ExamShortcut
🔍Log in

Mensuration (2D)

🔒 Log in to track
high importance~2 Q in Tier 123 formulas⚡ 12 shortcuts5 subtopics
OverviewLearn (5)Formulas (23)Examples (4)Practice (62)
Subtopic 3 of 5·← Areas of quadrilaterals·Regular polygons and inscribed figures →

Circles, sectors and rings

🔒 Log in to track

For a circle of radius rr: area πr2\pi r^2, circumference 2πr2\pi r (use π=227\pi=\frac{22}{7} unless told otherwise).

  • Arc length: θ360×2πr\frac{\theta}{360}\times2\pi r; sector area: θ360×πr2\frac{\theta}{360}\times\pi r^2 — the sector is just that fraction of the full circle.
  • Segment: sector area minus the triangle formed by the two radii and the chord.
  • Ring (annulus): π(R2−r2)\pi(R^2-r^2) — a path of width ww around a circle has outer radius r+wr+w.
  • Rolling wheel: distance == number of revolutions ×\times circumference =nπd=n\pi d.
📖 Notes🧩 4 question types⚡ Formulas🎯 12 practice Q

Detailed notes

The π machinery (use π=227\pi = \frac{22}{7} unless told otherwise)

C=2πr,K=πr2C = 2\pi r,\qquad K = \pi r^2 Reverse routes appear constantly: from circumference r=C2πr = \frac{C}{2\pi}; from area r=Kπr = \sqrt{\frac{K}{\pi}}. Keep the favourite radii at hand: r=7r = 7 → C=44C = 44, K=154K = 154; r=14r = 14 → C=88C = 88, K=616K = 616; r=21r = 21 → C=132C = 132, K=1386K = 1386.

Sector and arc

For angle θ\theta at the centre: sector area=θ360×πr2,arc length=θ360×2πr\text{sector area} = \frac{\theta}{360} \times \pi r^2,\qquad \text{arc length} = \frac{\theta}{360} \times 2\pi r Sector area is just the full area scaled by θ360\frac{\theta}{360} — the same fraction scales the circumference into the arc. Semi-circle, quarter-circle, 60∘60^\circ slice: halve, quarter, sixth.

Ring (annulus) and wheel

  • Ring: K=π(R2−r2)=π(R+r)(R−r)K = \pi(R^2 - r^2) = \pi(R+r)(R-r) — factorise before multiplying.
  • Wheel: one revolution covers one circumference. Revolutions =distancecircumference=dπD= \frac{\text{distance}}{\text{circumference}} = \frac{d}{\pi D} where DD is the wheel's diameter. Convert units first (cm vs m) — the classic trap.

Circles inside and outside squares

  • Circle inscribed in a square of side aa: diameter =a= a, so r=a2r = \frac{a}{2}.
  • Circle circumscribed about the square: diameter == diagonal =a2= a\sqrt2, so r=a2r = \frac{a}{\sqrt2} and K=π⋅a22=πa22K = \pi \cdot \frac{a^2}{2} = \frac{\pi a^2}{2} — exactly double the inscribed circle's area (πa24\frac{\pi a^2}{4}). That 2 : 1 ratio is asked as often as the areas themselves.

Composite shapes and costs

Compound figures are assembled from these pieces: a window of rectangle plus semicircle has perimeter == (two widths + length) + half-circumference, and area == rectangle + 12πr2\frac12\pi r^2 — always add PERIMETER pieces for fencing/border costs and AREA pieces for carpeting/painting costs. Fencing a circular park: cost =2πr×= 2\pi r \times rate per metre. Ploughing/circular path questions use the ring area with a rate: cost =π(R2−r2)×= \pi(R^2 - r^2) \times rate per sq m. Keep the pieces separate until the last step — merging them early is where sign and factor errors happen. A worked minute-example: a circular park of radius 35 m with a 7 m wide path outside → ring area =227×42×28=3696= \frac{22}{7} \times 42 \times 28 = 3696 sq m; at Rs 10 per sq m the path costs Rs 36,960.

Quick revision

  • C=2πrC = 2\pi r, K=πr2K = \pi r^2; (7,44,154)(7, 44, 154), (14,88,616)(14, 88, 616), (21,132,1386)(21, 132, 1386).
  • Sector/arc: multiply by θ360\frac{\theta}{360}.
  • Ring: π(R+r)(R−r)\pi(R+r)(R-r); wheel revolutions =distanceπD= \frac{\text{distance}}{\pi D}.
  • Inscribed in square: r=a/2r = a/2; circumscribed: r=a/2r = a/\sqrt2 (areas in ratio 1:21:2).

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Area and circumference, forward and reversevery common2 practice Q
How to spot it:

Radius/diameter given — or circumference/area given with the other quantity asked.

C=2πr,K=πr2C = 2\pi r,\quad K = \pi r^2
  1. Forward: substitute rr (halve the diameter first — the classic slip).
  2. Reverse: r=C2πr = \frac{C}{2\pi} from circumference, r=K/πr = \sqrt{K/\pi} from area.
  3. Use the memorised set: (7→44, 154), (14→88, 616), (21→132, 1386).

Why: both formulas are functions of rr alone, so every question is a one-step conversion through rr.

Example: The circumference of a circle is 132 cm. Its area is:

r=132×744=21r = \frac{132 \times 7}{44} = 21 cm; K=227×441=1386K = \frac{22}{7} \times 441 = 1386 sq cm.

Type 2: Sector area and arc lengthvery common2 practice Q
How to spot it:

An angle at the centre with the radius given — sector area or arc length asked (or reversed).

θ360πr2,θ3602πr\frac{\theta}{360}\pi r^2,\quad \frac{\theta}{360} 2\pi r
  1. Reduce the fraction θ360\frac{\theta}{360} first: 90°90° → 14\frac14, 60°60° → 16\frac16, 72°72° → 15\frac15.
  2. Sector area: the fraction × full area; arc: the fraction × full circumference.
  3. Reverse: θ=arc2πr×360°\theta = \frac{\text{arc}}{2\pi r} \times 360°.

Why: the sector is a proportional slice of the circle — one fraction does everything.

Example: Find the arc length of a sector of angle 72∘72^\circ in a circle of radius 3535 cm.

72360=15\frac{72}{360} = \frac15; arc =15×2×227×35=15×220=44= \frac15 \times 2 \times \frac{22}{7} \times 35 = \frac15 \times 220 = 44 cm.

Type 3: Ring (annulus) area and wheel revolutionscommon2 practice Q
How to spot it:

Two concentric circles (path/border) — area of the ring; or a wheel covering a distance — number of revolutions.

K=π(R2−r2)=π(R+r)(R−r);revs=dπDK = \pi(R^2 - r^2) = \pi(R+r)(R-r);\quad \text{revs} = \frac{d}{\pi D}
  1. Ring: factorise R2−r2=(R+r)(R−r)R^2 - r^2 = (R+r)(R-r) before multiplying by π\pi.
  2. Wheel: one revolution = one circumference πD\pi D; convert distance and diameter to the SAME units.
  3. Revolutions =distanceπD= \frac{\text{distance}}{\pi D}.

Why: the ring is the difference of two circle areas; a rolling wheel lays down its circumference per turn.

Example: A wheel of diameter 70 cm rolls a distance of 231 m. The number of revolutions it makes is:

C=227×0.7=2.2C = \frac{22}{7} \times 0.7 = 2.2 m; revs =2312.2=105= \frac{231}{2.2} = 105.

Type 4: Circles inscribed in / circumscribed about squarescommon2 practice Q
How to spot it:

A circle inside a square (touching sides) or around it (through corners); radius or area asked.

rin=a2,rout=a2r_{\text{in}} = \frac{a}{2},\quad r_{\text{out}} = \frac{a}{\sqrt{2}}
  1. Inscribed: diameter = side of square → r=a2r = \frac{a}{2}, K=πa24K = \frac{\pi a^2}{4}.
  2. Circumscribed: diameter = diagonal → r=a22=a2r = \frac{a\sqrt2}{2} = \frac{a}{\sqrt2}, K=πa22K = \frac{\pi a^2}{2}.
  3. Ratio fact: circumscribed area : inscribed area =2:1= 2 : 1.

Why: the inscribed diameter spans opposite sides; the circumscribed diameter spans opposite corners (the diagonal).

Example: A circle is inscribed in a square of side 14 cm. The area of the circle is:

r=7r = 7 cm; K=227×49=154K = \frac{22}{7} \times 49 = 154 sq cm.

Formulas

Circle
K=πr2,C=2πr=πdK=\pi r^2,\quad C=2\pi r=\pi d
Sector
K=θ360πr2,ℓ=θ360 2πrK=\frac{\theta}{360}\pi r^2,\quad \ell=\frac{\theta}{360}\,2\pi r
Segment
K=θ360πr2−12r2sin⁡θK=\frac{\theta}{360}\pi r^2-\frac12 r^2\sin\theta
Ring / path
K=π(R2−r2)=π(R+r)(R−r)K=\pi(R^2-r^2)=\pi(R+r)(R-r)
Rolling wheel
distance=nπd\text{distance}=n\pi d

Shortcut tricks

⚡ Sector as a fraction of the circle

θ=60∘→16\theta=60^\circ\to\frac16, 90∘→1490^\circ\to\frac14, 45∘→1845^\circ\to\frac18, 120∘→13120^\circ\to\frac13 of the circle. No separate formula needed.

Example: Find the area of a sector of angle 45∘45^\circ in a circle of radius 14 cm.

18×227×142=18×616=77\frac18\times\frac{22}{7}\times14^2=\frac18\times616=77 sq cm.

⚡ Ring: difference of squares

R2−r2=(R+r)(R−r)R^2-r^2=(R+r)(R-r): with a path of width ww, R−r=wR-r=w and R+r=2r+wR+r=2r+w — often a one-line answer.

Example: A path 7 m wide runs around a circular park of radius 14 m. Find the area of the path.

227×(212−142)=227×245=770\frac{22}{7}\times(21^2-14^2)=\frac{22}{7}\times245=770 sq m.

⚡ Wheel revolutions = distance / circumference

Convert the diameter to metres first: d=70d=70 cm =0.7=0.7 m gives a 2.2 m circumference with π=227\pi=\frac{22}{7}.

Example: A wheel of diameter 70 cm makes 300 revolutions. How far does it travel?

300×227×0.7=300×2.2=660300\times\frac{22}{7}\times0.7=300\times2.2=660 m.

Where students lose marks

  • Forgetting to add the path width when a path runs around a circle (R=r+wR=r+w).

  • Sector area taken as a fraction of the circumference, or arc length as a fraction of the area.

  • Mixing units: radius in cm vs answer demanded in m.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.