Cut a Slice, Then Cut the Triangle Out of It
Learn to find the area of a circle and work backwards to its radius, calculate sector and segment areas for a given angle, apply Brahmagupta's formula to a cyclic quadrilateral, and solve composite area problems.
How do you find the area of a segment of a circle?
A sector is a slice cut from the centre outward, like a piece of cake. A segment is what you get when you slice straight across with a chord, missing the middle.
There is no separate formula for a segment, and none is needed:
Take a circle of radius cm and a chord subtending at the centre. The whole circle has area
The sector is a quarter of that, . Inside it sits a right triangle with both legs equal to the radius:
So the segment is
That subtraction is the pattern for this whole page. A shape you have no formula for is almost always a difference or a sum of shapes you do, and the skill being built is the decomposition rather than the arithmetic.
This page covers the third part of the CBSE Class 9 Mathematics chapter on measuring space — the area of a circle, sectors and segments, the area of a cyclic quadrilateral from its four sides, and composite figures.
There is no separate formula for a segment, and none is needed:
Take a circle of radius cm and a chord subtending at the centre. The whole circle has area
The sector is a quarter of that, . Inside it sits a right triangle with both legs equal to the radius:
So the segment is
That subtraction is the pattern for this whole page. A shape you have no formula for is almost always a difference or a sum of shapes you do, and the skill being built is the decomposition rather than the arithmetic.
This page covers the third part of the CBSE Class 9 Mathematics chapter on measuring space — the area of a circle, sectors and segments, the area of a cyclic quadrilateral from its four sides, and composite figures.
Formula
How do you find a circle's area, and its radius from the area?
**The area is times the radius squared, and the radius comes back by dividing and taking a square root:**
Take when the radius is a multiple of , and otherwise.
Worked example 1 — area from the radius. cm:
Worked example 2 — from the diameter. cm gives cm:
Worked example 3 — radius from the area. A circular lawn has area :
Check: . Correct.
Worked example 4 — a larger one. Area gives , so cm.
Worked example 5 — a ring, or annulus. A circular path has outer radius m and inner radius m. Subtract:
Paving it at per square metre costs
Squaring the radius first matters more than it looks. Doubling the radius quadruples the area: gives and gives , which is four times as much, not twice. The circumference merely doubles over the same change, because it depends on and not .
So area and circumference answer different questions and must not be swapped. A pizza of twice the diameter feeds four people, not two — and the standard error in this chapter is using where belongs, which the units catch: an area must come out in square units.
Take when the radius is a multiple of , and otherwise.
Worked example 1 — area from the radius. cm:
Worked example 2 — from the diameter. cm gives cm:
Worked example 3 — radius from the area. A circular lawn has area :
Check: . Correct.
Worked example 4 — a larger one. Area gives , so cm.
Worked example 5 — a ring, or annulus. A circular path has outer radius m and inner radius m. Subtract:
Paving it at per square metre costs
Squaring the radius first matters more than it looks. Doubling the radius quadruples the area: gives and gives , which is four times as much, not twice. The circumference merely doubles over the same change, because it depends on and not .
So area and circumference answer different questions and must not be swapped. A pizza of twice the diameter feeds four people, not two — and the standard error in this chapter is using where belongs, which the units catch: an area must come out in square units.
How do you calculate the area of a sector and a segment?
A sector is the same fraction of the circle's area that its angle is of a full turn, and a segment is that sector with its triangle removed.
Worked example 1 — a sixth of a circle. cm, . The whole circle is , so
Worked example 2 — a quadrant. cm, . The circle is :
Worked example 3 — the segment of that quadrant. The triangle inside a sector is right-angled with both legs equal to :
**Worked example 4 — a segment.** For the triangle has two sides equal to the radius and the angle between them is , so the third side is also — it is equilateral of side cm:
The minor and major segments together make the whole circle. For the quadrant case, the minor segment is , so the major segment is
Check: the major sector is , and adding the triangle rather than subtracting gives . The same answer twice.
That sign change is worth pausing on. For the minor segment the triangle is outside the region, so it is subtracted. For the major segment the triangle is inside, so it is added. Reading which segment the question means, then sketching it and shading, decides the sign before any number is written.
A sector's area and its perimeter are different quantities. The sector of a radius- circle has area and perimeter cm, from the arc-length work in the first part of this chapter. One is in square centimetres and the other in centimetres, so the units settle which has been asked for.
Worked example 1 — a sixth of a circle. cm, . The whole circle is , so
Worked example 2 — a quadrant. cm, . The circle is :
Worked example 3 — the segment of that quadrant. The triangle inside a sector is right-angled with both legs equal to :
**Worked example 4 — a segment.** For the triangle has two sides equal to the radius and the angle between them is , so the third side is also — it is equilateral of side cm:
The minor and major segments together make the whole circle. For the quadrant case, the minor segment is , so the major segment is
Check: the major sector is , and adding the triangle rather than subtracting gives . The same answer twice.
That sign change is worth pausing on. For the minor segment the triangle is outside the region, so it is subtracted. For the major segment the triangle is inside, so it is added. Reading which segment the question means, then sketching it and shading, decides the sign before any number is written.
A sector's area and its perimeter are different quantities. The sector of a radius- circle has area and perimeter cm, from the arc-length work in the first part of this chapter. One is in square centimetres and the other in centimetres, so the units settle which has been asked for.
How does Brahmagupta's formula give the area of a cyclic quadrilateral?
Take half the perimeter, subtract each side in turn, and multiply the four results under a square root. No angles and no diagonals are needed.
For a cyclic quadrilateral with sides , , , :
This is Brahmagupta's formula, and the resemblance to Heron's formula in the previous part of this chapter is not accidental.
Worked example 1. A cyclic quadrilateral has sides , , and cm. Then , and the brackets are , , , :
Worked example 2 — a rectangle, as a check. A rectangle is cyclic, since its opposite angles are each and sum to . For a cm by cm rectangle the sides are , so and the brackets are , , , :
And directly. The formula agrees with the obvious answer, which is the reassurance worth having before trusting it on a shape you cannot check.
Worked example 3 — a square. Sides give and four brackets of :
Correct again.
Worked example 4. Sides , , , cm give , with brackets , , , :
Why it looks like Heron's formula. Let one side shrink to nothing, say . The quadrilateral collapses into a triangle with sides , , , and now while the bracket becomes just :
That is Heron's formula exactly. So Heron's is the special case of Brahmagupta's with a vanished fourth side, and knowing one means knowing both.
The quadrilateral must be cyclic, and that condition does real work. Four given lengths can be hinged into many different quadrilaterals, all with the same perimeter and different areas. Brahmagupta's formula returns the area of the one arrangement whose vertices lie on a circle — and that arrangement has the largest area of all of them. So applying the formula to a non-cyclic quadrilateral overstates its area, which is why a question must say cyclic before the formula may be used, and why the previous chapter's test for a cyclic quadrilateral matters here.
For a cyclic quadrilateral with sides , , , :
This is Brahmagupta's formula, and the resemblance to Heron's formula in the previous part of this chapter is not accidental.
Worked example 1. A cyclic quadrilateral has sides , , and cm. Then , and the brackets are , , , :
Worked example 2 — a rectangle, as a check. A rectangle is cyclic, since its opposite angles are each and sum to . For a cm by cm rectangle the sides are , so and the brackets are , , , :
And directly. The formula agrees with the obvious answer, which is the reassurance worth having before trusting it on a shape you cannot check.
Worked example 3 — a square. Sides give and four brackets of :
Correct again.
Worked example 4. Sides , , , cm give , with brackets , , , :
Why it looks like Heron's formula. Let one side shrink to nothing, say . The quadrilateral collapses into a triangle with sides , , , and now while the bracket becomes just :
That is Heron's formula exactly. So Heron's is the special case of Brahmagupta's with a vanished fourth side, and knowing one means knowing both.
The quadrilateral must be cyclic, and that condition does real work. Four given lengths can be hinged into many different quadrilaterals, all with the same perimeter and different areas. Brahmagupta's formula returns the area of the one arrangement whose vertices lie on a circle — and that arrangement has the largest area of all of them. So applying the formula to a non-cyclic quadrilateral overstates its area, which is why a question must say cyclic before the formula may be used, and why the previous chapter's test for a cyclic quadrilateral matters here.
How do you find the area of a shape made from several pieces?
Cut the figure into rectangles, triangles and circular parts, find each area, then add or subtract. A sketch with the pieces labelled is most of the work.
Worked example 1 — a rectangle with a semicircular end. A hall floor is a m by m rectangle with a semicircle of diameter m attached to one short side.
- rectangle:
- semicircle, :
Worked example 2 — a square with the corners rounded off. A square of side cm has a quadrant of radius cm removed from each of the four corners. Four quadrants of the same radius make one full circle:
- square:
- circle removed, :
Noticing that the four quadrants make one circle saves three calculations, and that kind of recombining is what makes composite problems quick.
Worked example 3 — a flower bed in a lawn. A square lawn of side m has a circular bed of radius m at its centre. The grass area is
At per square metre, turfing the grass costs .
Worked example 4 — a triangle on a rectangle. A pennant is a cm by cm rectangle with a triangle of base cm and height cm attached to one short side:
Worked example 5 — a shaded region needing two subtractions. A semicircle of radius cm has a smaller semicircle of radius cm removed from one half of its diameter:
- large semicircle:
- small semicircle:
Areas add and subtract; perimeters do not follow along. In worked example 2 the square lost of area, and its boundary changed from four straight sides into four short sides plus four arcs — a completely separate calculation. So a composite question asking for both wants two independent pieces of working, and using the area decomposition for the perimeter is the trap the first part of this chapter warned about.
Choose the decomposition that gives whole numbers. Worked example 5 could be done by splitting the region into thin strips, and the answer would be the same and the arithmetic far worse. **A good split is the one where every radius is a multiple of **, so that cancels and no decimals appear.
Worked example 1 — a rectangle with a semicircular end. A hall floor is a m by m rectangle with a semicircle of diameter m attached to one short side.
- rectangle:
- semicircle, :
Worked example 2 — a square with the corners rounded off. A square of side cm has a quadrant of radius cm removed from each of the four corners. Four quadrants of the same radius make one full circle:
- square:
- circle removed, :
Noticing that the four quadrants make one circle saves three calculations, and that kind of recombining is what makes composite problems quick.
Worked example 3 — a flower bed in a lawn. A square lawn of side m has a circular bed of radius m at its centre. The grass area is
At per square metre, turfing the grass costs .
Worked example 4 — a triangle on a rectangle. A pennant is a cm by cm rectangle with a triangle of base cm and height cm attached to one short side:
Worked example 5 — a shaded region needing two subtractions. A semicircle of radius cm has a smaller semicircle of radius cm removed from one half of its diameter:
- large semicircle:
- small semicircle:
Areas add and subtract; perimeters do not follow along. In worked example 2 the square lost of area, and its boundary changed from four straight sides into four short sides plus four arcs — a completely separate calculation. So a composite question asking for both wants two independent pieces of working, and using the area decomposition for the perimeter is the trap the first part of this chapter warned about.
Choose the decomposition that gives whole numbers. Worked example 5 could be done by splitting the region into thin strips, and the answer would be the same and the arithmetic far worse. **A good split is the one where every radius is a multiple of **, so that cancels and no decimals appear.
Exam tip
Exam tip: sketch and shade before you subtract anything
Draw the figure and shade the region asked for. Shading decides every sign in a composite problem, and it takes ten seconds.
**State your **: write *taking *. Use it whenever the radius is a multiple of , and otherwise.
**Area is , circumference is . The units tell you which was wanted — an area answer must carry square units.
Halve the diameter first.** A circle of diameter cm has , and using gives four times the area.
**Working backwards: , then take the root.** Area gives , so — and check by substituting back.
**Sector: . Sector area and sector perimeter are different quantities in different units.
Segment = sector - triangle for the minor segment, and sector + triangle for the major one. Shade to see which.
The triangle in a sector is ; in a sector it is equilateral**, area .
**For Brahmagupta, write on its own line, then the four brackets — and only use it when the question says cyclic.
Recombine repeated pieces: four quadrants of equal radius make one circle.
And never carry a perimeter method into an area question** — they need separate working even in the same problem.
**State your **: write *taking *. Use it whenever the radius is a multiple of , and otherwise.
**Area is , circumference is . The units tell you which was wanted — an area answer must carry square units.
Halve the diameter first.** A circle of diameter cm has , and using gives four times the area.
**Working backwards: , then take the root.** Area gives , so — and check by substituting back.
**Sector: . Sector area and sector perimeter are different quantities in different units.
Segment = sector - triangle for the minor segment, and sector + triangle for the major one. Shade to see which.
The triangle in a sector is ; in a sector it is equilateral**, area .
**For Brahmagupta, write on its own line, then the four brackets — and only use it when the question says cyclic.
Recombine repeated pieces: four quadrants of equal radius make one circle.
And never carry a perimeter method into an area question** — they need separate working even in the same problem.
Did you know
Why four sticks of fixed length can enclose different amounts of space
Hinge four sticks of lengths , , and cm into a loop. The perimeter is cm and cannot change. The area can.
Push the shape flat and it encloses almost nothing. Open it out and the enclosed space grows. Keep going and it starts to shrink again. Somewhere in between it reaches a maximum — and that maximum is exactly where the four corners happen to lie on a circle.
That is what Brahmagupta's formula computes: , the largest area those four sticks can ever enclose.
The same effect explains something about triangles, by contrast. Three sticks hinged into a loop cannot flex at all — the shape is rigid, which is why a triangle is used to brace a gate or a bridge. So three lengths determine one area, and Heron's formula has no choosing to do. Four lengths determine a family of shapes, and Brahmagupta's formula picks out one member of it.
Push this further and the pattern continues. Five sticks enclose the most when all five corners lie on a circle, and so on for any number. Let the number of sticks grow without limit, each one tiny, and the shape that encloses the most becomes the circle itself — which is the result the first part of this chapter reached from the other end, comparing m of fence as a circle, a square and a thin rectangle.
So the two statements are one statement. Among shapes with a given boundary, the circle holds the most, and among quadrilaterals with given sides, the cyclic one holds the most, are the same principle at different levels of freedom — and Brahmagupta's formula is what the second one looks like when you write it down.
Push the shape flat and it encloses almost nothing. Open it out and the enclosed space grows. Keep going and it starts to shrink again. Somewhere in between it reaches a maximum — and that maximum is exactly where the four corners happen to lie on a circle.
That is what Brahmagupta's formula computes: , the largest area those four sticks can ever enclose.
The same effect explains something about triangles, by contrast. Three sticks hinged into a loop cannot flex at all — the shape is rigid, which is why a triangle is used to brace a gate or a bridge. So three lengths determine one area, and Heron's formula has no choosing to do. Four lengths determine a family of shapes, and Brahmagupta's formula picks out one member of it.
Push this further and the pattern continues. Five sticks enclose the most when all five corners lie on a circle, and so on for any number. Let the number of sticks grow without limit, each one tiny, and the shape that encloses the most becomes the circle itself — which is the result the first part of this chapter reached from the other end, comparing m of fence as a circle, a square and a thin rectangle.
So the two statements are one statement. Among shapes with a given boundary, the circle holds the most, and among quadrilaterals with given sides, the cyclic one holds the most, are the same principle at different levels of freedom — and Brahmagupta's formula is what the second one looks like when you write it down.
Exam relevance
How do circle and sector areas feed into JEE Main?
Because the sector area becomes the radian-measure formula, and the decomposition habit is what integration formalises.
This is the foundation for Class 11 Mathematics Trigonometric Functions and Class 12 Application of Integrals, examined in JEE Main. The sector formula becomes
which is the Class 9 version with replaced by a unit chosen to make the clutter vanish. A student who has computed a sector as one sixth of has already done the arithmetic that formula encodes.
The segment calculation survives almost unchanged. Class 11 gives the triangle inside a sector as , so the segment becomes
and the two Class 9 cases are its values at and . The subtraction is the same subtraction; only the triangle's formula has been generalised to any angle.
Composite areas become definite integrals. Class 12 Application of Integrals finds the area between curves by decomposing a region and adding signed pieces — precisely the shade-and-subtract discipline of this page, with a limit process replacing the formulas. Questions on the area enclosed between a circle and a line, or a circle and a parabola, are recurring JEE Main material and are usually solved as sector minus triangle or integral minus triangle.
The maximisation in the previous section is treated properly in Class 12 Application of Derivatives, where fixed-perimeter problems are standard, and the result that the circle encloses the most is what those questions verify.
Where the circle's equation comes in. Class 11 Conic Sections gives , and area questions there reduce to the sector and segment work above once the chord is located — which uses the chord-distance relation from the circles chapter.
What the questions look like. For board work, expect area of a circle or a ring, radius from a given area, sector and segment areas for a stated angle, Brahmagupta's formula on four given sides, and a shaded composite region with a cost calculation. For JEE Main, the direct forms are , the segment formula, and areas bounded by a circle and a straight line.
How board and competitive emphasis differ. A board paper rewards the **stated **, the shaded sketch and the square units. A competitive paper works in radians with exact multiples of , so an answer there reads rather than , and never appears.
The single trap that costs the most marks. Subtracting the triangle when the major segment was asked for. The minor segment is sector - triangle and the major one is sector + triangle, and both answers look reasonable in isolation. Shading the region on the diagram settles it before the arithmetic starts — and the free check is that the two segments must add to the whole circle, in the worked case.
This is the foundation for Class 11 Mathematics Trigonometric Functions and Class 12 Application of Integrals, examined in JEE Main. The sector formula becomes
which is the Class 9 version with replaced by a unit chosen to make the clutter vanish. A student who has computed a sector as one sixth of has already done the arithmetic that formula encodes.
The segment calculation survives almost unchanged. Class 11 gives the triangle inside a sector as , so the segment becomes
and the two Class 9 cases are its values at and . The subtraction is the same subtraction; only the triangle's formula has been generalised to any angle.
Composite areas become definite integrals. Class 12 Application of Integrals finds the area between curves by decomposing a region and adding signed pieces — precisely the shade-and-subtract discipline of this page, with a limit process replacing the formulas. Questions on the area enclosed between a circle and a line, or a circle and a parabola, are recurring JEE Main material and are usually solved as sector minus triangle or integral minus triangle.
The maximisation in the previous section is treated properly in Class 12 Application of Derivatives, where fixed-perimeter problems are standard, and the result that the circle encloses the most is what those questions verify.
Where the circle's equation comes in. Class 11 Conic Sections gives , and area questions there reduce to the sector and segment work above once the chord is located — which uses the chord-distance relation from the circles chapter.
What the questions look like. For board work, expect area of a circle or a ring, radius from a given area, sector and segment areas for a stated angle, Brahmagupta's formula on four given sides, and a shaded composite region with a cost calculation. For JEE Main, the direct forms are , the segment formula, and areas bounded by a circle and a straight line.
How board and competitive emphasis differ. A board paper rewards the **stated **, the shaded sketch and the square units. A competitive paper works in radians with exact multiples of , so an answer there reads rather than , and never appears.
The single trap that costs the most marks. Subtracting the triangle when the major segment was asked for. The minor segment is sector - triangle and the major one is sector + triangle, and both answers look reasonable in isolation. Shading the region on the diagram settles it before the arithmetic starts — and the free check is that the two segments must add to the whole circle, in the worked case.
Key takeaways
Circle, sector, segment and composite areas: quick revision
- Circle: , and backwards .
- gives ; gives and .
- Area gives , so ; area gives .
- Ring: outer m, inner m gives , costing at per square metre.
- Doubling the radius quadruples the area ( to ) while only doubling the circumference.
- **Area is , circumference is — the square units tell them apart.
- Sector**: . , gives ; , gives .
- Minor segment = sector - triangle. For , : , with the triangle .
- For , the triangle is equilateral, , so the segment is about .
- Major segment = sector + triangle: , and rebuilds the circle.
- Brahmagupta's formula for a cyclic quadrilateral: and .
- Sides : , brackets , so .
- A rectangle gives , matching ; a square of side gives .
- Sides : , brackets , so .
- **Setting turns Brahmagupta's formula into Heron's — so the triangle case is contained in it.
- The quadrilateral must be cyclic: four fixed sides hinge into many shapes, and the cyclic one has the largest area.
- Composite areas**: rectangle plus a semicircle of gives .
- Square of side minus four corner quadrants of — which make one circle — gives .
- A m square lawn minus a circular bed of leaves , costing at per square metre.
- Semicircle minus semicircle gives .
- Shade the region before subtracting, and **choose a split where every radius is a multiple of .
- Areas and perimeters need separate working**, even within one question.
Trace a bangle on paper, measure its radius, then work out the area of the quarter you would cut off and check it against the whole — the fraction should be exactly one quarter.
- gives ; gives and .
- Area gives , so ; area gives .
- Ring: outer m, inner m gives , costing at per square metre.
- Doubling the radius quadruples the area ( to ) while only doubling the circumference.
- **Area is , circumference is — the square units tell them apart.
- Sector**: . , gives ; , gives .
- Minor segment = sector - triangle. For , : , with the triangle .
- For , the triangle is equilateral, , so the segment is about .
- Major segment = sector + triangle: , and rebuilds the circle.
- Brahmagupta's formula for a cyclic quadrilateral: and .
- Sides : , brackets , so .
- A rectangle gives , matching ; a square of side gives .
- Sides : , brackets , so .
- **Setting turns Brahmagupta's formula into Heron's — so the triangle case is contained in it.
- The quadrilateral must be cyclic: four fixed sides hinge into many shapes, and the cyclic one has the largest area.
- Composite areas**: rectangle plus a semicircle of gives .
- Square of side minus four corner quadrants of — which make one circle — gives .
- A m square lawn minus a circular bed of leaves , costing at per square metre.
- Semicircle minus semicircle gives .
- Shade the region before subtracting, and **choose a split where every radius is a multiple of .
- Areas and perimeters need separate working**, even within one question.
Trace a bangle on paper, measure its radius, then work out the area of the quarter you would cut off and check it against the whole — the fraction should be exactly one quarter.