MathsCBSE Class 10 · Areas Related to Circles

Shaded Region Problems with Sectors and Segments

Updated 13 October 2026 · Fact-checked

A shaded region problem asks for the area or perimeter of part of a figure built from sectors, segments, squares, rectangles and triangles. You split the region into known pieces, then add or subtract their areas. Boards ask these as 2, 3 and 5 mark questions and as case studies, often with corner quadrants, equilateral triangles and leaf shapes.

Add the two quadrants, then subtract the square once: the overlap counted twice is the leaf, so leaf = Q1 + Q2 − square.

Illustration drawn for QuestionBuddy.

Understand Shaded Region Problems with Sectors

A shaded region is any part of a figure whose boundary is made of arcs and straight lines. You almost never have a direct formula for it. Instead, you see it as a combination of shapes you already know: sectors, segments, semicircles, squares, rectangles and triangles.

The idea that makes it work is simple. Areas add when pieces do not overlap. If a region is a square with four quarter circles cut out, its area is the area of the square minus the four quadrants. If a leaf is formed where two quadrants overlap, its area is the sum of the two quadrants minus the square, because the square is counted once while the overlap is counted twice.

This topic uses everything else in the chapter. Each curved piece is a sector, $\frac{θ}{36 0 ^{\circ}} \times π r^{2}$ , or a segment, sector minus $△ O A B$ . The angle $θ$ comes from the figure: $9 0^{\circ}$ at a corner of a square or rectangle, $6 0^{\circ}$ at a vertex of an equilateral triangle, and $18 0^{\circ}$ for a semicircle. Three sectors at the vertices of any triangle always total $18 0^{\circ}$ , because the angles of a triangle add to $18 0^{\circ}$ .

Perimeter is different from area. For perimeter you do not add and subtract regions. You walk around the edge of the shaded part and add only the pieces you walk along: arc lengths $\frac{θ}{36 0 ^{\circ}} \times 2 π r$ and the straight parts that are actually on the boundary. Interior lines do not count.

The method holds under these conditions:

The pieces you add must not overlap, or you must subtract each overlap exactly once.
Every radius and angle must be read correctly from the figure (a diameter of $14$ cm means radius $7$ cm).
Segment areas in this syllabus use only central angles of $6 0^{\circ}$ , $9 0^{\circ}$ and $12 0^{\circ}$ , so the triangle $O A B$ always has a neat area.

Once you can write the shaded area in words, such as 'triangle minus three sectors', the arithmetic is routine.

Shaded Region Problems with Sectors: formulas to remember

Area of a sector: $A = \frac{θ}{36 0 ^{\circ}} \times π r^{2}$; Every curved piece of a shaded region starts here.
Length of an arc: $l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$; Use for the curved edges when perimeter is asked.
Quadrant and semicircle: $Quadrant = \frac{1}{4} π r^{2}, Semicircle = \frac{1}{2} π r^{2}$; Quadrants sit at corners of squares and rectangles; semicircles on their sides.
Area of a minor segment: $Segment = sector O A B - area of △ O A B$; Central angle 60°, 90° or 120° only.
Segment with central angle 90° (π = 22/7): $(\frac{π}{4} - \frac{1}{2}) r^{2} = \frac{2}{7} r^{2}$; Fast check for leaf and lune problems.
Leaf formed by two quadrants in a square of side a: $Leaf = 2 \times \frac{1}{4} π a^{2} - a^{2} = \frac{4}{7} a^{2}$; Two quadrants overlap; the square is counted once. Value uses π = 22/7.
Square minus four corner quadrants (radius half the side): $a^{2} - π (\frac{a}{2})^{2}$; Four quadrants of radius a/2 make one full circle.
Sectors at the three vertices of a triangle (same radius r): $\frac{18 0 ^{\circ}}{36 0 ^{\circ}} \times π r^{2} = \frac{1}{2} π r^{2}$; Angles of any triangle add to 180°, so the three sectors make a semicircle.
Area of an equilateral triangle of side a: $\frac{3}{4} a^{2}$; Needed whenever sectors are drawn at the vertices of an equilateral triangle.
Perimeter of a shaded region: $P = \sum (arc lengths on boundary) + \sum (straight edges on boundary)$; Never add interior lines or subtract lengths the way you subtract areas.

How to solve Shaded Region Problems with Sectors questions

Use this method on any shaded-region question, whether it is a 2-mark direct problem or a 5-mark figure. The first two steps decide your marks more than the arithmetic does.

1Read the figure description and list every piece: each sector with its centre, radius and angle; each square, rectangle or triangle with its sides.
2Find the angles from the shapes: 90° at a corner of a square or rectangle, 60° at a vertex of an equilateral triangle, 180° for a semicircle, sum 180° for sectors at all vertices of any triangle.
3Write the shaded area in words first, for example 'shaded area = area of square − 4 × area of quadrant'. This line often earns a mark.
4Check for overlap: if two pieces share a region, subtract the shared part once (the leaf rule: quadrant + quadrant − square).
5Calculate each piece separately with π = 22/7 unless told otherwise; find the full circle area first, then take the fraction.
6Combine the pieces as your word equation says, and write the answer with square units.
7If perimeter is asked, trace the boundary of the shaded part and add only the arcs and straight edges you actually walk along.
8Check: the shaded area must be positive and smaller than the outer shape; for a 90° segment, use the shortcut 2r²/7 as a second check.

Shaded Region Problems with Sectors 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.

Shaded regions built from sectors and segments

Suppose your school is marking a new games field shaped like a stadium: a rectangle 70 m long and 42 m wide, with a semicircle attached on each of the two 42 m sides. The whole field is to be grassed, and a running lane is painted along its outer boundary.

Find (i) the area of the field to be grassed, (ii) the length of the outer boundary, and (iii) how many complete laps of the boundary a student must run to cover 1.36 km. (Take π = 22/7)

Try it first, then show the solutionSolution

Each semicircle has diameter 42 m, so radius r = 21 m.
Area of rectangle = 70 × 42 = 2940 m².
Two semicircles together make one full circle: area = (22/7) × 21 × 21 = 1386 m².
Field area = 2940 + 1386 = 4326 m².
Boundary = two straight 70 m sides + two semicircular arcs; the 42 m sides are inside, so they are not counted.
Two semicircular arcs = one circumference = 2 × (22/7) × 21 = 132 m.
Boundary = 2 × 70 + 132 = 272 m.
1.36 km = 1360 m, so laps = 1360 ÷ 272 = 5.

Answer: (i) 4326 m² (ii) 272 m (iii) 5 laps

Memory hook: Stadium field: two semicircles make one circle, and the joining sides are inside, so they never go in the perimeter.

Tricks to remember Shaded Region Problems with Sectors

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

Say it in words first
Before any number, write 'shaded = big shape − pieces' or 'shaded = piece + piece − overlap'.
Use it when: Stops you adding when you should subtract, and earns the method mark.
Corners give the angle
Square corner 90°, equilateral corner 60°, any triangle's three corners together 180°.
Use it when: Read sector angles instantly in figures with sectors at vertices.
Two-sevenths for a quarter-segment
With π = 22/7, a 90° segment = (2/7)r², so a leaf in a square of side a = (4/7)a².
Use it when: Quick check on leaf, lune and flower-petal problems.
Walk the edge
For perimeter, imagine an ant walking along the boundary of the shaded part. Count only what it walks on.
Use it when: Prevents adding interior radii or sides that are covered by arcs.

Shaded Region Problems with Sectors: solved examples

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

Example 1: Square with four corner quadrants · 2 marks · Very short answerBoard level

$A B C D$ is a square of side 14 cm. With each vertex as centre, a quadrant of radius 7 cm is drawn inside the square. The region of the square not covered by the four quadrants is shaded. Find the area of the shaded region. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Each corner angle of a square is $9 0^{\circ}$ , so each piece is a quadrant of radius 7 cm.
Shaded area $=$ area of square $- 4 \times$ area of quadrant. [½ mark]
Four quadrants of equal radius make one full circle: $4 \times \frac{1}{4} π r^{2} = \frac{22}{7} \times 7 \times 7 = 154$ cm $^{2}$ . [½ mark]
Area of square $= 14 \times 14 = 196$ cm $^{2}$ . [½ mark]
Shaded area $= 196 - 154 = 42$ cm $^{2}$ . [½ mark]

Answer: The shaded area is 42 cm².

Example 2: Equilateral triangle with sectors at its vertices · 3 marks · Short answerBoard level

$A B C$ is an equilateral triangle of side 14 cm. With each of $A$ , $B$ and $C$ as centre, a sector of radius 7 cm is drawn inside the triangle, so that each arc meets the two sides at their midpoints. Find the area of the part of the triangle not covered by the three sectors. (Take $π = \frac{22}{7}$ and $3 = 1.73$ )

Try it first, then show the solutionSolution

Each angle of an equilateral triangle is $6 0^{\circ}$ , so each sector has angle $6 0^{\circ}$ and radius 7 cm. [½ mark]
Shaded area $=$ area of $△ A B C - 3 \times$ area of sector.
Area of $△ A B C = \frac{3}{4} \times 1 4^{2} = 493 = 49 \times 1.73 = 84.77$ cm $^{2}$ . [1 mark]
Three $6 0^{\circ}$ sectors total $18 0^{\circ}$ , so together they make a semicircle: $\frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77$ cm $^{2}$ . [1 mark]
Shaded area $= 84.77 - 77 = 7.77$ cm $^{2}$ . [½ mark]

Answer: The area not covered is 7.77 cm².

Example 3: Area and perimeter after cutting a quadrant · 3 marks · Short answerBoard level

$A B C D$ is a square of side 14 cm. A quadrant with centre $A$ and radius 14 cm is drawn inside the square, its arc running from $B$ to $D$ . The part of the square outside the quadrant (near corner $C$ ) is shaded. Find (i) the area and (ii) the perimeter of the shaded region. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Angle $A = 9 0^{\circ}$ , so the quadrant has radius 14 cm and angle $9 0^{\circ}$ .
Area of quadrant $= \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154$ cm $^{2}$ . [½ mark]
Shaded area $= 1 4^{2} - 154 = 196 - 154 = 42$ cm $^{2}$ . [1 mark]
The boundary of the shaded region is side $B C$ , side $C D$ and arc $B D$ ; sides $A B$ and $A D$ are not on its edge. [½ mark]
Arc $B D = \frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22$ cm. [½ mark]
Perimeter $= 14 + 14 + 22 = 50$ cm. [½ mark]

Answer: (i) 42 cm² (ii) 50 cm

Example 4: The leaf rule: two overlapping quadrants · 3 marks · Short answerHarder board level

$A B C D$ is a square of side 14 cm. Two quadrants of radius 14 cm are drawn inside the square, one with centre $A$ and one with centre $C$ . Both arcs run from $B$ to $D$ , and the region where the two quadrants overlap is a leaf shape lying along the diagonal $B D$ . This leaf is shaded. Find (i) the area and (ii) the perimeter of the leaf. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Each angle of a square is $9 0^{\circ}$ , so both pieces are quadrants of radius 14 cm. Together they cover the whole square, and the leaf is covered by both of them. [½ mark]
So quadrant at $A$ $+$ quadrant at $C$ $=$ square $+$ leaf, which gives leaf $= 2 \times$ quadrant $-$ square. [½ mark]
Area of one quadrant $= \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154$ cm $^{2}$ . [½ mark]
Area of leaf $= 2 \times 154 - 196 = 308 - 196 = 112$ cm $^{2}$ . [½ mark]
The boundary of the leaf is only the two arcs $B D$ ; no side of the square lies on it. Each arc $= \frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22$ cm. [½ mark]
Perimeter of leaf $= 22 + 22 = 44$ cm. Check on the area: the leaf is two $9 0^{\circ}$ segments on chord $B D$ , so $2 \times \frac{2}{7} \times 196 = 112$ cm $^{2}$ . [½ mark]

Answer: (i) 112 cm² (ii) 44 cm

Example 5: Segment with central angle 120° · 3 marks · Short answerHarder board level

A chord $A B$ of a circle with centre $O$ and radius 21 cm subtends $∠ A O B = 12 0^{\circ}$ at the centre. Find the area of (i) the minor segment and (ii) the major segment cut off by $A B$ . (Take $π = \frac{22}{7}$ and $3 = 1.73$ )

Try it first, then show the solutionSolution

Area of sector $O A B = \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{3} \times 1386 = 462$ cm $^{2}$ . [½ mark]
Draw $O M ⊥ A B$ . In $△ O M A$ and $△ O M B$ : $O A = O B$ , $O M$ is common and $∠ O M A = ∠ O M B = 9 0^{\circ}$ , so they are congruent (RHS). Hence $A M = M B$ and $∠ A O M = ∠ B O M = 6 0^{\circ}$ . [½ mark]
In right $△ O M A$ : $O M = 21 cos 6 0^{\circ} = \frac{21}{2}$ cm and $A M = 21 sin 6 0^{\circ} = \frac{21 3}{2}$ cm, so $A B = 213$ cm. [½ mark]
Area of $△ O A B = \frac{1}{2} \times A B \times O M = \frac{1}{2} \times 213 \times \frac{21}{2} = \frac{441 3}{4} = 110.25 \times 1.73 = 190.73$ cm $^{2}$ (approx). [½ mark]
Minor segment $=$ sector $-$ triangle $= 462 - 190.73 = 271.27$ cm $^{2}$ . [½ mark]
Area of circle $= \frac{22}{7} \times 441 = 1386$ cm $^{2}$ , so major segment $= 1386 - 271.27 = 1114.73$ cm $^{2}$ . [½ mark]

Answer: (i) Minor segment ≈ 271.27 cm² (ii) Major segment ≈ 1114.73 cm²

Example 6: Crescents on a square inscribed in a circle · 5 marks · Long answerHarder board level

A square $A B C D$ is inscribed in a circle with centre $O$ and radius 14 cm, so all four vertices lie on the circle. On each side of the square, a semicircle is drawn outwards, with that side as diameter. Each semicircle pokes outside the big circle, forming a crescent-shaped region between the semicircle's arc and the big circle's arc. Find the total area of the four crescents, and show that it equals the area of the square. (Take $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

The diagonal of the square is a diameter of the circle, so $A C = 28$ cm. [½ mark]
Each side subtends $9 0^{\circ}$ at $O$ (the diagonals of a square are perpendicular), so $A B = 142$ cm and the area of the square $= \frac{1}{2} \times 28 \times 28 = 392$ cm $^{2}$ . [1 mark]
Each semicircle has radius $72$ cm, so $r^{2} = 98$ . Area of one semicircle $= \frac{1}{2} \times \frac{22}{7} \times 98 = 154$ cm $^{2}$ ; four give 616 cm $^{2}$ . [1 mark]
Each crescent $=$ semicircle $-$ the minor segment of the big circle cut off by that side (central angle $9 0^{\circ}$ ). [½ mark]
Area of one $9 0^{\circ}$ segment $=$ sector $-$ triangle $= \frac{1}{4} \times 616 - \frac{1}{2} \times 14 \times 14 = 154 - 98 = 56$ cm $^{2}$ ; four give 224 cm $^{2}$ . [1 mark]
Total area of crescents $= 616 - 224 = 392$ cm $^{2}$ , which equals the area of the square. [1 mark]

Answer: The four crescents have total area 392 cm², exactly the area of the square.

Example 7: Case study: planning a school garden on a budget · 4 marks · Case-basedHOTS / competency

Imagine your school plans a rectangular garden $A B C D$ with $A B = 28$ m. The diagonal $A C$ is measured as 35 m, but side $B C$ is not marked on the plan. Flower beds shaped as quadrants of radius 7 m are made inside the garden at corners $A$ and $B$ . A semicircular pond of diameter 14 m is dug inside the garden, with its diameter along side $C D$ and its centre at the midpoint of $C D$ . The beds and the pond do not touch. The rest of the garden is to be turfed at Rs 20 per m², and the school has a budget of Rs 9000 for turfing. (Take $π = \frac{22}{7}$ )
(i) Find the length of side $B C$ . (1 mark)
(ii) Find the total area of the two flower beds. (1 mark)
(iii) Find the area to be turfed and decide whether the budget is enough. If it is, how much money is left? (2 marks)

Try it first, then show the solutionSolution

(i) $∠ A B C = 9 0^{\circ}$ , so by Pythagoras $B C^{2} = A C^{2} - A B^{2} = 1225 - 784 = 441$ , giving $B C = 21$ m. [1 mark]
(ii) Each bed is a quadrant of radius 7 m (corner angle $9 0^{\circ}$ ): two beds $= 2 \times \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = 77$ m $^{2}$ . [1 mark]
(iii) Area of garden $= 28 \times 21 = 588$ m $^{2}$ . [½ mark]
Pond radius $= 7$ m, so pond $= \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77$ m $^{2}$ . [½ mark]
Area to be turfed $= 588 - 77 - 77 = 434$ m $^{2}$ . [½ mark]
Cost $= 434 \times 20 = 8680$ , that is Rs 8680, which is less than Rs 9000. The budget is enough and Rs 320 is left. [½ mark]

Answer: (i) 21 m (ii) 77 m² (iii) 434 m²; cost Rs 8680, so the budget is enough and Rs 320 is left.

Where students lose marks in Shaded Region Problems with Sectors

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

Mistake: Using the diameter as the radius
Figures often give the side of a square or the diameter of a semicircle, and students plug it straight into πr².
Wrong working
Semicircle on a 14 cm side: area = (1/2) × (22/7) × 14 × 14 = 308 cm²
Right working
Radius = 7 cm, so area = (1/2) × (22/7) × 7 × 7 = 77 cm²
Do this: Before any formula, write 'r = ...' for every curved piece.
Usually costs: Usually all calculation marks for that piece and every step after it.
Mistake: Treating perimeter like area
Students subtract lengths the way they subtract regions.
Wrong working
Perimeter of square minus quadrant (side 14) = 56 − 22 = 34 cm
Right working
Boundary is BC + CD + arc BD = 14 + 14 + 22 = 50 cm
Do this: Trace the edge of the shaded region and add only what lies on it.
Usually costs: 1 to 2 marks in a 3-mark question.
Mistake: Forgetting to remove an overlap
In leaf and petal figures, two shapes cover the same region and it gets counted twice.
Wrong working
Leaf in square of side 14 = quadrant = 154 cm²
Right working
Leaf = 2 × 154 − 196 = 112 cm²
Do this: Ask: which part has been counted twice? Subtract it once.
Usually costs: Most of the marks, as the method itself is wrong.
Mistake: Wrong sector angle at a vertex
Students take 90° everywhere, even at the vertices of an equilateral or other triangle.
Wrong working
Three sectors of radius 7 at vertices of a triangle: 3 × 38.5 = 115.5 cm²
Right working
Angles total 180°, so sectors = (1/2) × (22/7) × 49 = 77 cm²
Do this: Read the angle from the shape; for any triangle the three sectors add to a semicircle.
Usually costs: 1 to 2 marks.

Exam tips

Write the shaded area as a word equation before calculating; examiners award the method mark for it even if arithmetic slips later.
Keep √3 as a symbol until the last line, then substitute the value the question gives, to avoid rounding errors.
State each angle with its reason, such as 'each angle of an equilateral triangle is 60°', in 3 and 5 mark answers.
Look for groups that make a whole circle or semicircle, like four corner quadrants or three triangle-vertex sectors; it saves time and reduces errors.
In case-based questions, sub-parts often build on each other, so reuse earlier answers, like a side found by Pythagoras in part (i), in the final part.

Practice questions on Shaded Region Problems with Sectors

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
A circle is drawn inside a square, touching all four sides. The area of the region inside the square but outside the circle is 168 cm². Find the side of the square and the radius of the circle. (Take $π = \frac{22}{7}$ )
Hint
Let the side be a; the radius is a/2. Write the shaded area in terms of a.
Answer
Let side = a, so radius = a/2. Shaded area = a² − (22/7) × (a²/4) = a² − 11a²/14 = 3a²/14. So 3a²/14 = 168, a² = 784, a = 28 cm and radius = 14 cm. Check: 784 − (22/7) × 196 = 784 − 616 = 168 cm².
Q2 · 3 marks · Short answerHarder board level
$A B C D$ is a square of side 14 cm. On each side, a semicircle is drawn inside the square with that side as diameter. The four semicircles overlap in four leaf-shaped regions that meet at the centre of the square. Find the total area of the four leaves. (Take $π = \frac{22}{7}$ )
Hint
Add the four semicircles; each leaf is counted twice while the square is covered once.
Answer
Each semicircle has radius 7: area 77 cm²; four give 308 cm². Leaves = 308 − 196 = 112 cm². Check: each leaf is two 90° segments of radius 7, each (2/7) × 49 = 14, so a leaf is 28 and four leaves are 112 cm².
Q3 · 3 marks · Short answerBoard level
A chord $A B$ of a circle with centre $O$ and radius 14 cm subtends $∠ A O B = 6 0^{\circ}$ . The minor segment is shaded. Find the area and the perimeter of the shaded region. (Take $π = \frac{22}{7}$ and $3 = 1.73$ )
Hint
With OA = OB and a 60° angle between them, triangle OAB is equilateral.
Answer
Sector = (60/360) × (22/7) × 196 = 308/3 = 102.67 cm². Triangle OAB is equilateral (OA = OB and ∠AOB = 60°), so its area = (√3/4) × 196 = 49√3 = 84.77 cm². Segment = 102.67 − 84.77 = 17.90 cm² (approx). Arc AB = (1/6) × 2 × (22/7) × 14 = 44/3 = 14.67 cm and chord AB = 14 cm, so perimeter = 28.67 cm (approx).
Q4 · 5 marks · Long answerHarder board level
A regular hexagon is inscribed in a circle of radius 28 cm, so its six vertices lie on the circle. The six regions inside the circle but outside the hexagon are shaded. Find the shaded area. (Take $π = \frac{22}{7}$ and $3 = 1.73$ )
Hint
Each side subtends 60° at the centre, so the hexagon is six equilateral triangles of side 28 cm.
Answer
Circle = (22/7) × 784 = 2464 cm². Hexagon = 6 × (√3/4) × 784 = 1176√3 = 2034.48 cm². Shaded = 2464 − 2034.48 = 429.52 cm². Check: six 60° segments, each 410.67 − 339.08 = 71.59 cm², total 429.52 cm² (approx).
Q5 · 3 marks · Short answerHOTS / competency
In a square of side 14 cm, four quadrants of radius 7 cm are drawn inside it, one at each corner. Riya says the perimeter of the region left uncovered is $4 \times 14 + 44 = 100$ cm. Find her error and give the correct perimeter. (Take $π = \frac{22}{7}$ )
Hint
Trace the edge of the uncovered region. Does any part of a side of the square belong to it?
Answer
Each side of 14 cm is fully covered by two radii of 7 cm, so no straight edge is on the boundary. The boundary is only the four quarter arcs, which together make one circle of radius 7: 2 × (22/7) × 7 = 44 cm. Correct perimeter = 44 cm; Riya wrongly added the 56 cm of sides.
Q6 · 5 marks · Long answerHarder board level
An equilateral triangle $A B C$ is inscribed in a circle with centre $O$ and radius 14 cm, so $A$ , $B$ and $C$ lie on the circle. The three regions inside the circle but outside the triangle are shaded. Find (i) the side of the triangle and (ii) the area of the shaded region. (Take $π = \frac{22}{7}$ and $3 = 1.73$ )
Hint
The three equal sides are equal chords, so each subtends 360° ÷ 3 = 120° at O. Each shaded region is a 120° segment.
Answer
∠AOB = ∠BOC = ∠COA = 120°, since equal chords subtend equal angles at the centre. Draw OM ⊥ AB; it bisects ∠AOB, so ∠AOM = 60°. OM = 14 cos 60° = 7 cm and AM = 14 sin 60° = 7√3 cm, so AB = 14√3 = 24.22 cm. Area of triangle OAB = (1/2) × 14√3 × 7 = 49√3 = 84.77 cm². Sector OAB = (1/3) × (22/7) × 196 = 205.33 cm², so one segment ≈ 120.56 cm². Shaded area = circle − 3 × triangle OAB = 616 − 254.31 = 361.69 cm² (approx). (i) 24.22 cm (ii) 361.69 cm².
Q7 · 5 marks · Long answerHarder board level
$A B C$ is an equilateral triangle of side 28 cm. With $A$ , $B$ and $C$ as centres, sectors of radius 14 cm are drawn inside the triangle, so each arc joins the midpoints of two sides. A semicircle is also drawn outside the triangle with $B C$ as diameter. The shaded region is the part of the triangle not covered by the three sectors, together with the semicircle. Find (i) the area of the shaded region and (ii) the total length of the three arcs of the sectors. (Take $π = \frac{22}{7}$ and $3 = 1.73$ )
Hint
Three 60° sectors of radius 14 make a semicircle of radius 14, the same size as the semicircle on BC.
Answer
Area of triangle = (√3/4) × 784 = 196√3 = 339.08 cm². Each sector has angle 60°, so the three sectors together = (1/2) × (22/7) × 196 = 308 cm². Uncovered part of triangle = 339.08 − 308 = 31.08 cm². Semicircle on BC has radius 14 cm: area = 308 cm². Shaded area = 31.08 + 308 = 339.08 cm², which equals the area of the triangle. Each arc = (60/360) × 2 × (22/7) × 14 = 44/3 cm, so the three arcs total 44 cm. (i) 339.08 cm² (ii) 44 cm.
Q8 · 5 marks · Long answerHarder board level
The perimeter of a quadrant $O A B$ of a circle with centre $O$ is 50 cm. Find (i) the radius, (ii) the area of the minor segment cut off by chord $A B$ and (iii) the area of the major segment. (Take $π = \frac{22}{7}$ )
Hint
The perimeter of a quadrant is two radii plus a quarter of the circumference.
Answer
Perimeter = 2r + (1/4) × 2 × (22/7) × r = 2r + 11r/7 = 25r/7. So 25r/7 = 50 and r = 14 cm. Sector = (1/4) × (22/7) × 196 = 154 cm². Triangle OAB = (1/2) × 14 × 14 = 98 cm². Minor segment = 154 − 98 = 56 cm² (check: (2/7) × 196 = 56). Circle = 616 cm², so major segment = 616 − 56 = 560 cm². (i) 14 cm (ii) 56 cm² (iii) 560 cm².
Q9 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a bicycle wheel makes 500 complete revolutions while the bicycle covers 1.1 km on a straight road. The wheel has 6 equally spaced spokes running from the centre to the rim. A reflector sticker exactly fills the region between the rim and the straight chord joining the outer ends of two adjacent spokes. (Take $π = \frac{22}{7}$ and $3 = 1.73$ )
(i) Find the radius of the wheel. (1 mark)
(ii) Find the length of the rim between two adjacent spokes. (1 mark)
(iii) Find the area of the reflector sticker. (2 marks)
Hint
One revolution covers one circumference. Six equal spokes make 360° ÷ 6 = 60° between neighbours.
Answer
(i) 1.1 km = 110000 cm, so circumference = 110000 ÷ 500 = 220 cm. Then 2 × (22/7) × r = 220, so r = 35 cm. (ii) Angle between adjacent spokes = 60°, so rim length = 220 ÷ 6 = 36.67 cm (approx). (iii) Sector = (1/6) × (22/7) × 1225 = 1925/3 = 641.67 cm². The triangle formed by the two spokes and the chord is equilateral (two radii with a 60° angle), so its area = (√3/4) × 1225 = 306.25 × 1.73 = 529.81 cm². Sticker = 641.67 − 529.81 = 111.85 cm² (approx).

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.What angle does a sector at a corner of a square have?Show answer
90°, so it is a quadrant.
2.Three equal-radius sectors at the vertices of any triangle total what?Show answer
A semicircle, since the angles add to 180°.
3.Area of a 90° segment of radius r with π = 22/7?Show answer
(2/7)r².
4.Leaf formed by two quadrants in a square of side a?Show answer
Two quadrants minus the square: (4/7)a² with π = 22/7.
5.Area of an equilateral triangle of side a?Show answer
(√3/4)a².
6.What counts in the perimeter of a shaded region?Show answer
Only the arcs and straight edges that form its boundary.

Shaded Region Problems with Sectors: frequently asked questions

Is the old exercise 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 are still asked, but they are built from sectors and segments together with squares, rectangles and triangles, which is what this page covers.

How do I know whether to add or subtract areas?

Describe the shaded part in words first. If it is what remains after cutting pieces out of a shape, subtract. If it is made of pieces joined together, add, and if two pieces overlap, subtract the overlap once.

Which value of pi should I use?

Use 22/7 unless the question says 3.14 or something else. Radii that are multiples of 7 are chosen so the answer comes out cleanly with 22/7.

Can a segment question use any central angle?

No. In this syllabus segment areas are restricted to central angles of 60, 90 and 120 degrees. Sectors can have other angles, but segments will use only these three.

Do I need to draw the figure in the exam?

If a figure is given, you can mark radii and angles on it. A quick labelled sketch helps you see which edges form the boundary when perimeter is asked.