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Slices of a circle
A STORY
Folds from the centre
Bhavisha folds a large paper circle in half. Then Bhavisha folds the half into three equal parts and opens it out on the worktable. Straight creases now run from the centre to the rim.
“Every crease starts at the centre,” Bhavisha says. “The paper between two creases is a sector.”
Deo holds the scissors ready by one slice. “Three equal parts of a half,” Deo says. “So this slice is one sixth of the whole circle.”
“And the angle at its point?” Bhavisha asks.
“One sixth of $360^\circ$, so $60^\circ$,” Deo says. “And one sixth of the paper sits inside it.”
“Now suppose a slice had an angle of $50^\circ$,” Bhavisha says. “No way of folding gives you that one.”
Deo turns the scissors slowly. The creases do not help with this slice.
How would you find the area of a sector from its angle alone?
Fold in half, then in three: every crease starts at the centre, and one slice is one sixth.
A circle can be cut into a smaller piece in two ways. This chapter uses both.
Let us picture a circular park. Draw two radii from its centre to two points on the boundary. The piece they cut off is a sector. Its area is the same fraction of the park’s area as the angle between the radii is of a full turn. That fraction is $\theta/360^\circ$. So the sector’s area is $(\theta/360^\circ) \cdot \pi r^2$. The arc is the curved edge along the park’s own boundary. Its length is the same fraction, applied to the boundary instead of the area: $(\theta/360^\circ) \cdot 2 \pi r$.
Now let us draw a single straight fence across the park, joining two points on the boundary directly. No radii are drawn at all. That fence cuts off a segment, not a sector. A segment is bounded by a straight chord and an arc, never by two radii. Its area is never read off a formula of its own. It is the sector built on the same arc, with the triangle formed by the two radii and the chord taken away.
One circle, two kinds of piece: a sector is bounded by two radii, a segment by one chord. Let’s work out exactly how much of the circle each one covers.
Above, a sector is cut by two radii; below, a segment is cut by one chord, both closed by an arc.
A SCHOLAR INDIA REMEMBERS
Before any formula, let us estimate. A circle of radius $7$ cm fits inside a square of side $14$ cm. So the circle’s area is less than $14 \times 14 = 196$ cm². A quarter of the circle is a $90^\circ$ sector. Its area must be less than $196 ÷ 4 = 49$ cm². Keep that bound in mind as you work.
Aryabhata · the hoopoe
A circle of radius 7 sits inside a square of side 14, and its quarter sector sits inside a quarter of that square.
Check yourself
A sector of a circle is the part cut off by
a chord and the arc it cuts off
two chords meeting at a point on the circle
two tangents drawn from an external point
two radii and the arc between them
Check your answer
a chord and the arc it cuts off — A chord and its arc bound a segment — a sector’s boundary uses radii, not a chord.
two chords meeting at a point on the circle — Two chords meeting on the circle do not bound a sector; a sector always starts at the centre.
two tangents drawn from an external point — Tangents from an external point are a different construction entirely — a sector needs two radii.
✓ two radii and the arc between them — (D) A sector’s boundary is exactly the two radii and the arc they enclose.
A segment is the sector with the triangle formed by the two radii and the chord taken away.
Check yourself
A segment’s area follows from its sector’s area because a segment is
that sector with its triangle removed
that sector with its area doubled
twice the sector’s arc length
the sector with its central angle halved
Check your answer
✓ that sector with its triangle removed — (A) A segment is the sector minus the triangle its two radii make with the chord.
that sector with its area doubled — Nothing in the definition doubles the sector’s area — a piece is removed, not added.
twice the sector’s arc length — Arc length and area are different quantities — the segment’s AREA comes from the sector’s AREA.
the sector with its central angle halved — The central angle stays the same; it is the triangle’s area that is taken away, not the angle.
Why can a segment’s area be written as its sector’s area minus a triangle’s area?
the triangle sits inside that very same sector
both shapes happen to use the same radius r
the chord and the arc always have equal length
the central angle is always 90° in this chapter
Check your answer
✓ the triangle sits inside that very same sector — (A) The triangle sits inside the sector on the same two radii and chord, so subtracting it leaves precisely the segment’s region.
both shapes happen to use the same radius r — Sharing a radius is necessary but not sufficient — the reason is that the triangle is the exact leftover piece inside the sector.
the chord and the arc always have equal length — A chord and its arc are almost never equal in length — that equality plays no part in the subtraction.
the central angle is always 90° in this chapter — The subtraction holds for any central angle θ, not only 90°.
The four options for what cuts off a sector, each drawn on its own circle.
A whole circle, which you can already measure, and one slice of it, which this chapter measures.
You already know how to measure a whole circle. Now measure a slice of one. Try each check below.
Area of a circle (Class 9, Ganita Manjari, page 143). Here: a sector’s area is a fraction of $\pi r^2$, the whole circle’s area. Check: A circle of radius $4$ has area $\pi \cdot 4^2 = 16 \pi$.
Length of an arc (Class 9, Ganita Manjari, page 125). Here: an arc’s length is the same fraction, applied to $2 \pi r$ instead. Check: A circle of radius $14$ has circumference $2 \cdot \pi \cdot 14 = 28 \pi$.
Area of a triangle (Class 8, Ganita Prakash Part 2, page 154). Here: a segment’s area subtracts this triangle from its sector. Check: A triangle with base $6$ and height $4$ has area $1/2 \cdot 6 \cdot 4 = 12$.
Perimeter of a shape (Class 9, Ganita Manjari, page 119). Here: a sector’s perimeter adds its two radii to the arc’s length. Check: A shape with sides $3, 4, 5$ has perimeter $3 + 4 + 5 = 12$.
*$\pi$ is irrational* (Class 9, Ganita Manjari, page 123). Here: every area here uses $\pi \approx 3.14$, a rounded value. Check: $22/7 \approx 3.14$, both approximations to $\pi$.
If any of these felt new, read the page named before going on.
Four panels show a circle’s area, an arc’s length, a triangle’s area and a shape’s perimeter.
A sector of a circle is the region enclosed by two radii and the arc between them: two straight cuts from the centre, plus the curved edge trapped between them.
Let us draw two radii on a circle: they divide its interior into two regions, not one. The smaller of the two is the minor sector, and the larger is the major sector. The same two radii, read round the circle in two directions, give both.
The two sectors together make up the whole circle, so their angles always add to $360^\circ$. We can check this: a minor sector’s angle of $90^\circ$ leaves exactly $270^\circ$ for the major sector cut by the same two radii.
Unless the two radii sit exactly opposite each other, giving two equal semicircular sectors, one of the two is always smaller. There is no tie to resolve.
Your turn: a minor sector’s angle is $130^\circ$. What is the major sector’s angle? (Answer: $230^\circ$, since $360^\circ - 130^\circ = 230^\circ$.)
The shaded wedge is a sector, closed off by an arc between two tick-marked radii from the circle’s centre.
Check yourself
Which of these is a sector of a circle?
two radii and the arc between them
the region enclosed by a chord and an arc
the straight line joining two points on the circle
a line touching the circle at exactly one point
Check your answer
✓ two radii and the arc between them — (A) A sector is precisely the region two radii and the arc between them enclose.
the region enclosed by a chord and an arc — That description is a segment’s — a sector’s boundary uses radii, not a chord.
the straight line joining two points on the circle — A chord is a line segment, not a two-dimensional region — a sector is an area.
a line touching the circle at exactly one point — That is a tangent’s definition — it touches the circle once and encloses no region at all.
The same two radii cut a circle into two sectors, a minor one and a major one.
Check yourself
Of the two sectors the same two radii cut from a circle, the minor sector is told apart from the major sector by
having a different radius from the major sector
being the smaller of the two regions
always containing the circle’s centre
having the shorter perimeter, whichever sector that is
Check your answer
having a different radius from the major sector — Both sectors share the very same radius r — only their angle and area differ.
✓ being the smaller of the two regions — (B) Minor and major simply name the smaller and larger of the two regions the same two radii cut off.
always containing the circle’s centre — Which sector contains the centre is not how minor and major are defined — size alone decides it.
having the shorter perimeter, whichever sector that is — Minor and major are defined by area, and a shorter perimeter is not what the terms mean.
A circle’s sector has angle 40°. The major sector cut by the same two radii has angle
80°
140°
400°
320°
Check your answer
80° — Doubling 40° gives 80°, but the major sector’s angle is 360° minus 40°, not twice it.
140° — The two sectors’ angles add to a full 360°, not to a straight 180°.
400° — 360° and 40° are added here by mistake — the major angle is the DIFFERENCE, 360°−40°.
✓ 320° — (D) The minor and major sectors’ angles always add to 360°, so 360°−40°=320°.
A sector of angle 40° and the major sector cut by the same two radii.
A segment of a circle is the region enclosed by a chord and the arc it cuts off. That is one straight cut, not two. No radii are drawn anywhere.
Naming the chord is the fastest way to tell a segment from a sector at a glance. A sector always has two straight edges meeting at the centre. A segment has exactly one straight edge. That edge touches the centre only when the chord happens to be a diameter.
Let us draw a chord on a circle. It divides the interior into two segments: the smaller is the minor segment, the larger the major segment. A chord through the centre is different: it cuts the circle into two equal semicircles, and neither one is minor or major. This is the same naming pattern a sector uses, carried over to a different pair of shapes.
A sector question names two radii. A segment question names one chord. Read the word before you draw anything.
Your turn: a straight fence crosses a field, with no line to the centre. What shape does it cut off? (Answer: a segment, since only a chord and an arc bound it, with no radii at all.)
The shaded region is a segment, closed off by a chord and the arc above it, with no line drawn to the circle’s centre.
Check yourself
A segment of a circle is the region enclosed by
two radii and an arc
two chords crossing inside the circle
a tangent and a radius drawn to the point of contact
a chord and the arc it cuts off
Check your answer
two radii and an arc — Two radii and an arc bound a sector — a segment’s boundary is a chord, not radii.
two chords crossing inside the circle — A segment needs only one chord and its arc — no second chord is involved.
a tangent and a radius drawn to the point of contact — A tangent and its radius meet at a right angle in a different construction — no segment is formed there.
✓ a chord and the arc it cuts off — (D) A segment is bounded by a chord and the arc that chord cuts off.
A single chord cuts a circle into two segments, a minor one and a major one.
Check yourself
Of the two regions a single chord cuts from a circle, the smaller one is called the
minor segment
major segment
minor sector
triangle
Check your answer
✓ minor segment — (A) The smaller of the two chord-cut regions is, by definition, the minor segment.
major segment — The major segment is the LARGER of the two regions — the question asks for the smaller.
minor sector — A sector is bounded by radii, not a chord — a single chord only ever produces segments.
triangle — The triangle is a separate piece used to compute the segment’s area, not the segment itself.
Unlike a sector’s perimeter, a segment’s perimeter never includes
the chord
two radii
the arc
the circle’s centre
Check your answer
the chord — The chord IS part of a segment’s boundary — it is the radii that are missing, not the chord.
✓ two radii — (B) A segment’s boundary is the chord and the arc — no radius is ever part of it.
the arc — The arc IS part of a segment’s boundary too — only the radii are absent.
the circle’s centre — The centre is a point, not a boundary length, and no perimeter — sector or segment — ever includes it.
The four options for what encloses a segment, each drawn on its own circle.
A sector’s area is a fraction of the whole circle’s area. The sector’s own angle decides which fraction, and nothing else.
A full turn measures $360^\circ$, and a circle of radius $r$ has area $\pi r^2$. *A sector of angle $\theta$ takes the same fraction of that area as $\theta$ is of $360^\circ$.* This gives sector area $(\theta/360^\circ) \cdot \pi r^2$.
Let us try it on a quarter turn: $90^\circ$ takes $90/360 = 1/4$ of the circle’s area. A sector with $\theta = 180^\circ$ is a semicircle. It needs no separate rule: the general formula already gives the expected $(1/2) \cdot \pi r^2$.
The angle always sits over $360^\circ$ in this fraction, never over $180^\circ$ or any other number. Check it on any angle you like: the same fraction rule holds every time.
Your turn: a sector has radius $8$ cm and angle $45^\circ$. Using $\pi \approx 3.14$, what is its area? (Answer: $25.12$ cm², since $(45/360) \cdot 3.14 \cdot 8^2 = (1/8) \cdot 3.14 \cdot 64 = 25.12$.)
A circle cut into eight equal 45 degree sectors has one shaded sector of area 25.12 square centimetres, one eighth of the whole.
A 72 degree sector of radius 3 takes one fifth of the circle, with its area and arc length shown below.
A SCHOLAR INDIA REMEMBERS
A $60^\circ$ sector is what part of the whole circle? Count it. $360 ÷ 60 = 6$, so it is one sixth. With radius $7$ cm, the circle’s area is $22/7 \times 7 \times 7 = 154$ cm². One sixth of $154$ is about $25.7$ cm². Check your formula gives the same.
Aryabhata · the hoopoe
Check yourself
The area of a sector with angle θ in a circle of radius r is
$(\theta/360) \cdot 2 \pi r$
$(\theta/180) \pi r^2$
$\pi r^2 - \theta$
$(\theta/360) \pi r^2$
Check your answer
$(\theta/360) \cdot 2 \pi r$ — $(\theta/360) \cdot 2 \pi r$ gives the arc LENGTH, not the sector’s area.
$(\theta/180) \pi r^2$ — A full circle is 360°, not 180°, so the fraction must be $\theta/360$.
$\pi r^2 - \theta$ — The sector’s area is a FRACTION of $\pi r^2$, not the full circle’s area with $\theta$ subtracted.
✓ $(\theta/360) \pi r^2$ — (D) A sector’s area is the fraction $\theta/360$ of the whole circle’s area $\pi r^2$.
A sector’s area is the same fraction of the circle’s area as its angle is of 360°.
Check yourself
A sector has radius 7 cm and angle 90°. Using $\pi = 22/7$, its area is
154 cm²
19.25 cm²
38.5 cm²
77 cm²
Check your answer
154 cm² — 154 cm² is the WHOLE circle’s area — the sector is only one-quarter of it at 90°.
19.25 cm² — Dividing 154 by 8 treats the angle as 45° — at 90° the divisor is 4, not 8.
77 cm² — Dividing 154 by 2 treats the angle as 180°, not the given 90°.
Doubling a sector’s angle θ, while keeping the radius r fixed, does what to its area?
quadruples the area
doubles the area
leaves the area unchanged
halves the area
Check your answer
quadruples the area — Quadrupling happens when the RADIUS doubles (since area depends on $r^2$) — doubling $\theta$ only doubles the area.
✓ doubles the area — (B) Area is directly proportional to $\theta$ when r is fixed, so doubling $\theta$ doubles the area.
leaves the area unchanged — The sector’s area is directly proportional to $\theta$ — changing $\theta$ does change the area.
halves the area — A larger angle gives a LARGER sector, not a smaller one — the relationship is direct, not inverse.
A sector of radius 7 cm and angle 90°, and a sector with angle θ doubled, radius r fixed.
▲CONCEPT
An arc’s length follows the same fractional idea as a sector’s area. It applies to the circle’s boundary, not its interior.
The whole circle’s circumference is $2 \pi r$. *An arc of angle $\theta$ is the same fraction of that circumference as $\theta$ is of $360^\circ$.* So arc length works out to $(\theta/360^\circ) \cdot 2 \pi r$.
Sector area and arc length share the same fraction, $\theta/360^\circ$. Only what the fraction multiplies changes: the circle’s area for a sector, the circle’s circumference for an arc. Let us compare the two: one number carries across both.
A semicircle’s arc, at $\theta = 180^\circ$, comes to exactly half the circumference, $\pi r$. This is the familiar fact that a semicircle’s curved edge is half the boundary, reached here from the general formula, not memorised on its own.
Your turn: an arc has radius $7$ cm and angle $45^\circ$. Using $\pi \approx 22/7$, what is its length? (Answer: $5.5$ cm, since $(45/360) \cdot 2 \cdot (22/7) \cdot 7 = (1/8) \cdot 44 = 5.5$.)
Your turn: an arc of radius 7 cm and angle 45°.
A SCHOLAR INDIA REMEMBERS
Is the arc longer or shorter than the straight chord below it? The arc bends, so it is longer. Take a $60^\circ$ sector of radius $7$ cm. Its chord is $7$ cm, because the triangle is equilateral. Its arc is $1/6 \times 2 \times 22/7 \times 7 = 22/3$, about $7.33$ cm. The arc is a little longer, as it should be.
Aryabhata · the hoopoe
A 60 degree sector of radius 7 has a chord of 7 and an arc of 7.33, the arc longer than the chord.
Check yourself
The arc length of a sector with angle θ in a circle of radius r is
$(\theta/360) \pi r^2$
$(\theta/180) \cdot 2 \pi r$
$(\theta/360) \cdot 2 \pi r$
$2 \pi r - \theta$
Check your answer
$(\theta/360) \pi r^2$ — $(\theta/360) \pi r^2$ gives the sector’s AREA, not its arc length.
$(\theta/180) \cdot 2 \pi r$ — A full turn is 360°, so the fraction must be $\theta/360$, not $\theta/180$.
✓ $(\theta/360) \cdot 2 \pi r$ — (C) An arc’s length is the fraction $\theta/360$ of the whole circumference $2 \pi r$.
$2 \pi r - \theta$ — Arc length is a FRACTION of the circumference, not the circumference minus $\theta$.
A sector of radius 21 cm and angle 60°: its arc is 60/360 of the circumference.
Check yourself
A sector has radius 21 cm and angle 60°. Using $\pi = 22/7$, its arc length is
132 cm
44 cm
11 cm
22 cm
Check your answer
132 cm — 132 cm is the WHOLE circle’s circumference — the arc is only one-sixth of it at 60°.
44 cm — Dividing 132 by 3 treats the angle as 120°, double the angle actually given.
11 cm — Dividing 132 by 12 treats the angle as 30°, half the angle actually given.
✓ 22 cm — (D) $(60/360) \cdot 2 \cdot (22/7) \cdot 21 = (1/6) \cdot 132 = 22$ cm.
Two sectors share the same angle θ, but circle B’s radius is twice circle A’s. Circle B’s arc length is
twice circle A’s arc length
four times circle A’s arc length
the same as circle A’s arc length
half of circle A’s arc length
Check your answer
✓ twice circle A’s arc length — (A) Arc length is directly proportional to r when $\theta$ is fixed, so doubling r doubles the arc length.
four times circle A’s arc length — Quadrupling is how AREA scales with $r^2$ — arc length depends on r alone, so it only doubles.
the same as circle A’s arc length — Arc length depends directly on r — a larger radius at the same angle gives a longer arc.
half of circle A’s arc length — A LARGER radius gives a LONGER arc, not a shorter one — the relationship is direct, not inverse.
Two sectors with the same angle θ, circle B’s radius twice circle A’s.
A circle’s two sectors, cut by the same two radii, are not two separate calculations. The second is always one subtraction away from the first.
*The major sector’s area is the whole circle’s area, $\pi r^2$, minus the minor sector’s area.* Its angle is $360^\circ - \theta$, where $\theta$ is the minor sector’s own angle.
Let us take a minor sector of angle $72^\circ$. It leaves a major sector of angle $360^\circ - 72^\circ = 288^\circ$, cut by the identical two radii. Its area is $\pi r^2$ minus the minor sector’s own area.
Find the minor sector first, then subtract. This is quicker and safer than working the major sector’s own fraction, $(360^\circ - \theta)/360^\circ$, directly. We still get the same answer, with one fewer number to carry.
Your turn: a minor sector has area $45$ cm² inside a circle of radius $18$ cm. Using $\pi \approx 3.14$, what is the major sector’s area? (Answer: $972.36$ cm², since the circle’s area is $3.14 \cdot 18^2 = 1017.36$ cm², and $1017.36 - 45 = 972.36$.)
Two panels cut the same circle, one by two radii and one by a chord, each into a minor piece and a major piece.
Check yourself
If a sector’s angle is θ, the major sector cut by the same two radii has angle
$180^\circ - \theta$
$2 \cdot \theta$
$360^\circ + \theta$
$360^\circ - \theta$
Check your answer
$180^\circ - \theta$ — A full circle is $360^\circ$, not $180^\circ$, so the complement is $360^\circ - \theta$.
$2 \cdot \theta$ — Doubling $\theta$ has no basis here — the major angle is $360^\circ$ minus $\theta$.
$360^\circ + \theta$ — The major angle is the full circle MINUS the minor angle, not plus it.
✓ $360^\circ - \theta$ — (D) The minor and major sectors’ angles always add to a full $360^\circ$.
If a sector’s angle is θ, the major sector cut by the same two radii has 360° − θ.
Check yourself
A circle of radius 7 cm (using $\pi = 22/7$) has a sector of angle 90°. The major sector’s area is
38.5 cm²
154.0 cm²
115.5 cm²
77.0 cm²
Check your answer
38.5 cm² — 38.5 cm² is the MINOR sector’s area — the major sector is the rest of the circle.
154.0 cm² — 154 cm² is the WHOLE circle; the major sector still needs the minor sector subtracted from it.
✓ 115.5 cm² — (C) Full circle $= 154$ cm²; minor sector $= 38.5$ cm²; major sector $= 154 - 38.5 = 115.5$ cm².
77.0 cm² — A major sector is only half the circle when the minor angle is 180° — here it must be computed from the actual 90°.
Why does the major sector’s area always equal the full circle’s area minus the minor sector’s area, whatever θ is?
the two sectors together make up the whole circle
the minor and major sectors are always congruent
a chord always splits the circle exactly in half
area is always proportional to the radius, not the angle
Check your answer
✓ the two sectors together make up the whole circle — (A) The minor and major sectors partition the circle completely — together, once, with nothing left over and nothing double-counted.
the minor and major sectors are always congruent — The two sectors are congruent only in the special case $\theta = 180^\circ$ — in general they have different areas.
a chord always splits the circle exactly in half — No chord is involved in a sector at all — its boundary is two radii and an arc.
area is always proportional to the radius, not the angle — Sector area is proportional to the ANGLE $\theta$ (for a fixed radius) — that proportionality is exactly why the two areas add to the full circle.
A circle of radius 7 cm with a 90° sector; the major sector is the rest.
A sector and a segment are bounded by different edges. Let us look at each in turn. Their perimeters are built from different pieces entirely.
A sector’s boundary is two straight radii plus one curved arc. *Its perimeter is $2 r + (\theta/360^\circ) \cdot 2 \pi r$*: two radii, each of length $r$, plus the arc length already found.
A segment’s boundary looks similar at a glance: one straight edge, one curved edge. Let us look again: the straight edge is a chord, never a radius, and there is only one of it. A segment’s perimeter is the chord’s length plus the arc’s length, with no radii counted at all.
Counting two radii into a segment’s perimeter is a common slip. So is leaving the two radii out of a sector’s perimeter. Check which figure the question names before writing down any lengths: that one habit catches both mistakes.
Your turn: a sector has radius $5$ cm and angle $144^\circ$. Using $\pi \approx 3.14$, what is its perimeter? (Answer: $22.56$ cm, since the arc is $(144/360) \cdot 2 \cdot 3.14 \cdot 5 = 12.56$ cm, and $2 \cdot 5 + 12.56 = 22.56$.)
A sector’s perimeter is two radii plus an arc, while a segment’s perimeter is just a chord plus an arc, with radii struck through.
Check yourself
The perimeter of a sector consists of
the chord plus the arc
the two radii only
the arc only
two radii plus the arc
Check your answer
the chord plus the arc — Chord plus arc is a SEGMENT’s perimeter — a sector’s perimeter uses radii, not a chord.
the two radii only — The arc is part of the boundary too — leaving it out understates the perimeter.
the arc only — The two radii are straight sides of the sector’s boundary and must be included.
✓ two radii plus the arc — (D) A sector’s perimeter is its two straight radii plus the curved arc joining their ends.
A student says a segment’s perimeter equals its chord plus its two radii. What is wrong with this claim?
nothing is wrong — that is the correct perimeter
the perimeter should use one radius, not two
a segment has no radii — just the chord and arc
the chord should be counted twice
Check your answer
nothing is wrong — that is the correct perimeter — A segment’s perimeter never includes a radius — the claim mixes up a sector’s perimeter with a segment’s.
the perimeter should use one radius, not two — Even ONE radius does not belong in a segment’s perimeter — the boundary is chord plus arc only.
✓ a segment has no radii — just the chord and arc — (C) A segment is bounded only by a chord and an arc; no radius is ever part of it.
the chord should be counted twice — The chord is counted once, like any straight side — doubling it has no basis.
A sector of radius 10 cm and arc length 12 cm.
Check yourself
A sector has radius 10 cm and arc length 12 cm. Its perimeter is
32 cm
22 cm
12 cm
20 cm
Check your answer
✓ 32 cm — (A) Perimeter $= 2r + \text{arc} = 2 \cdot 10 + 12 = 32$ cm.
22 cm — A sector’s perimeter needs BOTH radii — $10+12=22$ leaves one radius out.
12 cm — 12 cm is just the arc — the two radii must be added to it for the full perimeter.
20 cm — 20 cm is just the two radii — the arc must be added for the full perimeter.
The wedge splits into an outlined triangle, made by the two radii and the chord, and the shaded segment that remains beside it.
A segment sits inside its own sector, sharing the same arc. Its area is never found from a formula of its own. It is read off the sector’s area instead.
Draw the chord bounding the segment, and the two radii bounding the matching sector. Together, the two radii and the chord enclose a triangle, sitting inside the sector but outside the segment. A minor segment’s area is its sector’s area minus that triangle’s area. The major segment works differently — its own section, after this one, subtracts the minor segment from the whole circle instead. Picture the whole minor sector, with the triangular corner cut away.
For a radius $r$ and central angle $\theta$, this gives segment area equal to sector area minus triangle area. You work both out for the same $r$ and the same $\theta$. Skip the triangle, and the segment comes out too large, by exactly the triangle’s own area.
The triangle’s area still needs a formula of its own. We work that out next.
Your turn: a sector has area $60$ cm² and the triangle inside it has area $42$ cm². What is the segment’s area? (Answer: $18$ cm², since $60 - 42 = 18$.)
The brooch’s sector is 154 cm² and its triangle 84 cm²; the silver segment is the rest.
Check yourself
A segment’s area, for the same radius and central angle as its sector, equals
sector’s area minus triangle’s area
the sector’s area plus the triangle’s area
the sector’s area alone
the triangle’s area alone
Check your answer
✓ sector’s area minus triangle’s area — (A) A segment’s area is exactly its sector’s area with the triangle inside it taken away.
the sector’s area plus the triangle’s area — The triangle is REMOVED from the sector to leave the segment, not added to it.
the sector’s area alone — The sector alone still includes the triangle — the segment is what remains once that triangle is subtracted.
the triangle’s area alone — The triangle is the piece taken AWAY — the segment is the sector’s area after that removal, not the triangle itself.
A sector has area 28.26 cm² and its triangle has area 18 cm². The segment’s area is
46.26 cm² — $28.26+18=46.26$ adds the triangle instead of removing it — the segment needs a subtraction.
28.26 cm² — 28.26 cm² is the sector’s area — the triangle still needs to be subtracted from it.
18 cm² — 18 cm² is the triangle, the piece removed — not the segment, which is what remains.
A sector has area 28.26 cm² and its triangle has area 18 cm².
Check yourself
Why can a segment’s area never come out negative, given segment = sector − triangle?
area is always defined as a positive quantity, by convention
$\pi$ is always a positive number
the chord is always shorter than the arc it cuts off
the triangle is always smaller than its own sector
Check your answer
area is always defined as a positive quantity, by convention — That area is non-negative in general does not, by itself, explain why THIS particular subtraction stays non-negative — the reason is the triangle’s containment inside the sector.
$\pi$ is always a positive number — $\pi$’s sign has nothing to do with why the sector minus the triangle stays non-negative.
the chord is always shorter than the arc it cuts off — That the chord is shorter than its arc is true but irrelevant — the reason lies in the triangle sitting inside the sector, not in chord-versus-arc length.
✓ the triangle is always smaller than its own sector — (D) The triangle sits entirely inside the sector, so its area can never exceed the sector’s — the subtraction always leaves a non-negative result.
A decorative brooch’s silver segment is cut by a chord where the sector’s area is 154 cm² and the triangle’s area is 84 cm². The silver segment’s area is
238 cm²
154 cm²
84 cm²
70 cm²
Check your answer
238 cm² — $154+84=238$ adds the triangle rather than removing it from the sector.
154 cm² — 154 cm² is the full sector — the triangle piece must still be subtracted to get the silver segment.
84 cm² — 84 cm² is the triangle being cut away, not the silver segment left behind.
Halving the angle to 45 degrees gives a triangle of area 84.5, but using 90 degrees instead flattens it to a line.
The triangle that a segment’s area formula subtracts away has a fixed shape. It is worth working out once, not re-derived every time it appears.
Let us draw it: two of its sides are radii, each of length $r$, with the sector’s own angle $\theta$ trapped between them. This is an isosceles triangle by construction, for any angle $\theta$. *Drop a perpendicular from the centre straight down to the chord, and it bisects both the chord and the angle $\theta$.* This splits the isosceles triangle into two identical right triangles, each carrying the half-angle $\theta/2$.
In either right triangle, the perpendicular is the side adjacent to $\theta/2$, and the half-chord is the side opposite it. The radius $r$ is the hypotenuse. You now have perpendicular length $r \cdot \cos(\theta/2)$ and half-chord length $r \cdot \sin(\theta/2)$.
Doubling one right triangle’s area gives the whole triangle’s area: $r^2 \cdot \sin(\theta/2) \cdot \cos(\theta/2)$. Halve the angle first: every calculation with this triangle depends on that one step. Skip it, and $\theta$ itself ends up inside the ratios instead of $\theta/2$. That describes a different triangle altogether.
Your turn: a segment’s central angle is $90^\circ$ and its radius is $13$ cm. What is the triangle’s area? (Answer: $84.5$ cm², since the half-angle is $45^\circ$, and $13^2 \cdot \sin(45^\circ) \cdot \cos(45^\circ) = 169 \cdot 1/2 = 84.5$.)
A segment’s triangle with radius 12 cm, central angle 60° and half-angle 30°.
Check yourself
The triangle bounded by a segment’s two radii and its chord has
three unequal sides
a right angle between the two radii, always
two sides equal to the chord’s length
two radii as its equal sides
Check your answer
three unequal sides — Two of the three sides are both equal to the radius r — the triangle is isosceles, not scalene.
a right angle between the two radii, always — The angle between the two radii is whatever $\theta$ the sector has — 90° only when that particular sector is chosen.
two sides equal to the chord’s length — It is the two RADII that are equal, not two sides equal to the chord — the chord is the third, usually different, side.
✓ two radii as its equal sides — (D) The two radii are the triangle’s equal sides, with the sector’s own central angle θ held between them.
The perpendicular drawn from the centre to a segment’s chord
bisects the chord but not the angle θ
bisects both the chord and the angle θ
bisects the angle θ but not the chord
extends the chord into a full diameter
Check your answer
bisects the chord but not the angle θ — The same perpendicular bisects the angle too — it is not just the chord that splits evenly.
✓ bisects both the chord and the angle θ — (B) That single perpendicular splits the isosceles triangle into two congruent right triangles, bisecting both the chord and θ.
bisects the angle θ but not the chord — The same perpendicular bisects the chord too — it is not just the angle that splits evenly.
extends the chord into a full diameter — The perpendicular meets the chord at a right angle; it does not extend the chord into a diameter.
A perpendicular from centre O meets the chord AB at a right angle, splitting the triangle by the half-angle theta over two.
Check yourself
In a segment’s triangle with radius r = 12 cm and central angle θ = 60° (half-angle 30°), the half-chord has length
12 cm
6 cm
10.39 cm
3 cm
Check your answer
12 cm — 12 cm is just the radius — the half-chord needs it multiplied by $\sin(30^\circ) = 1/2$.
✓ 6 cm — (B) Half-chord $= r \cdot \sin(30^\circ) = 12 \cdot (1/2) = 6$ cm.
10.39 cm — 10.39 cm ($12 \cdot \cos(30^\circ)$) is the PERPENDICULAR’s length — the half-chord uses $\sin(30^\circ)$, not $\cos(30^\circ)$.
3 cm — $12 \cdot \sin(30^\circ) = 6$ cm already — halving it a second time to 3 cm is not part of the formula.
Check yourself
For a segment problem with central angle θ = 90°, the triangle’s half-angle used in the area formula is
90°
45°
180°
22.5°
Check your answer
90° — 90° is the FULL central angle — the half-angle used in the triangle is $90^\circ/2 = 45^\circ$.
✓ 45° — (B) The half-angle is always $\theta/2$, so at $\theta=90^\circ$ it is $45^\circ$.
180° — Doubling gives 180°, but the triangle needs the central angle HALVED, not doubled.
22.5° — Halving 90° once gives 45° — halving it again to 22.5° is one step too many.
At 30° and 60°, sine and cosine swap between them.
Check yourself
At half-angle 60°, the standard values are
$\sin(60^\circ) = 1/2$ and $\cos(60^\circ) = \sqrt{3}/2$
$\sin(60^\circ) = \sqrt{2}/2$ and $\cos(60^\circ) = \sqrt{2}/2$
$\sin(60^\circ) = 1$ and $\cos(60^\circ) = 0$
$\sin(60^\circ) = \sqrt{3}/2$ and $\cos(60^\circ) = 1/2$
Check your answer
$\sin(60^\circ) = 1/2$ and $\cos(60^\circ) = \sqrt{3}/2$ — Those are the values at $30^\circ$, with sine and cosine swapped — at $60^\circ$ it is the other way round.
$\sin(60^\circ) = \sqrt{2}/2$ and $\cos(60^\circ) = \sqrt{2}/2$ — $\sqrt{2}/2$ for both is the pair at $45^\circ$, not $60^\circ$.
$\sin(60^\circ) = 1$ and $\cos(60^\circ) = 0$ — $\sin=1, \cos=0$ is the pair at $90^\circ$, not $60^\circ$.
✓ $\sin(60^\circ) = \sqrt{3}/2$ and $\cos(60^\circ) = 1/2$ — (D) These are the standard sine and cosine values at $60^\circ$.
This chapter’s segment problems only use central angles 60°, 90° and 120° because
those are the only angles for which a triangle can be drawn inside a circle
their half-angles have standard sine and cosine values already known
a segment does not exist for any other central angle
the chord is only defined at those three angles
Check your answer
those are the only angles for which a triangle can be drawn inside a circle — A triangle can be drawn inside a segment for ANY central angle — the restriction is about which half-angles have memorised trig values, not about geometry.
✓ their half-angles have standard sine and cosine values already known — (B) 30°, 45° and 60° are exactly the half-angles with standard trigonometric values, which is what keeps the triangle’s area computable by hand.
a segment does not exist for any other central angle — Segments exist at every central angle — this chapter simply limits its NUMERIC problems to the three with clean trig values.
the chord is only defined at those three angles — A chord can be drawn for any central angle — the restriction here is only about which angles keep the arithmetic clean.
Three panels hold the two radii equal and open the angle between them to 60, 90 and 120 degrees, halving each for the triangle.
Finding this triangle’s area means finding a sine and a cosine at the segment’s half-angle.
Every segment problem here uses a central angle of $60^\circ$, $90^\circ$ or $120^\circ$. So the half-angle is always $30^\circ$, $45^\circ$ or $60^\circ$: three angles worth knowing outright, rather than looked up each time.
At $30^\circ$, $\sin(30^\circ) = 1/2$ and $\cos(30^\circ) = \sqrt{3}/2$. At $45^\circ$, $\sin(45^\circ) = \cos(45^\circ) = \sqrt{2}/2$. At $60^\circ$, $\sin(60^\circ) = \sqrt{3}/2$ and $\cos(60^\circ) = 1/2$. Let us look at the two outer pairs: sine and cosine swap between them, and that pattern makes the three pairs easier to hold as one rule.
A central angle outside $60^\circ$, $90^\circ$ and $120^\circ$ would need a calculator rather than memory, and none appears here. Once these three pairs are known, no calculator is ever needed for a segment like these.
Your turn: a segment’s central angle is $120^\circ$ and its radius is $9$ cm. Using $\sqrt{3} \approx 1.73$, what is the triangle’s area? (Answer: $\approx 35.03$ cm², since the half-angle is $60^\circ$, and $9^2 \cdot \sin(60^\circ) \cdot \cos(60^\circ) \approx 81 \cdot 0.865 \cdot 0.5 \approx 35.03$.)
Your turn: a segment’s central angle is 120° and its radius is 9 cm.
A student claims: “A segment’s area equals its sector’s area — subtracting the triangle is optional.” At r = 10 cm, θ = 90° (sector = 78.5 cm², triangle = 50 cm²), what is wrong?
the segment’s area is 28.5 cm², not 78.5 cm² — the triangle must always be subtracted from a minor sector
nothing is wrong — 78.5 cm² is correct
the segment’s area is 128.5 cm², since the triangle should be added instead
the segment’s area is 285.5 cm², the major segment’s area for this same circle, not the minor segment asked about
Check your answer
✓ the segment’s area is 28.5 cm², not 78.5 cm² — the triangle must always be subtracted from a minor sector — (A) A minor segment is always its sector minus its triangle: $78.5 - 50 = 28.5$ cm², never the sector’s own 78.5 cm².
nothing is wrong — 78.5 cm² is correct — 78.5 cm² is the SECTOR’s area — the segment always needs the triangle subtracted from it first.
the segment’s area is 128.5 cm², since the triangle should be added instead — Adding the triangle is the rule for a MAJOR segment — this is a MINOR segment, which always subtracts the triangle.
the segment’s area is 285.5 cm², the major segment’s area for this same circle, not the minor segment asked about — 285.5 cm² is the MAJOR segment’s area for this same circle — the question asks about the MINOR segment, which is 28.5 cm².
A 120° minor sector and its 240° major sector share the same chord and the same triangle T.
Check yourself
A sector has angle 120° (minor) and its complementary major sector has angle 240°, both sharing the same chord and the same triangle of area T. To get the MINOR segment’s area from the MINOR sector’s area, a student should
add T to the minor sector’s area
subtract T from the minor sector’s area
subtract T from the major sector’s area
add T to the major sector’s area
Check your answer
add T to the minor sector’s area — Adding T is the rule for a MAJOR segment from a major sector — the minor segment always subtracts T.
✓ subtract T from the minor sector’s area — (B) A minor segment is always its own minor sector’s area minus the triangle T.
subtract T from the major sector’s area — The question asks about the MINOR segment, which comes from the MINOR sector, not the major one.
add T to the major sector’s area — Adding T to the major sector correctly gives the MAJOR segment — but this question asks for the MINOR segment, found the opposite way.
Radius 21 cm, chord at 120°: major sector 924 cm², triangle about 190.73 cm².
Check yourself
For a circle of radius 21 cm and a chord subtending 120°, the minor sector’s area is 462 cm², the major sector’s area is 924 cm² ($\pi = 22/7$), and the triangle’s area is about 190.73 cm². The MAJOR segment’s area, in cm², is closest to
733.27
1114.73
462.00
1386.00
Check your answer
733.27 — $924 - 190.73 = 733.27$ blindly subtracts the triangle from the major sector — the major segment ADDS the triangle instead, since that triangle already belongs to the minor sector’s side.
✓ 1114.73 — (B) A major segment is its major sector PLUS the triangle: $924 + 190.73 = 1114.73$ cm².
462.00 — 462 cm² is the MINOR sector — a different quantity entirely from the major segment being asked for.
1386.00 — 1386 cm² is the WHOLE circle — the major segment is smaller than the full circle by the minor segment’s area.
Radius 10 cm and angle 90°: major sector 235.5 cm², triangle 50 cm².
Check yourself
Asked for the major segment of a circle with radius 10 cm and angle 90° (sector = 78.5 cm², triangle = 50 cm²), a student blindly subtracts the triangle from the major sector (235.5 cm²), getting 185.5 cm². What is the actual major segment’s area?
185.5 cm²
235.5 cm²
285.5 cm²
78.5 cm²
Check your answer
185.5 cm² — 185.5 cm² came from wrongly SUBTRACTING the triangle from the major sector — the correct operation is to ADD it.
235.5 cm² — 235.5 cm² is the major SECTOR — the major SEGMENT still needs the triangle added to it.
✓ 285.5 cm² — (C) The major segment ADDS the triangle to the major sector: $235.5 + 50 = 285.5$ cm², not subtracts it.
78.5 cm² — 78.5 cm² is the MINOR sector’s area — a different quantity from the major segment asked for.
A circular tablecloth has a decorative segment marked off by a chord. To find the area OUTSIDE that segment (the rest of the tablecloth), the correct method is that this area equals
the major sector’s area minus the triangle’s area
the minor segment’s area
the full circle’s area minus the sector’s area
the major sector’s area plus the triangle’s area
Check your answer
the major sector’s area minus the triangle’s area — Subtracting the triangle from the major sector undercounts the region — a major segment always ADDS that same triangle back in.
the minor segment’s area — The minor segment IS the marked-off decoration — the rest of the tablecloth is the major segment, a different region.
the full circle’s area minus the sector’s area — Full circle minus the sector gives the major SECTOR — the major SEGMENT needs the triangle added on top of that.
✓ the major sector’s area plus the triangle’s area — (D) The area outside a minor segment is the major segment, which equals the major sector PLUS the same triangle, not minus it.
A circular tablecloth has a decorative segment marked off by a chord.
Check yourself
A designer wants the area of the leftover cloth beyond a decorative segment cut by a chord. Which quantity must be found FIRST, before the major segment’s area can be computed?
the triangle’s area, which the major segment adds to the sector
the arc length, since the major segment’s boundary needs it
the chord’s length, since area depends directly on it
nothing extra — the major sector’s area already equals the major segment’s area
Check your answer
✓ the triangle’s area, which the major segment adds to the sector — (A) The major segment’s formula needs the triangle’s area as an ingredient, added to the major sector’s area.
the arc length, since the major segment’s boundary needs it — Arc length matters for the segment’s PERIMETER, not for its area — the area calculation needs the triangle, not the arc.
the chord’s length, since area depends directly on it — The chord’s length is not itself part of the area formula — the triangle’s AREA is, computed from r and θ.
nothing extra — the major sector’s area already equals the major segment’s area — The major sector and major segment differ by exactly the triangle’s area — they are not the same quantity.
The two shapes share the same arc, but they are not the same slice.
It is tempting to read a sector’s area and call that the segment’s area too. The two shapes share the same arc. At a glance, they look like the same slice of the circle.
Let us look again: they are not the same slice. The triangle bounded by the two radii and the chord has to be subtracted first. Skip that subtraction, and the sector’s own area gets reported as the segment’s: an answer too large by exactly the triangle’s area.
A segment lies inside its own sector, so its area must come out smaller than the sector’s. An answer equal to the sector’s area means the triangle was never taken away.
Your turn: a circle has radius $20$ cm and a central angle of $90^\circ$. Using $\pi \approx 3.14$, what is the segment’s area? (Answer: $114$ cm², since the sector’s area is $314$ cm², the triangle’s area is $200$ cm², and $314 - 200 = 114$.)
Reporting the whole sector, 314, skips subtracting the triangle, 200, leaving the true segment, 114.
First try
For a chord at $90^\circ$ in a circle of radius $10$ cm, I found the sector area $= (90/360) \cdot 3.14 \cdot 10^2 = 78.5$ cm squared, and called that the segment’s area.
Second look
The segment is the sector minus the triangle the two radii make with the chord. The triangle’s area is $10^2 \cdot \sin(45^\circ) \cdot \cos(45^\circ) = 50$ cm squared, so the segment’s area is $78.5 - 50 = 28.5$ cm squared.
A segment is always the sector minus its triangle. Never stop at the sector.
The wedge is filled in two colours, one for the triangle and one for the segment, showing both pieces that make up the whole shape.
Is a segment’s area ever the same as its sector’s area?
Weaker. A learner works out the sector’s area: $(90/360) \cdot 3.14 \cdot 10^2 = 78.5$ cm². They report $78.5$ cm² as the segment’s area too.
Then they notice something wrong. A segment sits inside its sector, with the triangle sitting outside it. So the segment must come out smaller than $78.5$ cm². An answer equal to the sector’s area means the triangle was never taken away.
Stronger. The segment is the sector minus the triangle. The triangle’s area is $10^2 \cdot \sin(45^\circ) \cdot \cos(45^\circ) = 50$ cm².
So the segment’s area is $78.5 - 50 = 28.5$ cm², smaller than the sector, exactly as it must be.
The segment sits inside its sector, so its area, 28.5 cm², is smaller than 78.5 cm².
A SCHOLAR INDIA REMEMBERS
Which is bigger, the sector or the segment? Picture them. The segment is the sector with the triangle cut away. So the segment must be smaller than the sector. If your segment comes out equal to the sector, the triangle has not been taken away.
A circle of radius 28 with area 2464 has a sliver of 64 cut off, leaving the major segment, 2400.
A circle’s two sectors from the same two radii add to the whole circle. Its two segments from the same chord add to the whole circle too.
The major segment’s area is the whole circle’s area, $\pi r^2$, minus the minor segment’s area. It is the same subtraction we already used for the major sector, run here on segments instead.
Let us run the numbers: a minor segment of area $20$ cm² inside a circle of area $154$ cm² leaves a major segment of area $154 - 20 = 134$ cm². No fresh calculation is needed for the larger piece.
Finding the minor segment first stays the practical order every time. Its own formula, sector minus triangle, is direct. The major segment’s boundary, most of the circle plus the same chord, is far harder to measure directly.
Your turn: a minor segment has area $64$ cm² inside a circle of radius $28$ cm. Using $\pi \approx 22/7$, what is the major segment’s area? (Answer: $2400$ cm², since the circle’s area is $(22/7) \cdot 28^2 = 2464$ cm², and $2464 - 64 = 2400$.)
A circle of radius 10 cm has a minor segment of area 28.5 cm².
Check yourself
The major segment cut by a chord equals
circle’s area minus the minor segment’s area
the minor segment’s area plus the sector’s area
the minor segment’s area doubled
the full circle’s area minus the sector’s area
Check your answer
✓ circle’s area minus the minor segment’s area — (A) Major and minor segments cut by the same chord together make up the whole circle.
the minor segment’s area plus the sector’s area — Mixing the minor segment with the sector’s area is not the relation — the major segment comes from the FULL CIRCLE minus the minor segment.
the minor segment’s area doubled — The two segments are equal only when the chord is a diameter — in general the major segment is the circle minus the minor one, not double it.
the full circle’s area minus the sector’s area — Full circle minus the SECTOR gives the major SECTOR — the major SEGMENT needs the minor segment subtracted, not the sector.
A circle of radius 10 cm (using $\pi = 3.14$) has a minor segment of area 28.5 cm². The major segment’s area is
78.5 cm²
285.5 cm²
342.5 cm²
28.5 cm²
Check your answer
78.5 cm² — 78.5 cm² is that circle’s SECTOR area at 90° — a different quantity from the major SEGMENT being asked for.
✓ 285.5 cm² — (B) Full circle $= 314$ cm²; major segment $= 314 - 28.5 = 285.5$ cm².
342.5 cm² — $314+28.5=342.5$ adds the minor segment rather than removing it from the full circle.
28.5 cm² — 28.5 cm² is the MINOR segment — the major segment still needs to be found from the full circle.
The minor segment and the major segment cut by the same chord make the whole circle.
Check yourself
Why do the minor segment and major segment cut by the same chord always add up to exactly the circle’s full area?
segments cut by a chord are always congruent to each other
a chord always passes through the circle’s centre
the two segments cover the whole circle exactly
a sector and a segment cut from the same figure always have equal area
Check your answer
segments cut by a chord are always congruent to each other — The two segments are congruent only when the chord is a diameter — in general they have different areas.
a chord always passes through the circle’s centre — Only a diameter passes through the centre — an ordinary chord does not, and the partition argument does not depend on it.
✓ the two segments cover the whole circle exactly — (C) A single chord splits the circle into exactly two regions that share no area and leave nothing uncovered.
a sector and a segment cut from the same figure always have equal area — A sector and its own segment are never equal in area — they differ by exactly the triangle’s area.
One straight cut: 28.5 cm² and 285.5 cm² make the whole 314 cm² roti.
A sector of radius 4 cm and angle 30° has area 4.19 cm²; the major sector is the rest.
Check yourself
Continuing that sector (radius 4 cm, angle 30°, area 4.19 cm², $\pi = 3.14$), the major sector’s area is closest to
4.19 cm²
46.05 cm²
54.43 cm²
25.12 cm²
Check your answer
4.19 cm² — 4.19 cm² is the MINOR sector — the major sector is the rest of the circle.
✓ 46.05 cm² — (B) Full circle $= 50.24$ cm²; major sector $= 50.24 - 4.19 = 46.05$ cm².
54.43 cm² — $50.24+4.19=54.43$ adds the minor sector rather than removing it from the full circle.
25.12 cm² — Halving the circle assumes a 180° split — the true minor angle here is only 30°.
If that same circle’s sector angle were doubled to 60° (radius still 4 cm, $\pi = 3.14$), the new sector’s area would be
4.19 cm²
8.37 cm²
16.75 cm²
25.12 cm²
Check your answer
4.19 cm² — Doubling the angle must double the area — 4.19 cm² was the area at the ORIGINAL 30°.
✓ 8.37 cm² — (B) $(60/360) \cdot 3.14 \cdot 16 \approx 8.37$ cm², double the 4.19 cm² at 30°, before rounding.
16.75 cm² — Quadrupling is how area scales when the RADIUS doubles — doubling the ANGLE only doubles the area.
25.12 cm² — 25.12 cm² is half the full circle, as if the angle were 180°, not the 60° actually given.
A sector of radius 4 cm at 30°, then the same radius with the angle doubled to 60°.
Worked example
A sector and its major sector
sector area $\approx (30/360) \cdot 3.14 \cdot 4^2$ Substitute $\theta = 30^\circ$, $\pi \approx 3.14$ and $r = 4$ cm into the sector-area formula.
$= (30/360) \cdot 3.14 \cdot 16 \approx 4.19$ cm² $4^2 = 16$, and $30/360$ simplifies to $1/12$.
whole circle’s area $\approx 3.14 \cdot 16 = 50.24$ cm² This is the same $\pi r^2$, with no angle fraction, since the whole circle is the sector at $\theta = 360^\circ$.
major sector’s area $\approx 50.24 - 4.19 = 46.05 \approx 46.1$ cm² The major sector is the whole circle’s area minus the minor sector already found.
minor sector plus major sector $= 4.19 + 46.05 = 50.24$ cm² Check by adding the two sectors back together. The total should land on the whole circle’s area, $50.24$ cm², already found above.
30/360 is 1/12, and the major sector is the whole circle minus the minor sector.
Find a sector’s area and its major sector, from the radius and the angle.
Find the angle’s fraction Divide the angle by $360^\circ$. For $\theta = 30^\circ$, that fraction is $30/360 = 1/12$.
Scale the circle’s area Multiply that fraction by $\pi r^2$. For $r = 4$ cm and $\pi \approx 3.14$, the sector’s area is $\approx 4.19$ cm².
Find the whole circle’s area This is the same $\pi r^2$ with no fraction taken. For $r = 4$ cm, that is $3.14 \cdot 16 = 50.24$ cm².
Subtract for the major sector The major sector is the whole circle minus the minor sector, $50.24 - 4.19 = 46.05$ cm².
one radius
the other radius
the sector
the rest is the major sector
A girl kneels in a round flower bed where two radii from the centre peg mark off a sector and the major sector.
A clock’s minute hand, 14 cm long, sweeps through 30° in 5 minutes. Using $\pi = 22/7$, the arc length it traces is closest to
88 cm
44 cm
14.67 cm
7.33 cm
Check your answer
88 cm — 88 cm is the WHOLE circle’s circumference — the swept arc is a much smaller fraction of it.
44 cm — 44 cm assumes the hand swept 180° — the actual sweep here is only 30°.
14.67 cm — 14.67 cm is the arc for a 60° sweep — double the 30° sweep actually described here.
✓ 7.33 cm — (D) $(30/360) \cdot 2 \cdot (22/7) \cdot 14 = 22/3 \approx 7.33$ cm.
A sector’s perimeter is two radii plus the arc.
Check yourself
For that same sweep (radius 14 cm, arc length about 7.33 cm), the sector’s perimeter is closest to
7.33 cm
21.33 cm
35.33 cm
28 cm
Check your answer
7.33 cm — 7.33 cm is just the arc — the two radii still need to be added for the perimeter.
21.33 cm — $14+7.33=21.33$ leaves one radius out — a sector’s perimeter needs BOTH radii.
✓ 35.33 cm — (C) Perimeter $= 2 \cdot 14 + 7.33 = 35.33$ cm.
28 cm — 28 cm is just the two radii — the arc must be added for the full perimeter.
In 20 minutes, the same minute hand sweeps through 120°, four times the 5-minute sweep. Its arc length for this longer sweep is closest to
7.33 cm
29.33 cm
88 cm
14.67 cm
Check your answer
7.33 cm — 7.33 cm was the arc for the ORIGINAL 30° sweep — a 120° sweep must trace a longer arc.
✓ 29.33 cm — (B) Arc length is directly proportional to the angle swept, so $88/3 \approx 29.33$ cm.
88 cm — 88 cm is the WHOLE circumference — 120° is only one-third of the full circle, not all of it.
14.67 cm — 14.67 cm doubles the original arc — but the sweep is FOUR times as large (120° vs 30°), so the arc should quadruple.
In 20 minutes the 14 cm minute hand sweeps through 120°.
Worked example
Arc length and perimeter of a sector
$\theta = (5/60) \cdot 360^\circ = 30^\circ$ A clock face is one full turn, $360^\circ$, and $5$ minutes is $5/60$ of the $60$-minute face.
arc length $\approx (30/360) \cdot 2 \cdot (22/7) \cdot 14$ Substitute $\theta = 30^\circ$, $\pi \approx 22/7$ and $r = 14$ cm into the arc-length formula.
$= 22/3 \approx 7.33$ cm $30/360$ simplifies to $1/12$, and $2 \cdot (22/7) \cdot 14$ simplifies to $88$, giving $88/12 = 22/3$.
perimeter $\approx 2 \cdot 14 + 22/3 = 106/3 \approx 35.33$ cm A sector’s perimeter is two radii plus the arc already found.
full circumference $\approx 2 \cdot (22/7) \cdot 14 = 88$ cm, one-twelfth of it $\approx 7.33$ cm Check the arc length against the full circumference. One-twelfth of $88$ cm matches the $7.33$ cm already found, confirming the swept angle really is $30^\circ$.
A clock’s hands from 12 to 1 mark a 30 degree sector of radius 14, with an arc of 7.33 and perimeter 35.33.
Find an arc’s length and a sector’s perimeter, from the radius and the swept angle.
Find the angle from minutes A clock face is $360^\circ$ for $60$ minutes. For $5$ minutes, the angle swept is $(5/60) \cdot 360^\circ = 30^\circ$.
Find the arc’s length Multiply $\theta/360^\circ$ by $2 \pi r$. For $r = 14$ cm and $\pi \approx 22/7$, the arc is $\approx 7.33$ cm.
Add two radii A sector’s perimeter is two radii plus the arc, $2 \cdot 14 + 7.33 \approx 35.33$ cm.
Check against the circumference The full circumference is $88$ cm. One-twelfth of it is $\approx 7.33$ cm, matching the arc found above.
A small central angle leaves only a thin segment, because its triangle takes almost all of the sector.
Worked example
A segment at 60 degrees
sector area $\approx (60/360) \cdot 3.14 \cdot 15^2$ Substitute $\theta = 60^\circ$, $\pi \approx 3.14$ and $r = 15$ cm into the sector-area formula.
$= (1/6) \cdot 3.14 \cdot 225 = 117.75$ cm² $60/360$ simplifies to $1/6$, and $15^2 = 225$.
triangle area $= 15^2 \cdot \sin(30^\circ) \cdot \cos(30^\circ)$ The half-angle of $60^\circ$ is $30^\circ$, so use $r^2 \cdot \sin(\theta/2) \cdot \cos(\theta/2)$.
$= 225 \cdot (1/2) \cdot (1.73/2) \approx 97.31$ cm² $\sin(30^\circ) = 1/2$ and $\cos(30^\circ) = \sqrt{3}/2 \approx 1.73/2$, from the standard pair.
minor segment area $\approx 117.75 - 97.31 = 20.44$ cm² Segment area is sector area minus triangle area.
major segment area $\approx 3.14 \cdot 225 - 20.44 = 706.5 - 20.44 = 686.06$ cm² The major segment is the whole circle’s area minus the minor segment already found.
The half-angle of 60° is 30°, so the triangle’s area is 15² · sin 30° · cos 30°.
Worked example
minor segment plus major segment $= 20.44 + 686.06 = 706.5$ cm² Check by adding the two segments back together. The total should land on the whole circle’s area, $3.14 \cdot 225 = 706.5$ cm², the same value already used to find the major segment.
100 cm² — 100 cm² is just $r^2$ — the triangle’s area still needs the $\sin(45^\circ) \cdot \cos(45^\circ)$ factor applied.
70.7 cm² — 70.7 cm² comes from $100 \cdot \sin(45^\circ)$ alone — the formula needs BOTH $\sin(45^\circ)$ and $\cos(45^\circ)$ multiplied in.
78.5 cm² — 78.5 cm² is the SECTOR’s area — the triangle inside it is a smaller, separate quantity.
Using sector = 78.5 cm² and triangle = 50 cm² from that same figure, the MAJOR SECTOR’s area (not the segment) is closest to
264 cm²
28.5 cm²
235.5 cm²
185.5 cm²
Check your answer
264 cm² — $314-50=264$ subtracts the TRIANGLE — a major SECTOR is the full circle minus the minor SECTOR, and the triangle is not involved at all.
28.5 cm² — 28.5 cm² is the MINOR SEGMENT — a completely different quantity from the major SECTOR being asked for.
✓ 235.5 cm² — (C) The major sector needs only the full circle minus the minor sector: $314 - 78.5 = 235.5$ cm² — the triangle plays no part in it.
185.5 cm² — $314-78.5-50=185.5$ removes the triangle too — but a major SECTOR only ever needs the minor sector removed, not the triangle as well.
A sector and its complement always add back to the full circle, so find whichever is easier and subtract.
Worked example
A segment at 90 degrees
sector area $\approx (90/360) \cdot 3.14 \cdot 10^2$ Substitute $\theta = 90^\circ$, $\pi \approx 3.14$ and $r = 10$ cm into the sector-area formula.
$= (1/4) \cdot 3.14 \cdot 100 = 78.5$ cm² $90/360$ simplifies to $1/4$, and $10^2 = 100$.
triangle area $= 10^2 \cdot \sin(45^\circ) \cdot \cos(45^\circ)$ The half-angle of $90^\circ$ is $45^\circ$.
$= 100 \cdot (\sqrt{2}/2) \cdot (\sqrt{2}/2) = 100 \cdot (1/2) = 50$ cm² $\sin(45^\circ) = \cos(45^\circ) = \sqrt{2}/2$, and their product is exactly $1/2$.
minor segment area $\approx 78.5 - 50 = 28.5$ cm² Segment area is sector area minus triangle area.
major sector area $\approx 3.14 \cdot 100 - 78.5 = 314 - 78.5 = 235.5$ cm² This step pairs with the major-sector relation instead of the major-segment one: the whole circle’s area minus the minor sector.
minor sector plus major sector $= 78.5 + 235.5 = 314$ cm² Check by adding the two sectors back together. The total should land on the whole circle’s area, $3.14 \cdot 100 = 314$ cm², confirming both areas above.
sin 45° and cos 45° are both √2/2, and their product is exactly 1/2: 50 cm².
Find a segment’s area, from the sector and the triangle inside it.
Find the sector’s area Use $(\theta/360^\circ) \cdot \pi r^2$. For $\theta = 90^\circ$, $r = 10$ cm and $\pi \approx 3.14$, that is $78.5$ cm².
Halve the angle The triangle’s half-angle is $\theta/2$. For $\theta = 90^\circ$, that is $45^\circ$.
Find the triangle’s area Use $r^2 \cdot \sin(\theta/2) \cdot \cos(\theta/2)$. With $\sin(45^\circ) = \cos(45^\circ) = \sqrt{2}/2$, that is $100 \cdot (1/2) = 50$ cm².
Subtract the triangle Segment area is sector area minus triangle area, $78.5 - 50 = 28.5$ cm².
Check the segment is smaller A segment always sits inside its sector, so $28.5$ cm² must be less than $78.5$ cm², and it is.
462 cm² — 462 cm² is the sector’s area — the triangle must still be subtracted to get the segment.
190.73 cm² — 190.73 cm² is the triangle removed, not the segment left behind.
A second circle also has a chord subtending 120°, but its radius is 42 cm — double the first circle’s 21 cm. Compared with the first circle’s 271.27 cm² segment, the second circle’s segment area will be
twice as large, since only the radius doubled
the same, since the angle is unchanged
eight times as large, since both the sector and the triangle grow with the doubled radius
four times as large
Check your answer
twice as large, since only the radius doubled — A LENGTH like arc length doubles with r — an AREA like a segment’s scales with $r^2$, so it quadruples instead.
the same, since the angle is unchanged — The angle staying fixed does not mean the area stays fixed — a larger radius at the same angle still gives a larger sector and triangle.
eight times as large, since both the sector and the triangle grow with the doubled radius — The sector and triangle both scale with $r^2$ TOGETHER, as one combined segment — their shared $r^2$ factor gives four times, not eight.
✓ four times as large — (D) Both the sector’s and the triangle’s areas scale with $r^2$ at a fixed angle, so doubling r makes the whole segment four times as large.
A chord subtending 120° in a circle of radius 21 cm, and in one of radius 42 cm.
Worked example
A segment at 120 degrees
sector area $\approx (120/360) \cdot (22/7) \cdot 21^2$ Substitute $\theta = 120^\circ$, $\pi \approx 22/7$ and $r = 21$ cm into the sector-area formula.
$= (1/3) \cdot (22/7) \cdot 441 = 462$ cm² $120/360$ simplifies to $1/3$, and $441/7 = 63$, so $(22/7) \cdot 441 = 22 \cdot 63 = 1386$, then $1386/3 = 462$.
triangle area $= 21^2 \cdot \sin(60^\circ) \cdot \cos(60^\circ)$ The half-angle of $120^\circ$ is $60^\circ$.
segment area $\approx 462 - 190.73 \approx 271.27$ cm² Segment area is sector area minus triangle area.
segment area $271.27$ cm² is less than sector area $462$ cm² Check the size of the answer. A segment sits inside its sector with a triangle removed, so it can never be larger than the sector, and $271.27$ is less than $462$, as it must be.
A 120 degree sector of a 21 cm circle has a triangle of 190.73 and a segment of 271.27 square centimetres.
Every sector’s area and arc length are both the same fraction of the circle’s area and circumference as
the radius r is of the diameter
the chord is of the circumference
the triangle is of the sector
the angle θ is of 360°
Check your answer
the radius r is of the diameter — Radius to diameter is always a fixed 1:2 ratio — it is the ANGLE’s fraction of 360° that scales area and arc length.
the chord is of the circumference — A chord’s length compared with the circumference plays no part in the sector’s area or arc-length formula.
the triangle is of the sector — The triangle only matters when finding a SEGMENT’s area — the sector’s own area and arc length depend on $\theta/360$, not on any triangle.
✓ the angle θ is of 360° — (D) Both quantities scale with the fraction $\theta/360$ of the whole circle.
A sector’s area and arc length are the same fraction of the circle: θ out of 360°.
Check yourself
A segment is best described, in one line, as
a sector doubled in area
an arc without a chord
a triangle without a sector
a sector minus its triangle
Check your answer
a sector doubled in area — Nothing about a segment doubles the sector’s area — a piece is removed, not added.
an arc without a chord — An arc alone is a curve, not a region — a segment is bounded BY a chord and its arc together.
a triangle without a sector — The triangle is the piece taken away — the segment is what is left of the sector after that removal.
✓ a sector minus its triangle — (D) That one line captures the whole chapter’s segment idea: sector area minus triangle area.
The sector’s area and its triangle’s area are both found from the same θ and the same r.
Check yourself
What is the one central-angle-and-radius pair that both a sector’s area formula and its segment’s area formula always need?
the same angle θ and radius r for both shapes
two different radii, one for the sector and one for the triangle
the angle θ for the sector but the angle 360°−θ for the triangle
only the radius, since the angle cancels out in both formulas
Check your answer
✓ the same angle θ and radius r for both shapes — (A) The sector, the triangle, and therefore the segment all use exactly the same r and θ — nothing changes between the two calculations.
two different radii, one for the sector and one for the triangle — The triangle’s two equal sides ARE the sector’s own radii — there is only one radius r in the whole figure.
the angle θ for the sector but the angle 360°−θ for the triangle — The triangle uses the very same angle $\theta$ as the sector, not its complement to 360°.
only the radius, since the angle cancels out in both formulas — The angle $\theta$ is essential to both the sector’s area and the triangle’s area — it never cancels out.
Check yourself: the whole chapter
For a sector of angle θ in a circle of radius r, which pair of formulas is correct?
area $= (\theta/360) \cdot 2 \pi r$; arc length $= (\theta/360) \cdot \pi r^2$ — $(\theta/360) \cdot 2 \pi r$ is the arc length formula, and $(\theta/360) \cdot \pi r^2$ is the area formula — this option has swapped which one belongs to which quantity.
area $= (\theta/180) \cdot \pi r^2$; arc length $= (\theta/180) \cdot 2 \pi r$ — A full circle is 360°, not 180° — using 180° as the reference gives exactly double the correct fraction in both formulas.
area $= (\theta/360) \cdot \pi r^2$; arc length $= (\theta/360) \cdot \pi r^2$ — The circle’s full circumference is $2 \pi r$, not $\pi r^2$ — copying the area formula for arc length does more than drop a factor of 2 — it also squares the radius, which arc length never does.
✓ area $= (\theta/360) \cdot \pi r^2$; arc length $= (\theta/360) \cdot 2 \pi r$ — (D) A sector’s area is $(\theta/360)$ of the circle’s area, $\pi r^2$; its arc length is $(\theta/360)$ of the circle’s circumference, $2 \pi r$.
A sector of central angle 90° and an arc of central angle 60°, each in a circle of radius r.
Check yourself: the whole chapter
A sector has a central angle of $90^\circ$ in a circle of radius r. What fraction of the whole circle’s area does the sector cover?
1/2
1/4
1/3
it depends on the value of r, not just on the angle
Check your answer
1/2 — One-half is the fraction for a $180^\circ$ sector. At $90^\circ$, $90/360$ reduces to one-quarter, not one-half.
✓ 1/4 — (B) $90/360$ reduces to $1/4$ — a quarter-turn sector always covers a quarter of the circle’s area, whatever the radius.
1/3 — One-third is the fraction for a $120^\circ$ sector. At $90^\circ$, $90/360$ reduces to one-quarter, not one-third.
it depends on the value of r, not just on the angle — The fraction $\theta/360$ depends only on the angle — a bigger or smaller radius changes the area itself, but never the fraction of the circle a given angle covers.
A sector has a central angle of $60^\circ$ in a circle of radius r. What fraction of the whole circle’s circumference does the arc cover?
1/2
1/4
1/6
it depends on the value of r, not just on the angle
Check your answer
1/2 — One-half is the fraction for a $180^\circ$ sector. At $60^\circ$, $60/360$ reduces to one-sixth, not one-half.
1/4 — One-quarter is the fraction for a $90^\circ$ sector. At $60^\circ$, $60/360$ reduces to one-sixth, not one-quarter.
✓ 1/6 — (C) $60/360$ reduces to $1/6$ — the arc covers one-sixth of the whole circumference, whatever the radius.
it depends on the value of r, not just on the angle — The fraction $\theta/360$ depends only on the angle — a bigger or smaller radius changes the actual arc length, but never the fraction of the circumference a given angle covers.
A sector’s perimeter is 2r plus the arc; a segment’s is the chord plus the arc.
Check yourself: the whole chapter
A segment’s area is obtained from its sector’s area by which single step?
adding the triangle that the two radii make with the chord
doubling the sector’s area
leaving the sector’s area unchanged, since a segment and its sector always have equal area
subtracting the triangle that the two radii make with the chord
Check your answer
adding the triangle that the two radii make with the chord — The triangle is removed from the sector to leave the segment, not added to it — adding it would make the region larger than the sector itself.
doubling the sector’s area — Nothing in the relationship doubles the sector’s area. One piece, the triangle, is taken away — nothing is added or doubled.
leaving the sector’s area unchanged, since a segment and its sector always have equal area — A segment’s area is smaller than its sector’s area by exactly the triangle’s area — reporting the sector’s area as the segment’s skips the one step that makes them different.
✓ subtracting the triangle that the two radii make with the chord — (D) A segment’s area is its sector’s area minus the triangle the two radii make with the chord — the one region inside the sector that the segment does not include.
Which statement about perimeter is correct?
a sector’s perimeter is its arc length alone, with no radii included
a sector’s perimeter includes its two radii plus its arc; a segment’s perimeter includes its chord plus its arc, with no radii
a segment’s perimeter includes its two radii plus its arc, the same as a sector’s
both a sector’s and a segment’s perimeter are just their arc lengths, since the straight sides do not count as part of the perimeter
Check your answer
a sector’s perimeter is its arc length alone, with no radii included — A sector’s boundary is its two radii and its arc — leaving out the radii only counts part of the boundary, not the whole perimeter.
✓ a sector’s perimeter includes its two radii plus its arc; a segment’s perimeter includes its chord plus its arc, with no radii — (B) A sector’s boundary is its two radii plus its arc; a segment’s boundary is its chord plus its arc — the two shapes’ perimeters follow different rules because their straight sides are different.
a segment’s perimeter includes its two radii plus its arc, the same as a sector’s — This swaps the two shapes’ rules. A sector’s perimeter includes two radii; a segment has no radii on its boundary at all — only a chord and an arc.
both a sector’s and a segment’s perimeter are just their arc lengths, since the straight sides do not count as part of the perimeter — A perimeter is the length all the way around a shape’s boundary. A sector’s straight radii and a segment’s straight chord are both part of that boundary, not excluded from it.
Two ideas drive every formula in this chapter: the θ/360 fraction, and a segment read as its sector minus a triangle.
Every sector and every segment is a slice cut from a circle, in one of two ways. A sector’s area and arc length are both the same fraction of the circle, set by how its angle $\theta$ compares to $360^\circ$. A segment shares that same sector’s arc, but trades the two radii for a single chord. We should reach for the sector first, even when the question asks for a segment. Its formula is direct, and the segment’s formula is always just one subtraction away. Let us use this order on the next problem: find the sector, then take the triangle away.
Read left to right, the strip is the whole method for any segment question: the sector first, because its formula is direct, then one subtraction.
You divide round things by angle, or cut them with a straight line, more often than you think. Here are seven places a sector or a segment decides the numbers.
Drawing a pie chart. A class survey asks $40$ students their favourite sport. $10$ choose cricket. The maths: cricket’s slice is a sector, and every sector’s angle is a share of the full $360^\circ$. $10$ out of $40$ is $1/4$. So the slice’s angle is $1/4$ of $360^\circ$, which is $90^\circ$.
10 out of 40 students choose cricket: 1/4 of 360°, which is 90°.
Designing a round park garden. A round garden in a park, radius $7$ m, is split into $6$ equal sectors for $6$ flowers. The maths: each sector’s area is the same fraction of the bed’s area as its angle is of $360^\circ$. Each sector’s angle is $60^\circ$. Using $\pi \approx 22/7$, its area is $(60/360) \cdot (22/7) \cdot 7^2 \approx 25.67$ m².
Cutting a pizza into slices. A round pizza, radius $14$ cm, is cut into $8$ equal slices for a party. The maths: each slice is a sector, and its angle is $1/8$ of the full $360^\circ$. Each slice’s angle is $360 / 8 = 45^\circ$. Using $\pi \approx 22/7$, one slice’s area is $(45/360) \cdot (22/7) \cdot 14^2 = 77$ cm².
A pizza of radius 14 cm is cut into 8 equal slices of 45°; one slice is 77 cm².
Pricing a wedge-shaped plot. A wedge-shaped plot at a road junction is a sector of radius $21$ m and angle $60^\circ$. The maths: the plot’s area is the same fraction of a full circle’s area as its angle is of $360^\circ$. Using $\pi \approx 22/7$, its area is $(60/360) \cdot (22/7) \cdot 21^2 = 231$ m². At ₹$2000$ per m², it costs ₹$462000$, or ₹$4.62$ lakh.
Cutting a cake for a party. A round cake, radius $14$ cm, is cut so one slice has a $90^\circ$ angle. The maths: the rest of the cake is the major sector, the full circle’s angle minus the slice’s angle. Using $\pi \approx 22/7$, the slice’s area is $(90/360) \cdot (22/7) \cdot 14^2 = 154$ cm². The rest of the cake is $616 - 154 = 462$ cm².
Find the smaller sector and subtract it from the whole circle: the major sector needs no formula of its own.
Edging a quarter-circle flower bed. A flower bed in a corner is a quarter circle of radius $5$ m. You buy plastic edging for its curved side. The maths: the curved side is an arc. It is the same fraction of the whole circle’s edge as its angle is of $360^\circ$. A quarter circle has angle $90^\circ$. Using $\pi \approx 3.14$, the arc is $(90/360) \cdot 2 \cdot 3.14 \cdot 5 = 7.85$ m of edging.
A garden corner’s quarter circle of radius 5 needs edging only along its curved arc, 7.85 m, not the straight sides.
A path across a circle always makes a segment, and pricing the grass beyond it is the chapter’s subtraction with real square metres attached.
Reseeding grass cut off by a path. You run a landscaping business. A client’s circular lawn, radius $5$ m, is crossed by a straight path. Its two ends and the centre make a right angle. The maths: the small patch cut off is a segment. Its area is the sector’s area minus the triangle’s area. Using $\pi \approx 3.14$, the sector’s area is $(90/360) \cdot 3.14 \cdot 5^2 = 19.625$ m². The triangle’s area is $12.5$ m². The patch is $19.625 - 12.5 = 7.125$ m².
Your turn. A round mural is painted with one coloured wedge, radius $6$ m and angle $120^\circ$. Using $\pi \approx 3.14$, what is the wedge’s area? Answer: Its area is $(120/360) \cdot 3.14 \cdot 6^2 = 37.68$ m².
Your turn: a round mural has one coloured wedge, radius 6 m and angle 120°.
practice A sector is cut from a circle of centre $O$ and radius $7$ cm, with an angle of $90^\circ$ at the centre. Find the area of the sector. Use $\pi \approx 22/7$. (Worked in full below — read it, then do the next two the same way.)
practice The same circle, centre $O$ and radius $7$ cm, and the same $90^\circ$ sector. Find the arc length of the sector. Use $\pi \approx 22/7$. (Same circle, same fraction — but an arc is a fraction of the circumference $2 \pi r$, not of the area.)
practice The minute hand of a clock is $7$ cm long. Find the area it sweeps in $15$ minutes. Use $\pi \approx 22/7$. (The angle is not given. Ask what fraction of an hour $15$ minutes is, then turn that fraction into degrees.)
$15$ minutes is a quarter of an hour, so $\theta = 90^\circ$ and the area swept is $38.5$ cm² — the same quarter circle as question 1.
In 15 minutes the minute hand turns through a quarter of an hour: 90°.
Exercise 11.1 — continued
practice Find the area of a sector of a circle with radius $6$ cm if the angle of the sector is $60^\circ$. Use $\pi \approx 22/7$.
practice A sector has radius $21$ cm and angle $60^\circ$. Find its arc length. Use $\pi \approx 22/7$.
practice The minute hand of a clock is $7$ cm long. Find the area swept by it in $10$ minutes. Use $\pi \approx 22/7$.
Answers
$132/7 \approx 18.86$ cm².
$22$ cm.
$77/3 \approx 25.67$ cm².
A goat tied by a 28 m rope to a peg at the centre grazes a 90° sector.
Exercise 11.1 — continued
practice A goat is tied to a peg at the centre of a circular field of radius $28$ m by a rope, and can graze over a sector of angle $90^\circ$. Find the area it can graze. Use $\pi \approx 22/7$.
practice A brooch is designed as a circle of radius $17.5$ mm, divided into $9$ equal sectors by wire radii from the centre. Find the angle of each sector, and the arc length of one sector. Use $\pi \approx 22/7$.
practice An umbrella has $8$ ribs, equally spaced, each $45$ cm long. Treating the umbrella as a flat circle, find the area between two consecutive ribs. Use $\pi \approx 22/7$.
practice A table cover is a circle of radius $35$ cm. A chord cutting off a design segment subtends an angle of $90^\circ$ at the centre. Find the area of the minor segment. Use $\pi \approx 22/7$.
practice A chord of a circle of radius $12$ cm subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding minor segment. Use $\pi = 3.14$ and $\sqrt{3} = 1.73$.
Answers
$616$ m².
Each sector’s angle is $40^\circ$; arc length $= 110/9 \approx 12.22$ mm.
$22275/28 \approx 795.54$ cm².
$350$ cm².
$88.44$ cm².
Two wiper blades, each 14 cm long, sweep through 90° without overlapping.
Exercise 11.1 — continued
practice A car has two wiper blades, each $14$ cm long, sweeping through an angle of $90^\circ$ without overlapping. Find the total area cleaned by the two blades. Use $\pi \approx 22/7$.
practice A lighthouse’s rotating warning beam lights up a sector of angle $72^\circ$ out to a radius of $17.5$ m. Find the area lit up by the beam. Use $\pi \approx 22/7$.
practice A circle has radius $10.5$ cm. A minor sector cut from it has angle $60^\circ$. Find the area of the major sector. Use $\pi \approx 22/7$.
Answers
$308$ cm² in total, from two sectors of $154$ cm² each.
$192.5$ m².
Minor sector $= 57.75$ cm², so major sector $= 346.5 - 57.75 = 288.75$ cm².
A regular hexagon sits inside a circle, leaving six equal curved pieces between its sides and the circle, one of them labelled.
A circle is divided into eight equal wedges in alternating colours, with one wedge’s arc labelled 360 degrees divided by 8.
Exercise 11.1 — further practice
practice Find the area of a sector of a circle with radius $42$ cm if the angle of the sector is $60^\circ$. Use $\pi \approx 22/7$.
practice A sector of a circle has radius $63$ cm and angle $40^\circ$. Find the length of its arc. Use $\pi \approx 22/7$.
practice A circular lawn has radius $14$ m. A gardener marks out a sector-shaped flower bed with a central angle of $108^\circ$. Find the area of the flower bed. Use $\pi \approx 22/7$.
Answers
$924$ cm².
$44$ cm.
$184.8$ m².
A 56 cm fan blade turning through 45° about the centre.
Exercise 11.1 — further practice (continued)
practice A ceiling fan has a single blade of length $56$ cm, fixed at the centre of rotation. As the blade turns through an angle of $45^\circ$, find the distance travelled by the tip of the blade. Use $\pi \approx 22/7$.
practice The minor sector of a circle has a central angle of $80^\circ$. What is the angle of the major sector cut off by the same two radii?
$80^\circ$
$180^\circ$
$280^\circ$
$300^\circ$
practice A sector is cut from a circle of radius $28$ cm, with an angle of $45^\circ$ at the centre. Find the perimeter of the sector. Use $\pi \approx 22/7$.
Answers
$44$ cm.
C — $280^\circ$.
$78$ cm.
A chord of a circle of radius 20 cm subtends a right angle at the centre.
Exercise 11.1 — further practice (continued)
practice A chord of a circle of radius $20$ cm subtends a right angle at the centre. Find the perimeter of the minor segment cut off by the chord. Use $\pi = 3.14$ and $\sqrt{2} = 1.41$.
practice Which two lengths make up the perimeter of a segment cut off by a chord?
The two radii and the arc
The chord and the two radii
The chord and the arc
The arc alone
practice A chord of a circle of radius $40$ cm subtends a right angle at the centre. Find the area of the corresponding minor segment. Use $\pi = 3.14$.
Answers
$59.6$ cm.
C — The chord and the arc.
$456$ cm².
A chord of a circle of radius 30 cm subtends 60° at the centre.
Exercise 11.1 — further practice (continued)
practice A chord of a circle of radius $30$ cm subtends an angle of $60^\circ$ at the centre. Find the area of the corresponding minor segment. Use $\pi = 3.14$ and $\sqrt{3} = 1.73$.
practice A circle has radius $20$ cm. A chord subtends a right angle at the centre. The sector so formed has area $314$ cm², and the triangle formed by the two radii and the chord has area $200$ cm². What is the area of the minor segment cut off by the chord?
$314$ cm²
$200$ cm²
$114$ cm²
$514$ cm²
practice A minor sector cut from a circle of radius $49$ cm has an angle of $90^\circ$ at the centre. Find the area of the major sector cut off by the same two radii. Use $\pi \approx 22/7$.
Answers
$81.75$ cm².
C — $114$ cm².
$5659.5$ cm².
Radius 60 cm: the minor segment; radius 42 cm: the major segment. Both chords subtend 120°.
Exercise 11.1 — further practice (continued)
practice A chord of a circle of radius $60$ cm subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding minor segment. Use $\pi = 3.14$ and $\sqrt{3} = 1.73$.
practice A chord of a circle of radius $42$ cm subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding major segment. Use $\pi \approx 22/7$ and $\sqrt{3} = 1.73$.
Answers
$2211$ cm².
$4458.93$ cm².
A tablecloth of radius 70 cm with a sector-shaped patch of 144°.
Exercise 11.1 — further practice (continued)
practice A circular tablecloth has radius $70$ cm. A decorative sector-shaped patch stitched onto it covers an angle of $144^\circ$ at the centre. Find the area of the tablecloth not covered by the patch. Use $\pi \approx 22/7$.
practice A sector of a circle has radius $21$ cm. The length of its arc is $11$ cm. Find the area of the sector. Use $\pi \approx 22/7$.
