Areas of Sectors and Segments of Circles Practice Questions | Class 10 Math

Sectors and Segments of a Circle

Ten step-by-step practice problems for Grade 10, from foundational to challenging. Learn the logic, not just the formulas.

Areas of Sectors and Arc Lengths

The big idea: a sector is just a fraction of a circle. With central angle $\theta$, the fraction is $\frac{\theta}{360}$.

• Area of sector $=\dfrac{\theta}{360}\times\pi r^2$
• Arc length $=\dfrac{\theta}{360}\times2\pi r$
• Major sector $=\pi r^2-$ minor sector

1 · Foundational

Find the area of a sector and the length of its arc, if the radius is $21$ cm and the angle is $40^\circ$ (use $\pi=\frac{22}{7}$).

Step 1: Fraction of the circle: $\frac{40}{360}=\frac19$

Step 2: Area: $\frac19\times\frac{22}{7}\times21\times21=\frac19\times22\times63=\frac{1386}{9}=154$

Step 3: Arc: $\frac19\times2\times\frac{22}{7}\times21=\frac19\times132=\frac{44}{3}$

✅ Area $=154\text{ cm}^2$; arc length $=\frac{44}{3}\text{ cm}\approx14.67$ cm

🔍 Verification: $40^\circ$ is $\frac19$ of a circle, so nine such sectors rebuild the whole circle. Full area: $\frac{22}{7}\times441=1386$ and $9\times154=1386$ ✓. Full circumference: $2\times\frac{22}{7}\times21=132$ and $9\times\frac{44}{3}=132$ ✓.

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2 · Foundational+

An arc of length $11$ cm is cut from a circle of radius $14$ cm. Find the angle it subtends at the centre and the area of the sector (use $\pi=\frac{22}{7}$).

Step 1: $\frac{\theta}{360}\times2\times\frac{22}{7}\times14=11\Rightarrow\frac{\theta}{360}\times88=11$

Step 2: $\theta=\frac{11\times360}{88}=45^\circ$

Step 3: Area: $\frac{45}{360}\times\frac{22}{7}\times196=\frac18\times616=77$

✅ $\theta=45^\circ$; area $=77\text{ cm}^2$

3 · Intermediate

A pendulum $30$ cm long swings through $60^\circ$. Find (i) the distance travelled by its tip and (ii) the area swept (use $\pi=3.14$).

Step 1: The tip moves along an arc: $r=30$, fraction $\frac{60}{360}=\frac16$.

Step 2: Arc: $\frac16\times2\times3.14\times30=\frac{188.4}{6}=31.4$

Step 3: Area: $\frac16\times3.14\times900=\frac{2826}{6}=471$

✅ (i) $31.4$ cm; (ii) $471\text{ cm}^2$

4 · Upper-Intermediate

A circular garden has radius $10$ m. A sprinkler waters a sector of angle $144^\circ$. Find the watered area and the dry area (use $\pi=3.14$).

Step 1: Total area: $3.14\times100=314\text{ m}^2$

Step 2: Fraction watered: $\frac{144}{360}=\frac25$

Step 3: Watered: $\frac25\times314=125.6$

Step 4: The dry part is the major sector ($216^\circ$): $314-125.6=188.4$

✅ Watered $=125.6\text{ m}^2$; dry $=188.4\text{ m}^2$

5 · Challenging

The perimeter of a sector of a circle of radius $21$ cm is $64$ cm. Find the angle of the sector and its area (use $\pi=\frac{22}{7}$).

Step 1: Using the standard sector-perimeter formula: perimeter $=2r+$ arc.

Step 2: Arc $=64-42=22$ cm

Step 3: $\frac{\theta}{360}\times132=22\Rightarrow\theta=60^\circ$

Step 4: Area: $\frac{60}{360}\times\frac{22}{7}\times441=\frac{1386}{6}=231$

Cross-check: $\frac12\times r\times\text{arc}=\frac12\times21\times22=231$ ✓

✅ $\theta=60^\circ$; area $=231\text{ cm}^2$

Areas of Segments of a Circle

The big idea: a segment is what is left when you cut the triangle out of the sector. $$\text{Segment}=\frac{\theta}{360}\pi r^2-\text{Area of }\triangle OAB$$

• $\theta=90^\circ$: triangle area $=\frac12r^2$
• $\theta=60^\circ$: equilateral, area $=\frac{\sqrt3}{4}r^2$
• $\theta=120^\circ$: draw $OM\perp AB$; $OM=r\cos60^\circ$, $AM=r\sin60^\circ$

1 · Foundational

A chord of a circle of radius $7$ cm subtends a right angle at the centre. Find the area of the minor segment (use $\pi=\frac{22}{7}$).

Step 1: Sector: $\frac{90}{360}\times\frac{22}{7}\times49=\frac14\times154=38.5$

Step 2: Triangle: the radii are perpendicular, so $\frac12\times7\times7=24.5$

Step 3: Segment $=38.5-24.5=14$

✅ $14\text{ cm}^2$

🔍 Verification: triangle + segment must give the sector: $24.5+14=38.5$ ✓. The segment is also smaller than the sector, $14<38.5$ ✓.

2 · Foundational+

A chord of a circle of radius $12$ cm subtends $60^\circ$ at the centre. Find the area of the minor segment (use $\pi=3.14$, $\sqrt3=1.73$).

Step 1: Sector: $\frac16\times3.14\times144=75.36$

Step 2: $OA=OB$ and $\angle AOB=60^\circ$, so $\triangle OAB$ is equilateral. Using the standard equilateral triangle formula: $\frac{\sqrt3}{4}\times144=36\sqrt3=62.28$

Step 3: Segment $=75.36-62.28=13.08$

✅ $13.08\text{ cm}^2$

3 · Intermediate

A chord of a circle of radius $6$ cm subtends $120^\circ$ at the centre. Find the area of the minor segment (use $\pi=3.14$, $\sqrt3=1.73$).

Step 1: Sector: $\frac13\times3.14\times36=37.68$

Step 2: Draw $OM\perp AB$. By RHS congruence $M$ is the midpoint of $AB$, so $\angle AOM=60^\circ$.

Step 3: $OM=6\cos60^\circ=3$, $AM=6\sin60^\circ=3\sqrt3$, so $AB=6\sqrt3$.

Step 4: Triangle: $\frac12\times6\sqrt3\times3=9\sqrt3=15.57$

Step 5: Segment $=37.68-15.57=22.11$

✅ $22.11\text{ cm}^2$

4 · Upper-Intermediate

A square is inscribed in a circle of radius $14$ cm. Find the area of the circle lying outside the square (use $\pi=\frac{22}{7}$).

Step 1: The vertices split the circle into $4$ equal arcs, so each side subtends $90^\circ$. The regions outside the square are four equal segments.

Step 2: One sector: $\frac14\times\frac{22}{7}\times196=154$

Step 3: One triangle: $\frac12\times14\times14=98$

Step 4: One segment $=56$, so four $=224$

Cross-check: circle $=616$; square (diagonal $28$) $=\frac12\times28^2=392$; $616-392=224$ ✓

✅ $224\text{ cm}^2$

5 · Challenging

A chord of length $14\sqrt3$ cm lies in a circle of radius $14$ cm. Find the areas of the minor and major segments (use $\pi=\frac{22}{7}$, $\sqrt3=1.73$).

Step 1: Drop $OM\perp AB$, so $AM=7\sqrt3$ and $\sin\angle AOM=\frac{7\sqrt3}{14}=\frac{\sqrt3}{2}$. Hence $\angle AOM=60^\circ$ and $\theta=120^\circ$.

Step 2: Sector: $\frac13\times\frac{22}{7}\times196=\frac{616}{3}\approx205.33$

Step 3: $OM=14\cos60^\circ=7$; triangle $=\frac12\times14\sqrt3\times7=49\sqrt3=84.77$

Step 4: Minor segment $=205.33-84.77=120.56$

Step 5: Circle $=616$, so major segment $=616-120.56=495.44$

Cross-check: $120.56+495.44=616$ ✓

✅ Minor $\approx120.56\text{ cm}^2$; major $\approx495.44\text{ cm}^2$

Exam Tips

1. Always check whether you should use $\pi = 3.14$ or $\pi = \frac{22}{7}$ based on the numbers given in the question.
2. For equilateral triangles ($\theta = 60^\circ$), memorizing the formula $\frac{\sqrt{3}}{4}a^2$ saves valuable time.
3. Always cross-check major segment areas by subtracting the minor segment from the total circle area.

All problems are original practice questions, designed to go alongside NCERT Exercise 11.1.