MathsCBSE Class 10 · Areas Related to Circles

Area of a Sector of a Circle for Class 10

Q: What is the difference between a sector and a segment?

A sector is bounded by two radii and an arc, like a pizza slice. A segment is bounded by a chord and an arc. A segment's area is found by taking the sector and subtracting the triangle formed by the two radii and the chord.

Q: Is the perimeter of a sector the same as the arc length?

No. The perimeter includes the two radii as well as the arc, so it is the arc length plus twice the radius. Only the arc is the curved part.

Q: How do I find the area of a major sector?

Either use the angle 360 degrees minus the minor angle in the sector formula, or subtract the minor sector's area from the whole circle's area. Doing both is a good check.

Q: Can I use pi as 3.14 instead of 22/7?

Use 22/7 unless the question tells you to use 3.14. Questions with radii like 7, 14 or 21 are designed so that 22/7 cancels neatly.

Q: Is the section on areas of combinations of plane figures still in the syllabus?

That separate NCERT section was removed in the rationalisation and is not in the 2026-27 syllabus. Shaded-region questions built from sectors and segments with squares, rectangles and triangles are still asked.

Updated 13 October 2026 · Fact-checked

A sector is the slice of a circle between two radii and the arc joining them. Its area is the fraction theta over 360 of the full circle's area, pi r squared. Board questions ask you to find a sector's area or perimeter, work backwards to the angle or radius, and handle both minor and major sectors.

Understand Area of a Sector of a Circle

A sector is the region of a circle enclosed by two radii $O A$ and $O B$ and the arc $A B$ between them. Think of one slice of a round pizza cut from the centre. The angle $∠ A O B = θ$ at the centre is the central angle, and it decides how big the slice is.

Why the formula works. A full turn at the centre is $36 0^{\circ}$ and covers the whole area $π r^{2}$ . If you double the angle, you double the slice, so area is in direct proportion to the angle. A central angle $θ$ therefore gives the fraction $\frac{θ}{36 0 ^{\circ}}$ of the circle: $A = \frac{θ}{36 0 ^{\circ}} \times π r^{2}$ . A $9 0^{\circ}$ sector is a quarter (a quadrant), $6 0^{\circ}$ is one sixth, $12 0^{\circ}$ is one third.

Perimeter of a sector. The boundary of the slice has three parts: two straight radii and one curved arc. So $P = 2 r + l$ , where $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$ is the arc length. Forgetting the $2 r$ is the most common slip. When the arc is known but not the angle, the shortcut $A = \frac{1}{2} l r$ gives the area directly.

Minor and major sectors. Two radii split the circle into two sectors. The smaller one, with $θ < 18 0^{\circ}$ , is the minor sector; the rest, with angle $36 0^{\circ} - θ$ , is the major sector. You can find the major sector using its own angle, or by subtracting: major $= π r^{2} -$ minor. The two areas always add up to the whole circle, which is a free check.

Where it leads. The sector is the building block of the chapter. A segment is a sector minus a triangle, and shaded-region questions add and subtract sectors, squares and triangles. The formula holds for any central angle measured in degrees, as long as $r$ is the radius (halve a diameter first) and all lengths are in the same unit.

Area of a Sector of a Circle: formulas to remember

Area of a sector: $A = \frac{θ}{36 0 ^{\circ}} \times π r^{2}$; $θ$ is the central angle in degrees, $r$ the radius.
Length of the arc of a sector: $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$; Needed for the perimeter of a sector.
Perimeter of a sector: $P = 2 r + l$; Two radii plus the arc. Never just the arc.
Sector area from arc length: $A = \frac{1}{2} l r$; Use when arc or perimeter is given but the angle is not.
Area of a quadrant: $A = \frac{1}{4} π r^{2}$; A $9 0^{\circ}$ sector.
Area of a major sector: $A_{major} = \frac{36 0 ^{\circ} - θ}{36 0 ^{\circ}} \times π r^{2} = π r^{2} - A_{minor}$; Use one form, check with the other.
Perimeter of a major sector: $P_{major} = 2 r + (2 π r - l)$; Uses the major arc, not the minor arc.
Finding the angle from the area: $θ = \frac{A \times 36 0 ^{\circ}}{π r ^{2}}$; For reverse questions.

How to solve Area of a Sector of a Circle questions

Every sector question is the fraction idea plus careful reading. This method works for direct, reverse and major-sector questions.

1Read the radius. If a diameter is given, halve it. Convert all lengths to one unit.
2Identify the central angle $θ$ , and decide whether the question wants the minor sector ( $θ$ ) or the major sector ( $36 0^{\circ} - θ$ ).
3Find the full area $π r^{2}$ and the full circumference $2 π r$ first; with radii like 7, 14, 21 they are whole numbers.
4Multiply by the fraction $\frac{θ}{36 0 ^{\circ}}$ to get the sector area or arc length.
5For perimeter, add both radii: $P = 2 r + l$ .
6For reverse questions, write the formula as an equation in the unknown ( $θ$ or $r$ ) and solve; if the perimeter is given, use $l = P - 2 r$ and $A = \frac{1}{2} l r$ .
7Check: minor plus major should equal $π r^{2}$ , or area divided by arc should equal $\frac{r}{2}$ . Write units ( $cm^{2}$ for area, cm for perimeter).

Area of a Sector of a Circle in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Area of a sector

Suppose your family's monthly spending is Rs 30,000, and Rs 10,000 of it goes on food. For a school project you draw the budget as a pie chart on a circular card of radius 21 cm, and you want to stick a coloured ribbon all round the edge of the food slice.

Find the area of the food sector and the length of ribbon needed to go round its full boundary. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Fraction for food $= \frac{10000}{30000} = \frac{1}{3}$ , so central angle $= \frac{1}{3} \times 36 0^{\circ} = 12 0^{\circ}$ .
Full area $= \frac{22}{7} \times 21 \times 21 = 1386$ cm $^{2}$ .
Food sector area $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times 1386 = 462$ cm $^{2}$ .
Full circumference $= 2 \times \frac{22}{7} \times 21 = 132$ cm, so arc $= \frac{1}{3} \times 132 = 44$ cm.
Ribbon $= 2 r + l = 42 + 44 = 86$ cm.

Answer: Area of the food sector is 462 cm² and the ribbon needed is 86 cm.

Memory hook: Pie chart slice = sector: share of money = share of 360°.

Area of a sector

Imagine you are making a circular Diwali rangoli of radius 14 cm. You divide it from the centre into 8 equal slices and fill alternate slices with red powder. One gram of powder covers 4 cm² of floor.

Find the total area filled with red powder and how many grams of red powder you need. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Each slice has central angle $\frac{36 0 ^{\circ}}{8} = 4 5^{\circ}$ .
Alternate slices means 4 red slices, total angle $4 \times 4 5^{\circ} = 18 0^{\circ}$ .
Full area $= \frac{22}{7} \times 14 \times 14 = 616$ cm $^{2}$ .
Red area $= \frac{18 0 ^{\circ}}{36 0 ^{\circ}} \times 616 = 308$ cm $^{2}$ .
Powder needed $= \frac{308}{4} = 77$ g.

Answer: Red area is 308 cm² and 77 g of red powder is needed.

Memory hook: Rangoli slices: add the angles first, then take one fraction of the circle.

Tricks to remember Area of a Sector of a Circle

Short memory aids. Say each one aloud once, then use it on a question the same day.

Angle over 360, times the whole
Say it as: part of the turn times the whole thing. Whole area $π r^{2}$ for area, whole circumference $2 π r$ for arc.
Use it when: Stops you mixing area and arc formulas.
Pizza crust plus two cuts
The edge of a pizza slice is the crust (arc) plus the two cut edges (radii): $P = l + 2 r$ .
Use it when: Every perimeter-of-sector question.
Sector is a triangle in disguise
Unroll the arc as a base and the radius as height: $A = \frac{1}{2} \times l \times r$ , just like $\frac{1}{2} \times$ base $\times$ height.
Use it when: When arc length or perimeter is given instead of the angle.
Minor plus major is whole
Minor sector + major sector $= π r^{2}$ ; minor arc + major arc $= 2 π r$ .
Use it when: Quick check, and a fast way to find major sectors.

Area of a Sector of a Circle: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Area and perimeter of a 120° sector · 2 marks · Very short answerBoard level

Find the area and the perimeter of a sector of a circle of radius 10.5 cm whose central angle is $12 0^{\circ}$ . (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Area $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times \frac{22}{7} \times 10.5 \times 10.5 = \frac{1}{3} \times 346.5 = 115.5$ cm $^{2}$ . (1 mark)
Arc $l = \frac{1}{3} \times 2 \times \frac{22}{7} \times 10.5 = \frac{1}{3} \times 66 = 22$ cm. (½ mark)
Perimeter $= 2 r + l = 21 + 22 = 43$ cm. (½ mark)

Answer: Area = 115.5 cm², perimeter = 43 cm.

Example 2: Radius and angle from arc and area · 3 marks · Short answerBoard level

The arc of a sector is 44 cm long and the area of the sector is 462 cm $^{2}$ . Find the radius of the circle and the central angle of the sector. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Use $A = \frac{1}{2} l r$ because the angle is unknown. (½ mark)
$462 = \frac{1}{2} \times 44 \times r = 22 r$ , so $r = 21$ cm. (1 mark)
Arc: $44 = \frac{θ}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 21 = \frac{θ}{36 0 ^{\circ}} \times 132$ . (½ mark)
$θ = \frac{44}{132} \times 36 0^{\circ} = 12 0^{\circ}$ . (½ mark)
Check: $\frac{120}{360} \times \frac{22}{7} \times 441 = \frac{1}{3} \times 1386 = 462$ cm $^{2}$ . Correct. (½ mark)

Answer: Radius = 21 cm, central angle = 120°.

Example 3: Radius from sector area, then major sector · 3 marks · Short answerHarder board level

A sector of a circle with central angle $4 5^{\circ}$ has area 77 cm $^{2}$ . Find (i) the radius of the circle, (ii) the perimeter of the sector, (iii) the area of the major sector. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

$\frac{4 5 ^{\circ}}{36 0 ^{\circ}} \times \frac{22}{7} \times r^{2} = 77$ , so $\frac{1}{8} \times \frac{22}{7} r^{2} = 77$ . (½ mark)
$r^{2} = \frac{77 \times 8 \times 7}{22} = 196$ , so $r = 14$ cm. (½ mark)
Arc $= \frac{1}{8} \times 2 \times \frac{22}{7} \times 14 = \frac{88}{8} = 11$ cm. (½ mark)
Perimeter $= 2 \times 14 + 11 = 39$ cm. (½ mark)
Circle area $= \frac{22}{7} \times 196 = 616$ cm $^{2}$ . (½ mark)
Major sector $= 616 - 77 = 539$ cm $^{2}$ (check: $\frac{315}{360} \times 616 = \frac{7}{8} \times 616 = 539$ ). (½ mark)

Answer: (i) 14 cm (ii) 39 cm (iii) 539 cm².

Example 4: Minor and major sectors in a given ratio · 5 marks · Long answerHarder board level

Two radii $O A$ and $O B$ divide a circle of radius 21 cm into two sectors such that the area of the major sector is 5 times the area of the minor sector. Find (i) the central angle of the minor sector, (ii) the areas of both sectors, (iii) the perimeters of both sectors, (iv) how much larger the major sector is than the minor one. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Let minor angle $= θ$ ; major angle $= 36 0^{\circ} - θ$ . Areas are proportional to angles, so $36 0^{\circ} - θ = 5 θ$ . (1 mark)
$6 θ = 36 0^{\circ}$ , so $θ = 6 0^{\circ}$ . (½ mark)
Circle area $= \frac{22}{7} \times 441 = 1386$ cm $^{2}$ ; minor sector $= \frac{1}{6} \times 1386 = 231$ cm $^{2}$ . (½ mark)
Major sector $= 5 \times 231 = 1155$ cm $^{2}$ ; check $231 + 1155 = 1386$ . (½ mark)
Circumference $= 132$ cm; minor arc $= \frac{1}{6} \times 132 = 22$ cm, major arc $= 132 - 22 = 110$ cm. (1 mark)
Perimeter of minor sector $= 42 + 22 = 64$ cm. (½ mark)
Perimeter of major sector $= 42 + 110 = 152$ cm. (½ mark)
Difference in area $= 1155 - 231 = 924$ cm $^{2}$ . (½ mark)

Answer: (i) 60° (ii) minor 231 cm², major 1155 cm² (iii) minor 64 cm, major 152 cm (iv) 924 cm².

Example 5: Case study: a car windscreen wiper · 4 marks · Case-basedHOTS / competency

Suppose a car's windscreen wiper has its arm pivoted at a point $O$ . The arm $O P$ is 42 cm long. The rubber blade is fixed along the outer 21 cm of the arm only, from the point $Q$ on $O P$ (with $O Q = 21$ cm) to the tip $P$ . In one sweep the arm turns through $12 0^{\circ}$ about $O$ . The glass actually cleaned is the region between the arc traced by $Q$ and the arc traced by $P$ , bounded at the two ends by the starting and final positions of the blade. (Take $π = \frac{22}{7}$ )
(i) Find the area swept by the whole arm $O P$ in one sweep. (1 mark)
(ii) Find the area of glass cleaned by the blade in one sweep. (1 mark)
(iii) Find the perimeter of the cleaned region, and what fraction of the area swept by the arm is actually cleaned. Explain why this fraction does not depend on the angle of sweep. (2 marks)

Try it first, then show the solutionSolution

Outer radius $R = O P = 42$ cm, inner radius $r = O Q = 42 - 21 = 21$ cm, angle $θ = 12 0^{\circ}$ , so the fraction is $\frac{12 0 ^{\circ}}{36 0 ^{\circ}} = \frac{1}{3}$ .
(i) Area swept by arm $= \frac{1}{3} \times \frac{22}{7} \times 42 \times 42 = \frac{1}{3} \times 5544 = 1848$ cm $^{2}$ . (1 mark)
(ii) Inner sector (part of arm without blade) $= \frac{1}{3} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{3} \times 1386 = 462$ cm $^{2}$ , so cleaned area $= 1848 - 462 = 1386$ cm $^{2}$ . (1 mark)
(iii) Outer arc $= \frac{1}{3} \times 2 \times \frac{22}{7} \times 42 = 88$ cm, inner arc $= \frac{1}{3} \times 2 \times \frac{22}{7} \times 21 = 44$ cm, two straight ends of 21 cm each, so perimeter $= 88 + 44 + 21 + 21 = 174$ cm. (1 mark)
Fraction cleaned $= \frac{1386}{1848} = \frac{3}{4}$ . Both areas carry the same factor $\frac{θ}{36 0 ^{\circ}} \times π$ , which cancels, leaving $\frac{R ^{2} - r ^{2}}{R ^{2}} = \frac{1764 - 441}{1764} = \frac{3}{4}$ for any angle. (1 mark)

Answer: (i) 1848 cm² (ii) 1386 cm² (iii) perimeter 174 cm; fraction cleaned = 3/4, the same for any sweep angle because the angle factor cancels.

Where students lose marks in Area of a Sector of a Circle

The slips examiners see most often, and what to write instead.

Mistake: Using $2 π r$ in the area formula
Arc and area formulas look alike, so students mix them under pressure.
Wrong working
Area of $6 0^{\circ}$ sector, $r = 21$ : $\frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22$ cm $^{2}$
Right working
Area $= \frac{60}{360} \times \frac{22}{7} \times 21 \times 21 = 231$ cm $^{2}$
Do this: Area needs $r^{2}$ and square units; length needs $2 π r$ and plain units.
Usually costs: The whole calculation mark, often 1 to 2 marks.
Mistake: Perimeter of a sector written as only the arc
Students think of the curved edge and forget the two straight radii.
Wrong working
Perimeter of $9 0^{\circ}$ sector, $r = 14$ : $\frac{1}{4} \times 88 = 22$ cm
Right working
Perimeter $= 2 \times 14 + 22 = 50$ cm
Do this: Always write $P = 2 r + l$ as a line before substituting.
Usually costs: ½ to 1 mark, and wrong option in an MCQ.
Mistake: Using the minor arc for the perimeter of a major sector
The arc is calculated once and reused without thinking.
Wrong working
Major sector, $r = 21$ , minor angle $6 0^{\circ}$ : $P = 42 + 22 = 64$ cm
Right working
Major arc $= 132 - 22 = 110$ cm, so $P = 42 + 110 = 152$ cm
Do this: For a major sector, use angle $36 0^{\circ} - θ$ throughout.
Usually costs: 1 mark in a 3- or 5-mark question.
Mistake: Putting the diameter in for $r$
The question gives a diameter and it is copied straight in.
Wrong working
Diameter 28 cm, $9 0^{\circ}$ sector: $\frac{1}{4} \times \frac{22}{7} \times 2 8^{2} = 616$ cm $^{2}$
Right working
$r = 14$ cm: $\frac{1}{4} \times \frac{22}{7} \times 196 = 154$ cm $^{2}$
Do this: Write " $r = \dots$ " as your first line every time.
Usually costs: All marks after the first step.

Exam tips

Write the formula in letters before substituting; the formula line usually carries ½ mark even if arithmetic slips.
In reverse questions (find $θ$ or $r$ ), show the equation clearly and square-root only at the end; check by putting the answer back.
When the perimeter is given, think of $A = \frac{1}{2} l r$ first: it avoids finding an awkward angle.
Read for the words major, remaining or not swept; they signal angle $36 0^{\circ} - θ$ .
Give units in every final answer: cm $^{2}$ for area, cm for arc and perimeter.

Practice questions on Area of a Sector of a Circle

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
Find the area and perimeter of a sector of a circle of radius 28 cm with central angle $4 5^{\circ}$ . (Take $π = \frac{22}{7}$ )
Hint
A $4 5^{\circ}$ sector is one eighth of the circle.
Answer
Area $= \frac{1}{8} \times \frac{22}{7} \times 784 = \frac{2464}{8} = 308$ cm². Arc $= \frac{1}{8} \times 176 = 22$ cm, so perimeter $= 56 + 22 = 78$ cm.
Q2 · 3 marks · Short answerHarder board level
The perimeter of a quadrant of a circle is 25 cm. Find (i) the radius, (ii) the area of the quadrant, (iii) the area and perimeter of the remaining (major) sector of the circle. (Take $π = \frac{22}{7}$ )
Hint
Write the perimeter in terms of $r$ : $2 r + \frac{1}{4} \times 2 π r$ , and solve for $r$ first.
Answer
(i) $2 r + \frac{1}{4} \times 2 \times \frac{22}{7} r = 2 r + \frac{11}{7} r = \frac{25}{7} r = 25$ , so $r = 7$ cm. (ii) Area $= \frac{1}{4} \times \frac{22}{7} \times 49 = 38.5$ cm². (iii) Major sector area $= 154 - 38.5 = 115.5$ cm² (check $\frac{3}{4} \times 154 = 115.5$ ); major arc $= \frac{3}{4} \times 44 = 33$ cm, so perimeter $= 14 + 33 = 47$ cm.
Q3 · 3 marks · Short answerHarder board level
The minute hand of a wall clock is 21 cm long. Find (i) the area swept by the minute hand between 7:05 pm and 7:25 pm, (ii) the distance moved by the tip of the minute hand in that time, (iii) how many minutes the tip takes to travel 99 cm. (Take $π = \frac{22}{7}$ )
Hint
The minute hand turns $36 0^{\circ}$ in 60 minutes, so $6^{\circ}$ per minute.
Answer
(i) Time $= 20$ minutes, angle $= 20 \times 6^{\circ} = 12 0^{\circ}$ . Area $= \frac{1}{3} \times \frac{22}{7} \times 441 = \frac{1}{3} \times 1386 = 462$ cm². (ii) Distance $= \frac{1}{3} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{3} \times 132 = 44$ cm. (iii) $\frac{99}{132} = \frac{3}{4}$ of a full turn, so time $= \frac{3}{4} \times 60 = 45$ minutes.
Q4 · 5 marks · Long answerHarder board level
Imagine a square grassy field $A B C D$ of side 28 m. One cow is tied at corner $A$ and another at the opposite corner $C$ , each with a rope 14 m long, so each cow can graze a quadrant inside the field with centre at its corner. Find (i) the total area grazed by the two cows, (ii) the area of the field left ungrazed, (iii) the length of the boundary of the ungrazed region, (iv) the extra area the cow at $A$ can graze if its rope is lengthened to 21 m (the other rope stays 14 m). (Take $π = \frac{22}{7}$ )
Hint
Each side of the square is split into a 14 m grazed part and a 14 m ungrazed part. For (iv), the extra region lies between two quadrants with the same centre.
Answer
(i) Each quadrant $= \frac{1}{4} \times \frac{22}{7} \times 196 = 154$ m², so total $= 2 \times 154 = 308$ m² (1½ marks). (ii) Ungrazed $= 28 \times 28 - 308 = 784 - 308 = 476$ m² (1 mark). (iii) On each of the four sides, 14 m is not grazed: $4 \times 14 = 56$ m; the two arcs are each $\frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22$ m. Boundary $= 56 + 44 = 100$ m (1½ marks). (iv) Since $21 + 14 = 35$ m is less than the diagonal $A C = 282$ m (about 39.6 m) and $21 < 28$ , the regions do not overlap and stay inside the field. Extra area $= \frac{1}{4} \times \frac{22}{7} \times (441 - 196) = \frac{11}{14} \times 245 = 192.5$ m² (1 mark).
Q5 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): If the radius of a circle is halved and the central angle of a sector is doubled, the area of the sector becomes half.
Reason (R): The area of a sector is directly proportional to its central angle and to the square of the radius.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Area changes by (angle factor) × (radius factor)².
Answer
(a). New area $= \frac{2 θ}{36 0 ^{\circ}} \times π (\frac{r}{2})^{2} = 2 \times \frac{1}{4} \times$ old $= \frac{1}{2}$ old. A is true, R is true and it is exactly the reason.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Area of a sector of angle $θ$ , radius $r$ ?Show answer
$\frac{θ}{36 0 ^{\circ}} \times π r^{2}$
2.Perimeter of a sector?Show answer
$2 r + l$ , where $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$
3.Area of a quadrant of radius 7 cm?Show answer
$\frac{1}{4} \times 154 = 38.5$ cm $^{2}$
4.Angle of the major sector if the minor sector is $8 0^{\circ}$ ?Show answer
$28 0^{\circ}$
5.Sector area in terms of arc $l$ and radius $r$ ?Show answer
$\frac{1}{2} l r$

Area of a Sector of a Circle: frequently asked questions

What is the difference between a sector and a segment?

A sector is bounded by two radii and an arc, like a pizza slice. A segment is bounded by a chord and an arc. A segment's area is found by taking the sector and subtracting the triangle formed by the two radii and the chord.

Is the perimeter of a sector the same as the arc length?

No. The perimeter includes the two radii as well as the arc, so it is the arc length plus twice the radius. Only the arc is the curved part.

How do I find the area of a major sector?

Either use the angle 360 degrees minus the minor angle in the sector formula, or subtract the minor sector's area from the whole circle's area. Doing both is a good check.

Can I use pi as 3.14 instead of 22/7?

Use 22/7 unless the question tells you to use 3.14. Questions with radii like 7, 14 or 21 are designed so that 22/7 cancels neatly.

Is the section on areas of combinations of plane figures still in the syllabus?

That separate NCERT section was removed in the rationalisation and is not in the 2026-27 syllabus. Shaded-region questions built from sectors and segments with squares, rectangles and triangles are still asked.