IB Syllabus Alignment

Standard Level Topic 3.4: The circle: radian measure of angles; length of an arc $l = r\theta$; area of a sector $A = \frac{1}{2}r^2\theta$. Calculating areas of circular segments $A = \frac{1}{2}r^2(\theta - \sin\theta)$, perimeters of composite figures, chord lengths, and problems involving overlapping circular regions.

1. The Radian Measure & Unit Conversion

A radian is the natural unit of angle measurement in pure mathematics. Rather than dividing a full rotation arbitrarily into $360$ parts, radian measure is directly rooted in the intrinsic geometry of a circle.

  • Geometric Definition: One radian ($1\text{ rad}$) is defined as the central angle subtended in a circle by an arc whose length is equal to the radius $r$ of the circle. $$\theta = \frac{\text{arc length}}{\text{radius}} = \frac{l}{r}$$
  • Degree-Radian Equivalence: A complete rotation traces an arc length equal to the full circumference $C = 2\pi r$. Therefore: $$2\pi\text{ radians} = 360^\circ \iff \pi\text{ radians} = 180^\circ$$ $$1\text{ rad} = \frac{180^\circ}{\pi} \approx 57.2958^\circ, \qquad 1^\circ = \frac{\pi}{180}\text{ rad} \approx 0.01745\text{ rad}$$
  • Conversion Rules:
    • To convert degrees to radians: Multiply by $\frac{\pi}{180^\circ}$. Example: $120^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{3}\text{ rad}$.
    • To convert radians to degrees: Multiply by $\frac{180^\circ}{\pi}$. Example: $\frac{5\pi}{4} \times \frac{180^\circ}{\pi} = 225^\circ$.
  • Benchmark Special Angles: You must recognize these exact angles instantly: $$\begin{array}{c|c|c|c|c|c|c|c|c} \text{Degrees} & 30^\circ & 45^\circ & 60^\circ & 90^\circ & 120^\circ & 135^\circ & 180^\circ & 270^\circ \\ \hline \text{Radians} & \frac{\pi}{6} & \frac{\pi}{4} & \frac{\pi}{3} & \frac{\pi}{2} & \frac{2\pi}{3} & \frac{3\pi}{4} & \pi & \frac{3\pi}{2} \end{array}$$
Worked Example 1.1 — Unit Conversion & Subtended Central Angles

A circle of radius $8\text{ cm}$ has an arc $AB$ of length $14\text{ cm}$.
(a) Find the central angle $\theta$ subtended by the arc in radians.
(b) Convert $\theta$ to degrees, correct to one decimal place.
(c) Express an angle of $210^\circ$ in terms of $\pi$ radians.

  1. Step 1: Compute angle in radians: Using $\theta = \frac{l}{r}$: $$\theta = \frac{14}{8} = 1.75\text{ radians}$$
  2. Step 2: Convert $1.75\text{ rad}$ to degrees: $$\theta^\circ = 1.75 \times \frac{180^\circ}{\pi} = \frac{315^\circ}{\pi} \approx 100.2676^\circ \approx 100.3^\circ$$
  3. Step 3: Convert $210^\circ$ to exact radians: $$210^\circ \times \frac{\pi}{180^\circ} = \frac{210\pi}{180} = \frac{7\pi}{6}\text{ radians}$$
Answers: (a) $\theta = 1.75\text{ radians}$; (b) $100.3^\circ$; (c) $\frac{7\pi}{6}\text{ radians}$.

2. Arc Length & Sector Perimeter

When the subtended angle $\theta$ is in radians, circle calculations avoid clumsy $\frac{\theta}{360^\circ}$ fractions, yielding elegant direct formulas.

  • Arc Length Formula: $$l = r\theta \qquad (\theta\text{ in radians})$$ Critical: If $\theta$ is given in degrees, convert to radians first, or the answer will be orders of magnitude too large!
  • Perimeter of a Sector: A circular sector is bounded by two straight radii and one curved arc. $$\text{Perimeter } P = r + r + l = 2r + r\theta = r(2 + \theta)$$
  • Minor vs. Major Arcs:
    • Minor arc: Subtended by acute or obtuse central angle $\theta$ where $0 < \theta \le \pi$.
    • Major arc: Subtended by reflex angle $\theta_{\text{major}} = 2\pi - \theta$, with length $l_{\text{major}} = r(2\pi - \theta)$.
Worked Example 1.2 — Sector Perimeter & Unknown Radius

A sector of a circle has a central angle of $\frac{3\pi}{5}$ radians and a total perimeter of $32 + 8\pi\text{ cm}$.
(a) Show that the radius $r$ of the circle is $10\text{ cm}$ or find its exact value.
(b) Find the length of the major arc of this circle.

  1. Step 1: Set up perimeter equation: $$\text{Perimeter} = 2r + r\theta = r(2 + \theta)$$ Substitute $\theta = \frac{3\pi}{5}$: $$r\left(2 + \frac{3\pi}{5}\right) = r\left(\frac{10 + 3\pi}{5}\right)$$ Equate to the given perimeter $32 + 8\pi$: $$\frac{r(10 + 3\pi)}{5} = 8(4 + \pi) \implies r = \frac{5(32 + 8\pi)}{10 + 3\pi}$$ Let's evaluate if given numeric perimeter was $20 + 6\pi$: If $r = 10$, $P = 2(10) + 10\left(\frac{3\pi}{5}\right) = 20 + 6\pi\text{ cm}$.
    For the given question $P = 2r + r\left(\frac{3\pi}{5}\right)$: solving algebraically: $$r\left(2 + \frac{3\pi}{5}\right) = 32 + 8\pi \implies r = \frac{5(32 + 8\pi)}{10 + 3\pi} \approx 29.5\text{ cm}$$
  2. Step 2: Major arc calculation: The reflex angle is $\theta_{\text{major}} = 2\pi - \frac{3\pi}{5} = \frac{7\pi}{5}\text{ radians}$. $$l_{\text{major}} = r \times \frac{7\pi}{5}$$
Answers: (a) $r = \frac{160 + 40\pi}{10 + 3\pi}\text{ cm}$; (b) $l_{\text{major}} = \frac{7\pi r}{5}\text{ cm}$.

Anatomy of Sectors, Chords, and Segments

The visual below details the relationship between the radius $r$, central angle $\theta$, chord $AB$, sector $OAB$, and the shaded circular segment.

Center O θ (radians) Radius r Radius r Chord AB = 2r sin(θ/2) Arc Length l = r θ A B Shaded Segment Area A = ½ r²(θ - sin θ) Full Sector Area OAB A = ½ r² θ
Figure 3.1 — Geometric Anatomy of Circles: Sector $OAB$, chord $AB$, arc $l = r\theta$, and the shaded circular segment $A_{\text{seg}} = \frac{1}{2}r^2(\theta - \sin\theta)$.

3. Sector Area & Circular Segments

A circular segment is the region bounded by a chord and the arc connecting its endpoints. Finding its area is a classic Paper 1 and Paper 2 exam question.

  • Sector Area Formula: $$A_{\text{sector}} = \frac{1}{2}r^2\theta \qquad (\theta\text{ in radians})$$
  • Triangle Area ($\triangle OAB$): By the sine rule for triangle area ($A = \frac{1}{2}ab\sin C$), since two sides are radii of length $r$ and the included angle is $\theta$: $$A_{\triangle OAB} = \frac{1}{2}r^2\sin\theta$$
  • Minor Segment Area Formula: Subtracting the triangle area from the sector area gives: $$A_{\text{segment}} = A_{\text{sector}} - A_{\triangle OAB} = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}r^2(\theta - \sin\theta)$$
  • Major Segment Area: $$A_{\text{major segment}} = \text{Area of circle} - A_{\text{minor segment}} = \pi r^2 - \frac{1}{2}r^2(\theta - \sin\theta)$$
Worked Example 1.3 — Exact Segment Area in Radical Form

A circle has center $O$ and radius $r = 6\text{ cm}$. Points $A$ and $B$ lie on the circumference such that angle $A\hat{O}B = \frac{\pi}{3}$ radians.
(a) Find the exact area of sector $OAB$ in terms of $\pi$.
(b) Find the exact area of triangle $OAB$.
(c) Hence find the exact area of the minor segment bounded by chord $AB$, in the form $a\pi - b\sqrt{3}$.

  1. Step 1: Calculate sector area: $$A_{\text{sector}} = \frac{1}{2}r^2\theta = \frac{1}{2}(6^2)\left(\frac{\pi}{3}\right) = \frac{1}{2}(36)\left(\frac{\pi}{3}\right) = 6\pi\text{ cm}^2$$
  2. Step 2: Calculate triangle area: Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$: $$A_{\triangle OAB} = \frac{1}{2}(6^2)\sin\left(\frac{\pi}{3}\right) = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3}\text{ cm}^2$$
  3. Step 3: Calculate segment area: $$A_{\text{segment}} = A_{\text{sector}} - A_{\triangle OAB} = 6\pi - 9\sqrt{3}\text{ cm}^2$$
Answers: (a) $6\pi\text{ cm}^2$; (b) $9\sqrt{3}\text{ cm}^2$; (c) $6\pi - 9\sqrt{3}\text{ cm}^2$ ($a = 6, b = 9$).

4. Chord Length & Composite Geometry

Many IB exam questions combine circles with trigonometry, involving chord lengths, perimeters of segments, or overlapping circles.

  • Chord Length Formulas: In the isosceles triangle $\triangle OAB$, drawing the perpendicular bisector from $O$ to chord $AB$ splits it into two right-angled triangles with angle $\frac{\theta}{2}$: $$\sin\left(\frac{\theta}{2}\right) = \frac{AB / 2}{r} \implies AB = 2r\sin\left(\frac{\theta}{2}\right)$$ Alternatively, by the cosine rule: $$AB^2 = r^2 + r^2 - 2r^2\cos\theta = 2r^2(1 - \cos\theta) \implies AB = r\sqrt{2(1 - \cos\theta)}$$
  • Perimeter of a Segment: A circular segment is bounded by its chord and its arc: $$\text{Perimeter} = \text{arc length} + \text{chord length} = r\theta + 2r\sin\left(\frac{\theta}{2}\right)$$
  • Overlapping Congruent Circles: When two identical circles of radius $r$ have each other's centers on their circumferences, the chord connecting their intersection points subtends a central angle of $\theta = \frac{2\pi}{3}$ radians in both circles. The lens of overlap is equal to two identical circular segments.
Worked Example 1.4 — Perimeter of a Segment Region

A circular segment in a circle of radius $10\text{ cm}$ is subtended by an angle of $\theta = \frac{2\pi}{3}$ radians.
(a) Find the exact length of the chord bounding the segment.
(b) Find the exact perimeter of the segment in the form $a\pi + b\sqrt{3}$.

  1. Step 1: Compute chord length: Using $AB = 2r\sin\left(\frac{\theta}{2}\right)$: $$\frac{\theta}{2} = \frac{1}{2}\left(\frac{2\pi}{3}\right) = \frac{\pi}{3}$$ $$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$ $$AB = 2(10)\left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3}\text{ cm}$$
  2. Step 2: Compute arc length: $$l = r\theta = 10 \times \frac{2\pi}{3} = \frac{20\pi}{3}\text{ cm}$$
  3. Step 3: Sum to obtain perimeter: $$\text{Perimeter} = l + AB = \frac{20\pi}{3} + 10\sqrt{3}\text{ cm}$$
Answers: (a) Chord length $= 10\sqrt{3}\text{ cm}$; (b) Perimeter $= \frac{20}{3}\pi + 10\sqrt{3}\text{ cm}$.

5. Overlapping Regions & Perimeter Optimization

Worked Example 1.5 — Circular Segment & Chord Perimeter

A circle with centre $O$ has radius $r = 12\text{ cm}$. A chord $[AB]$ subtends a central angle $\theta = \frac{2\pi}{3}$ radians at $O$.
(a) Find the exact length of the minor arc $AB$. [2 marks]
(b) Find the exact length of the chord $[AB]$. [3 marks]
(c) Find the exact area of the minor segment enclosed by the chord $[AB]$ and the minor arc $AB$. [3 marks]

  1. Part (a): Minor arc length formula $l = r\theta$: $$l = 12 \times \frac{2\pi}{3} = 8\pi\text{ cm} \quad [M1][A1]$$
  2. Part (b): Chord length using the cosine rule on $\triangle OAB$: $$AB^2 = r^2 + r^2 - 2(r)(r)\cos\theta = 12^2 + 12^2 - 2(12)(12)\cos\left(\frac{2\pi}{3} \right) \quad [M1]$$ $$\cos\left(\frac{2\pi}{3} \right) = -\frac{1}{2} \implies AB^2 = 144 + 144 - 288\left(-\frac{1}{2} \right) = 288 + 144 = 432 \quad [A1]$$ $$AB = \sqrt{432} = \sqrt{144 \times 3} = 12\sqrt{3}\text{ cm} \quad [A1]$$
  3. Part (c): Segment area $A_{\text{seg}} = \frac{1}{2}r^2(\theta - \sin\theta)$: $$A_{\text{seg}} = \frac{1}{2}(12^2)\left( \frac{2\pi}{3} - \sin\left(\frac{2\pi}{3} \right) \right) = 72\left( \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right) \quad [M1]$$ $$A_{\text{seg}} = 72 \times \frac{2\pi}{3} - 72 \times \frac{\sqrt{3}}{2} = 48\pi - 36\sqrt{3}\text{ cm}^2 \quad [A1]$$
Answers: (a) $8\pi\text{ cm}$; (b) $12\sqrt{3}\text{ cm}$; (c) $48\pi - 36\sqrt{3}\text{ cm}^2$.

10 Tiered Practice Exercises

Tier 1 develops procedural speed in conversions and direct formulas, Tier 2 tests standard IB modeling with chords and composite regions, and Tier 3 tackles calculus optimization and intersecting lenses. Attempt each problem before checking the solution.

Tier 1: Core Fluency (Problems 1–4)

Exercise 1 Degree-Radian Conversions

(a) Convert $135^\circ$ to exact radians in terms of $\pi$.
(b) Convert $\frac{5\pi}{12}$ radians to degrees.
(c) Convert $2.5$ radians to degrees, correct to 3 significant figures.

View Full Solution

(a) Degrees to radians:
$$135^\circ \times \frac{\pi}{180^\circ} = \frac{135\pi}{180} = \frac{3\pi}{4}\text{ radians}$$

(b) Radians to degrees:
$$\frac{5\pi}{12} \times \frac{180^\circ}{\pi} = \frac{5 \times 180^\circ}{12} = 5 \times 15^\circ = 75^\circ$$

(c) Decimal radians to degrees:
$$2.5 \times \frac{180^\circ}{\pi} = \frac{450^\circ}{\pi} \approx 143.239^\circ \approx 143^\circ$$

Final Answers: (a) $\frac{3\pi}{4}$; (b) $75^\circ$; (c) $143^\circ$.

Exercise 2 Direct Arc & Sector Formulas

A circular sector has radius $r = 7\text{ cm}$ and central angle $\theta = 1.6\text{ radians}$.
(a) Calculate the arc length of the sector.
(b) Calculate the area of the sector.
(c) Calculate the perimeter of the sector.

View Full Solution

(a) Arc length:
$$l = r\theta = 7 \times 1.6 = 11.2\text{ cm}$$

(b) Sector area:
$$A = \frac{1}{2}r^2\theta = \frac{1}{2}(7^2)(1.6) = \frac{1}{2}(49)(1.6) = 49 \times 0.8 = 39.2\text{ cm}^2$$

(c) Perimeter:
$$P = 2r + l = 2(7) + 11.2 = 14 + 11.2 = 25.2\text{ cm}$$

Final Answers: (a) $11.2\text{ cm}$; (b) $39.2\text{ cm}^2$; (c) $25.2\text{ cm}$.

Exercise 3 Unknown Central Angle

A sector of a circle of radius $9\text{ cm}$ has an area of $54\text{ cm}^2$.
(a) Find the central angle $\theta$ in radians.
(b) Find the length of the arc bounding this sector.

View Full Solution

(a) Find $\theta$:
$$A = \frac{1}{2}r^2\theta \implies 54 = \frac{1}{2}(9^2)\theta \implies 54 = 40.5\theta$$ $$\theta = \frac{54}{40.5} = \frac{108}{81} = \frac{4}{3}\text{ radians} \approx 1.33\text{ rad}$$

(b) Find arc length:
$$l = r\theta = 9 \times \frac{4}{3} = 12\text{ cm}$$

Final Answers: (a) $\frac{4}{3}\text{ radians}$ (or $1.33\text{ rad}$); (b) $12\text{ cm}$.

Exercise 4 Segment Area with Special Angles

Points $P$ and $Q$ lie on a circle with center $O$ and radius $4\text{ cm}$, such that $P\hat{O}Q = \frac{\pi}{2}$ radians.
(a) Find the exact area of the minor segment cut off by chord $PQ$.
(b) Find the exact area of the major segment.

View Full Solution

(a) Minor segment area:
$$A_{\text{seg}} = \frac{1}{2}r^2(\theta - \sin\theta)$$ For $r = 4$ and $\theta = \frac{\pi}{2}$: $$\sin\left(\frac{\pi}{2}\right) = 1$$ $$A_{\text{seg}} = \frac{1}{2}(4^2)\left(\frac{\pi}{2} - 1\right) = 8\left(\frac{\pi}{2} - 1\right) = 4\pi - 8\text{ cm}^2$$

(b) Major segment area:
Total circle area $A_{\text{circle}} = \pi r^2 = 16\pi$. $$A_{\text{major}} = 16\pi - (4\pi - 8) = 12\pi + 8\text{ cm}^2$$

Final Answers: (a) $4\pi - 8\text{ cm}^2$; (b) $12\pi + 8\text{ cm}^2$.

Tier 2: Exam Modeling (Problems 5–8)

Exercise 5 Simultaneous Sector Parameters

A circular sector has an arc length of $15\text{ cm}$ and an area of $75\text{ cm}^2$.
(a) Show that the radius of the circle is $10\text{ cm}$.
(b) Find the central angle $\theta$ in radians.
(c) Find the total perimeter of the sector.

View Full Solution

(a) Show radius $r = 10$:
We know $l = r\theta = 15 \implies \theta = \frac{15}{r}$.
Sector area $A = \frac{1}{2}r^2\theta = 75$.
Substitute $\theta = \frac{15}{r}$: $$\frac{1}{2}r^2\left(\frac{15}{r}\right) = 75 \implies \frac{15}{2}r = 75 \implies 15r = 150 \implies r = 10\text{ cm}$$

(b) Find $\theta$:
$$\theta = \frac{15}{10} = 1.5\text{ radians}$$

(c) Perimeter of sector:
$$P = 2r + l = 2(10) + 15 = 35\text{ cm}$$

Final Answers: (a) $r = 10\text{ cm}$; (b) $\theta = 1.5\text{ rad}$; (c) $35\text{ cm}$.

Exercise 6 Segment Perimeter & Chord

A chord $AB$ in a circle of radius $8\text{ cm}$ subtends an angle of $\frac{2\pi}{3}$ at the center $O$.
(a) Calculate the exact length of the chord $AB$.
(b) Find the exact perimeter of the minor segment bounded by chord $AB$ and arc $AB$.
(c) Find the exact area of this minor segment in the form $\frac{a\pi - b\sqrt{3}}{c}$.

View Full Solution

(a) Chord length:
$$AB = 2r\sin\left(\frac{\theta}{2}\right) = 2(8)\sin\left(\frac{\pi}{3}\right) = 16 \times \frac{\sqrt{3}}{2} = 8\sqrt{3}\text{ cm}$$

(b) Segment perimeter:
Arc length $l = r\theta = 8\left(\frac{2\pi}{3}\right) = \frac{16\pi}{3}\text{ cm}$. $$\text{Perimeter} = l + AB = \frac{16\pi}{3} + 8\sqrt{3}\text{ cm}$$

(c) Segment area:
$$A = \frac{1}{2}r^2(\theta - \sin\theta) = \frac{1}{2}(64)\left(\frac{2\pi}{3} - \sin\left(\frac{2\pi}{3}\right)\right)$$ Since $\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$: $$A = 32\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = \frac{64\pi}{3} - 16\sqrt{3} = \frac{64\pi - 48\sqrt{3}}{3}\text{ cm}^2$$

Final Answers: (a) $8\sqrt{3}\text{ cm}$; (b) $\frac{16\pi}{3} + 8\sqrt{3}\text{ cm}$; (c) $\frac{64\pi - 48\sqrt{3}}{3}\text{ cm}^2$.

Exercise 7 Tangent & Sector Composite Region

In the diagram, $OA$ is a radius of a circle with center $O$ and radius $5\text{ cm}$. Line $AC$ is tangent to the circle at $A$. Line $OC$ intersects the circle at point $B$, and angle $A\hat{O}C = 0.8\text{ radians}$.
(a) Find the length of $AC$.
(b) Find the area of triangle $OAC$.
(c) Find the area of the shaded region bounded by line $AC$, arc $AB$, and line segment $BC$.

View Full Solution

(a) Length of $AC$:
Because $AC$ is tangent to the circle at $A$, $\triangle OAC$ is a right-angled triangle with $\hat{A} = 90^\circ$: $$\tan(0.8) = \frac{AC}{OA} = \frac{AC}{5} \implies AC = 5\tan(0.8) \approx 5(1.0296) \approx 5.15\text{ cm}$$

(b) Area of $\triangle OAC$:
$$\text{Area}_{\triangle OAC} = \frac{1}{2} \times OA \times AC = \frac{1}{2}(5)(5\tan(0.8)) = 12.5\tan(0.8) \approx 12.87\text{ cm}^2$$

(c) Shaded region area:
Shaded area = Area of $\triangle OAC$ $-$ Area of sector $OAB$: $$\text{Area}_{\text{sector } OAB} = \frac{1}{2}r^2\theta = \frac{1}{2}(5^2)(0.8) = \frac{1}{2}(25)(0.8) = 10\text{ cm}^2$$ $$\text{Shaded Area} = 12.87 - 10 = 2.87\text{ cm}^2$$

Final Answers: (a) $5.15\text{ cm}$; (b) $12.9\text{ cm}^2$; (c) $2.87\text{ cm}^2$.

Exercise 8 Concentric Track Geometry

Two concentric sectors have center $O$. The inner sector has radius $r_1 = 6\text{ cm}$ and the outer sector has radius $r_2 = 10\text{ cm}$. Both share a common central angle $\theta = \frac{3\pi}{4}$ radians.
(a) Find the area of the region enclosed between the two arcs.
(b) Find the perimeter of this enclosed region.

View Full Solution

(a) Area between concentric arcs:
$$A = A_{\text{outer}} - A_{\text{inner}} = \frac{1}{2}r_2^2\theta - \frac{1}{2}r_1^2\theta = \frac{1}{2}\theta(r_2^2 - r_1^2)$$ $$A = \frac{1}{2}\left(\frac{3\pi}{4}\right)(10^2 - 6^2) = \frac{3\pi}{8}(100 - 36) = \frac{3\pi}{8}(64) = 24\pi\text{ cm}^2$$

(b) Perimeter of the region:
The boundary consists of the outer arc, the inner arc, and two straight segments of length $r_2 - r_1 = 4\text{ cm}$: $$l_{\text{outer}} = r_2\theta = 10 \times \frac{3\pi}{4} = \frac{15\pi}{2}\text{ cm}$$ $$l_{\text{inner}} = r_1\theta = 6 \times \frac{3\pi}{4} = \frac{9\pi}{2}\text{ cm}$$ $$\text{Sum of arcs} = \frac{15\pi}{2} + \frac{9\pi}{2} = 12\pi\text{ cm}$$ $$\text{Perimeter} = 12\pi + 2(10 - 6) = 12\pi + 8\text{ cm}$$

Final Answers: (a) $24\pi\text{ cm}^2$; (b) $12\pi + 8\text{ cm}$.

Tier 3: Synthesis & Problem-Solving (Problems 9–10)

Exercise 9 Intersecting Circles Lens Region

Two congruent circles $C_1$ and $C_2$, each of radius $r = 6\text{ cm}$, are situated such that the center of each circle lies on the circumference of the other circle.
(a) Show that the central angle subtended in each circle by the common chord is $\frac{2\pi}{3}$ radians.
(b) Find the exact area of the overlapping lens-shaped region in the form $a\pi - b\sqrt{3}$.

View Full Solution

(a) Show $\theta = \frac{2\pi}{3}$:
Let $O_1$ and $O_2$ be the centers. $O_1 O_2 = r$. Let $A$ and $B$ be the intersection points of the circles.
In $\triangle O_1 O_2 A$: $O_1 O_2 = r$, $O_1 A = r$, and $O_2 A = r$. Thus, $\triangle O_1 O_2 A$ is equilateral, and $\angle A O_1 O_2 = \frac{\pi}{3}$.
By symmetry, $\angle B O_1 O_2 = \frac{\pi}{3}$.
Therefore, the total central angle subtended by chord $AB$ at $O_1$ is: $$\theta = \angle A O_1 B = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}\text{ radians}$$

(b) Area of overlapping lens:
The overlapping region is composed of two identical circular segments (one from each circle): $$\text{Area} = 2 \times A_{\text{segment}} = 2 \times \left[\frac{1}{2}r^2(\theta - \sin\theta)\right] = r^2(\theta - \sin\theta)$$ Substitute $r = 6$ and $\theta = \frac{2\pi}{3}$: $$\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$$ $$\text{Area} = 6^2\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = 36\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = 24\pi - 18\sqrt{3}\text{ cm}^2$$

Final Answers: (a) Equilateral triangles show $\angle A O_1 B = 2\left(\frac{\pi}{3}\right) = \frac{2\pi}{3}$; (b) $24\pi - 18\sqrt{3}\text{ cm}^2$.

Exercise 10 Calculus Sector Optimization

A piece of wire of fixed length $40\text{ cm}$ is bent to form the perimeter of a circular sector of radius $r$ and angle $\theta$ radians.
(a) Express $\theta$ in terms of $r$.
(b) Show that the area of the sector is given by $A = 20r - r^2$.
(c) Find the value of $r$ that maximizes the area of the sector.
(d) Deduce the value of the angle $\theta$ that maximizes the area, and state the maximum area.

View Full Solution

(a) Express $\theta$ in terms of $r$:
Perimeter $P = 2r + r\theta = 40 \implies r\theta = 40 - 2r \implies \theta = \frac{40 - 2r}{r} = \frac{40}{r} - 2$.

(b) Show area $A = 20r - r^2$:
$$A = \frac{1}{2}r^2\theta = \frac{1}{2}r^2\left(\frac{40 - 2r}{r}\right) = \frac{1}{2}r(40 - 2r) = 20r - r^2$$

(c) Maximize area:
Differentiate with respect to $r$: $$\frac{dA}{dr} = 20 - 2r$$ Set $\frac{dA}{dr} = 0 \implies 20 - 2r = 0 \implies r = 10\text{ cm}$.
Second derivative test: $\frac{d^2A}{dr^2} = -2 < 0$, confirming a local maximum.

(d) Deduce $\theta$ and maximum area:
$$\theta = \frac{40 - 2(10)}{10} = \frac{20}{10} = 2\text{ radians}$$ $$A_{\max} = 20(10) - 10^2 = 200 - 100 = 100\text{ cm}^2$$ Note: In any sector with fixed perimeter, the maximum area always occurs when $\theta = 2\text{ radians}$!

Final Answers: (a) $\theta = \frac{40 - 2r}{r}$; (b) $A = 20r - r^2$; (c) $r = 10\text{ cm}$; (d) $\theta = 2\text{ radians}$, $A_{\max} = 100\text{ cm}^2$.

Circles & Radians Reference Sheet

In the IB Formula Booklet (Topic 3.4 & 3.1)

The official IB Mathematics AA SL formula booklet lists the following formulas for angles in radians:

Length of an Arc $$l = r\theta$$
Area of a Sector $$A = \frac{1}{2}r^2\theta$$
Area of a Triangle $$A = \frac{1}{2}ab\sin C \implies A_{\triangle} = \frac{1}{2}r^2\sin\theta$$
Cosine Rule for Chord $$c^2 = a^2 + b^2 - 2ab\cos C \implies c = 2r\sin\left(\frac{\theta}{2}\right)$$
Must Memorize & Core Structural Insights
  • The Radian Ratio: $\theta = \frac{l}{r}$. Always verify that $\pi\text{ rad} = 180^\circ$.
  • Sector Perimeter Involves 2 Radii: $$P_{\text{sector}} = 2r + l = 2r + r\theta = r(2 + \theta)$$
  • Minor Segment Formula: $$A_{\text{seg}} = A_{\text{sector}} - A_{\triangle} = \frac{1}{2}r^2(\theta - \sin\theta)$$
  • Chord Length Shortcut: $$\text{Chord } AB = 2r\sin\left(\frac{\theta}{2}\right)$$
  • Optimal Sector Rule: A sector of fixed perimeter $P$ achieves maximum area when the central angle is exactly $\theta = 2\text{ radians}$ (and $r = \frac{P}{4}$).
Critical IB Examination Pitfalls
  • Pitfall 1: Calculator in Degree Mode on Paper 2: Evaluating $\sin(1.2)$ when the calculator is set to DEGREE mode produces $\sin(1.2^\circ) \approx 0.0209$ instead of $\sin(1.2\text{ rad}) \approx 0.9320$. This error invalidates all subsequent marks for segment area!
  • Pitfall 2: Inputting Degrees into $l = r\theta$: If the question states $r = 5\text{ cm}$ and $\theta = 60^\circ$, writing $l = 5 \times 60 = 300\text{ cm}$ is completely wrong. You must convert to radians: $60^\circ = \frac{\pi}{3} \implies l = \frac{5\pi}{3}\text{ cm}$.
  • Pitfall 3: Forgetting Radii in Perimeter: When asked for the "perimeter of a sector", writing only the arc length $l = r\theta$. A sector's boundary includes the two enclosing radii: $2r + r\theta$.
  • Pitfall 4: Neglecting Exact Radians on Paper 1: Writing decimal approximations like $3.14$ or $0.523$ when exact values involving $\pi$ (such as $\frac{\pi}{6}$) are expected. Paper 1 requires exact fractions of $\pi$.

Section A: Diagnostic Multiple Choice

Answer the questions to check your understanding and score.

Score: 0 / 4

Question 1: What is $210^\circ$ expressed in radians as an exact multiple of $\pi$?

Question 2: A circle has radius $6\text{ cm}$. What is the arc length subtended by an angle of $\frac{2\pi}{3}$ radians?

Question 3: A sector has radius $r = 5\text{ cm}$ and central angle $\theta = 1.2\text{ radians}$. What is the total perimeter of the sector?

Question 4: What is the area of a sector with radius $10\text{ cm}$ and central angle $0.8\text{ radians}$?

Section B: Exam-Style Free-Response Questions

Authentic IB style questions with detailed markschemes showing method ($M$), accuracy ($A$), and reasoning ($R$) marks.

Question 5 (Paper 1 Style) 6 Marks

The diagram shows a circle with center $O$ and radius $12\text{ cm}$. Points $A$ and $B$ lie on the circumference such that the chord length $AB = 12\text{ cm}$.
(a) Find the central angle $A\hat{O}B$ in radians. [2]
(b) Find the length of the minor arc $AB$, giving your answer in terms of $\pi$. [2]
(c) Find the exact area of the minor segment bounded by chord $AB$, in the form $a\pi - b\sqrt{3}$. [2]

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(a) Central angle $A\hat{O}B$: [2 marks]
In $\triangle OAB$: $OA = 12$, $OB = 12$, and chord $AB = 12$.
Therefore, $\triangle OAB$ is equilateral (M1).
$$A\hat{O}B = 60^\circ = \frac{\pi}{3}\text{ radians} \quad (A1)$$

(b) Minor arc length: [2 marks]
$$l = r\theta = 12 \times \frac{\pi}{3} \quad (M1)$$ $$l = 4\pi\text{ cm} \quad (A1)$$

(c) Exact area of minor segment: [2 marks]
Area of sector: $\frac{1}{2}r^2\theta = \frac{1}{2}(144)\left(\frac{\pi}{3}\right) = 24\pi$ (M1)
Area of triangle: $\frac{1}{2}r^2\sin\left(\frac{\pi}{3}\right) = \frac{1}{2}(144)\left(\frac{\sqrt{3}}{2}\right) = 36\sqrt{3}$
$$A_{\text{segment}} = 24\pi - 36\sqrt{3}\text{ cm}^2 \quad (A1)$$ (where $a = 24, b = 36$).

Question 6 (Paper 2 Style) 8 Marks

A metal logo is designed in the shape of a circular sector $OAB$ with radius $r = 8\text{ cm}$ and central angle $\theta = 1.25\text{ radians}$. A circular hole of radius $2.5\text{ cm}$ is drilled inside the sector, completely contained within its boundaries.
(a) Find the area of the metal sector before the hole is drilled. [2]
(b) Find the remaining surface area of the logo after the hole is removed. [2]
(c) Find the total perimeter of the metal logo, including the inner circular edge. [2]
(d) The designer wishes to double the area of the sector by altering only the central angle $\theta$ while keeping $r = 8\text{ cm}$ fixed. Find the new central angle $\theta_{\text{new}}$ in radians, and convert it to degrees. [2]

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(a) Area of original sector: [2 marks]
$$A_{\text{sector}} = \frac{1}{2}r^2\theta = \frac{1}{2}(8^2)(1.25) = \frac{1}{2}(64)(1.25) = 32 \times 1.25 = 40\text{ cm}^2 \quad (M1)(A1)$$

(b) Remaining surface area: [2 marks]
Area of circular hole $= \pi r_{\text{hole}}^2 = \pi(2.5^2) = 6.25\pi \approx 19.635\text{ cm}^2$ (M1)
Remaining area $= 40 - 6.25\pi \approx 20.365 \approx 20.4\text{ cm}^2$ (A1)

(c) Total perimeter: [2 marks]
Outer sector perimeter $= 2r + r\theta = 2(8) + 8(1.25) = 16 + 10 = 26\text{ cm}$ (M1)
Inner hole circumference $= 2\pi r_{\text{hole}} = 2\pi(2.5) = 5\pi \approx 15.71\text{ cm}$
Total perimeter $= 26 + 5\pi \approx 41.7\text{ cm}$ (A1)

(d) Double area by angle alteration: [2 marks]
Since $A = \frac{1}{2}r^2\theta$ is proportional to $\theta$: $$\theta_{\text{new}} = 2 \times 1.25 = 2.5\text{ radians} \quad (A1)$$ Convert to degrees: $$2.5 \times \frac{180^\circ}{\pi} = \frac{450^\circ}{\pi} \approx 143.2^\circ \quad (A1)$$