15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
The area of a sector of a circle of radius \(7\) cm with central angle \(60^\circ\) is (Take \(\pi = \dfrac{22}{7}\).)
Why: Area = (θ/360°) × πr² = (1/6) × (22/7) × 49.
\(\dfrac{60}{360} \times \dfrac{22}{7} \times 7^2 = \dfrac16 \times 154 = \dfrac{77}{3}\ \text{cm}^2\).
An arc of a circle of radius \(28\) cm subtends \(45^\circ\) at the centre. Its length is (Take \(\pi = \dfrac{22}{7}\).)
Why: Arc = (θ/360°) × 2πr = (1/8) × 176.
\(\dfrac{45}{360} \times 2 \times \dfrac{22}{7} \times 28 = \dfrac18 \times 176 = 22\) cm.
The area of a sector is \(\dfrac38\) of the area of its circle. The central angle of the sector is
Why: θ/360° = 3/8.
\(\dfrac{\theta}{360} \times \pi r^2 = \dfrac38 \pi r^2 \Rightarrow \theta = \dfrac38 \times 360^\circ = 135^\circ\).
The minute hand of a clock is \(10.5\) cm long. The area swept by it in \(20\) minutes is (Take \(\pi = \dfrac{22}{7}\).)
Why: In 20 minutes the minute hand turns 120°, one-third of the circle.
Angle \(= \dfrac{20}{60} \times 360^\circ = 120^\circ\). Area \(= \dfrac13 \times \dfrac{22}{7} \times 10.5^2 = \dfrac13 \times 346.5 = 115.5\ \text{cm}^2\).
The arc of a sector of central angle \(90^\circ\) is \(22\) cm long. The radius of the circle is (Take \(\pi = \dfrac{22}{7}\).)
Why: (90/360) × 2πr = πr/2 = 22.
\(\dfrac{90}{360} \times 2 \times \dfrac{22}{7} \times r = \dfrac{11r}{7} = 22 \Rightarrow r = 14\) cm.
A chord of a circle of radius \(10\) cm subtends an angle of \(60^\circ\) at the centre. The area of the minor segment is
Why: Segment = sector − triangle; the triangle is equilateral with area (√3/4) × 100.
Sector \(= \dfrac{60}{360}\pi(10)^2 = \dfrac{50\pi}{3}\). The triangle is equilateral (two radii, angle \(60^\circ\)): area \(= \dfrac{\sqrt3}{4} \times 100 = 25\sqrt3\). Segment \(= \dfrac{50\pi}{3} - 25\sqrt3\ \text{cm}^2\).
Two sectors have the same central angle, but their radii are \(3\) cm and \(6\) cm. The ratio of their areas is
Why: With the same angle, sector area is proportional to r².
\(\dfrac{\frac{\theta}{360}\pi(3)^2}{\frac{\theta}{360}\pi(6)^2} = \dfrac{9}{36} = \dfrac14\), i.e. \(1 : 4\).
A sector of a circle of radius \(12\) cm has an arc of length \(4\pi\) cm. Its central angle is
Why: (θ/360) × 24π = 4π gives θ/360 = 1/6.
\(\dfrac{\theta}{360} \times 2\pi(12) = 4\pi \Rightarrow \dfrac{\theta}{360} = \dfrac16 \Rightarrow \theta = 60^\circ\).
A sector cut from a circle of radius \(21\) cm has a central angle of \(60^\circ\). Its perimeter (both radii and the arc) is (Take \(\pi = \dfrac{22}{7}\).)
Why: Perimeter = two radii + arc = 42 + 22.
Arc \(= \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 21 = \dfrac16 \times 132 = 22\) cm. Perimeter \(= 21 + 21 + 22 = 64\) cm. (\(43\) counts only one radius.)
A sector of a circle of radius \(6\) cm has an arc of length \(5\) cm. The area of the sector is
Why: Sector area = ½ × arc length × radius.
Area \(= \dfrac{\theta}{360}\pi r^2 = \dfrac12 r \left(\dfrac{\theta}{360} \cdot 2\pi r\right) = \dfrac12 \times 6 \times 5 = 15\ \text{cm}^2\).
In a circle of radius \(7\) cm, a minor sector has central angle \(90^\circ\). The area of the corresponding major sector is (Take \(\pi = \dfrac{22}{7}\).)
Why: The major sector has angle 360° − 90° = 270°.
\(\dfrac{270}{360} \times \dfrac{22}{7} \times 49 = \dfrac34 \times 154 = 115.5\ \text{cm}^2\).
The minute hand of a clock is \(14\) cm long. The distance travelled by its tip in \(15\) minutes is (Take \(\pi = \dfrac{22}{7}\).)
Why: In 15 minutes the hand turns 90°: a quarter of the circumference.
\(\dfrac14 \times 2 \times \dfrac{22}{7} \times 14 = \dfrac14 \times 88 = 22\) cm.
The area of a circle is \(616\ \text{cm}^2\). The area of a sector of this circle with central angle \(45^\circ\) is
Why: A 45° sector is one-eighth of the circle.
\(\dfrac{45}{360} \times 616 = \dfrac{616}{8} = 77\ \text{cm}^2\).
A chord of a circle of radius \(6\) cm subtends an angle of \(120^\circ\) at the centre. The area of the minor segment is
Why: Sector 12π; the triangle has height 6 cos 60° = 3 on base 6√3, so its area is 9√3.
Sector \(= \dfrac{120}{360}\pi(36) = 12\pi\). The triangle \(OAB\) has height \(6\cos 60^\circ = 3\) on base \(AB = 2 \times 6\sin 60^\circ = 6\sqrt3\), so area \(= \dfrac12 \times 6\sqrt3 \times 3 = 9\sqrt3\). Segment \(= (12\pi - 9\sqrt3)\ \text{cm}^2\).
A sector of a circle of radius \(r\) has the same area as a whole circle of radius \(\dfrac{r}{2}\). The central angle of the sector is
Why: (θ/360)πr² = π(r/2)² = πr²/4 gives θ/360 = 1/4.
\(\dfrac{\theta}{360}\pi r^2 = \dfrac{\pi r^2}{4} \Rightarrow \theta = 90^\circ\).
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
A gardener places a rotating sprinkler at the corner \(O\) of a large lawn. The spray reaches \(14\) m, and the sprinkler is set to turn through an angle of \(90^\circ\), so it waters a sector of radius \(14\) m. (Take \(\pi = \dfrac{22}{7}\).)
In a physics practical, a pendulum with a thread \(21\) cm long swings from one extreme position \(A\) to the other extreme position \(B\), turning through an angle of \(60^\circ\) about the fixed point \(O\). The bob traces the arc \(AB\). (Take \(\pi = \dfrac{22}{7}\).)
This free set is separate from the chapter's question bank. On the Areas Related to Circles chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 37 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.