circumscribed radius = half the diagonal = 7√2, area = 22/7 × 98 = 308 cm². Ratio 1 : 2.
Answer: 1 : 2
60-Second Revision
- C = 2πr, A = πr²; radius multiples of 7 suit π = 22/7.
- Semicircle perimeter = πr + 2r; quadrant = πr/2 + 2r.
- Ring = π(R + r)(R − r).
- Revolutions = distance ÷ circumference.
- Circumscribed circle of a square has twice the inscribed circle's area.