Areas Related to Circles

This is a short, dense chapter with one big idea worked from two directions. It opens by recalling sector and segment as pieces of a circular region, and immediately splits each into its minor and major versions — the smaller piece and its complement in the full circle. It then applies the unitary method twice: once to the area $\pi r^2$ to get the area of a sector, and once to the circumference $2\pi r$ to get the length of an arc. The next move is the one genuinely new idea of the chapter — a segment's area is obtained by taking the sector that contains it and subtracting the straight-sided triangle formed by the two radii and the chord, which for standard angles like $60°$, $90°$, and $120°$ can be evaluated exactly using basic trigonometry or the properties of special triangles. Two fully-worked NCERT examples anchor the method: one on a plain sector and its major complement, one on a segment that needs trigonometry. The chapter closes with the compact Summary of its three formulas, and its exercise set is where the formulas meet real objects — clock hands, tethered horses, wiper blades, umbrella ribs, and decorated table covers.

Sector and segment: two ways to slice a circle

Take a circle with centre $O$ and radius $r$. Choose two points $A$ and $B$ on the circle and draw the two radii $OA$ and $OB$. These two radii, together with one of the two arcs joining $A$ and $B$, enclose a region called a sector of the circle; $\angle AOB$ is called the angle of the sector. Because $A$ and $B$ split the circle into two arcs — a shorter one and a longer one — there are two sectors: the one bounded by the shorter arc is the minor sector, and the one bounded by the longer arc is the major sector. If the minor sector's angle is $\theta$, the major sector's angle is automatically $360° - \theta$, because the two sectors together make up the whole circle. Now instead join $A$ and $B$ with a straight chord instead of two radii. The chord, together with an arc, encloses a region called a segment of the circle. Again there are two segments — the minor segment, cut off by the shorter arc, and the major segment, cut off by the longer arc. Unless a problem explicitly says "major sector" or "major segment," the plain words "sector" and "segment" always mean the minor ones — this convention is used throughout NCERT and in board papers, and forgetting it is a common source of lost marks.

From the whole circle to a slice: the unitary method

You already know that the area enclosed by a circle of radius $r$ is $\pi r^2$, and its circumference is $2\pi r$. Think of the full circular region as a sector too — one whose angle happens to be the entire $360°$ available at the centre. The unitary method now asks: if $360°$ worth of angle gives area $\pi r^2$, how much area does $1°$ give, and hence how much does $\theta°$ give? Scaling down from $360°$ to $1°$ divides by $360$, and scaling back up to $\theta°$ multiplies by $\theta$ — so a sector of angle $\theta$ carries exactly the fraction $\theta/360$ of the full circle's area. Precisely the same scaling argument, applied to the circumference $2\pi r$ instead of the area, gives the length of the arc that bounds that sector.

Area of a sector

A sector of angle $\theta$ (in degrees) cut from a circle of radius $r$ has area

where: $r$ is the radius of the circle and $\theta$ is the angle of the sector at the centre, measured in degrees, with $0° < \theta < 360°$. Setting $\theta = 360°$ recovers the full circle's area $\pi r^2$, exactly as it should; setting $\theta = 90°$ gives one quadrant, area $\pi r^2/4$; and $\theta = 180°$ gives a semicircle, area $\pi r^2/2$ — both useful checkpoints for sanity-testing an answer.

Length of an arc

The curved boundary of that same sector — the arc joining $A$ and $B$ on the near side — has length

where: the symbols mean the same as in equation $(1)$, and the arc length is a length (units cm, m, and so on), never to be confused with the sector's area (units cm$^2$, m$^2$). A neat consistency check links the two formulas: Area of sector $= \tfrac{1}{2} \times$ (arc length) $\times r$, the same relation that connects a triangle's area to its base and height, because a very thin sector looks almost like a thin triangle with the arc as its "base."

Area of a segment

The segment cut off by chord $AB$ sits inside the sector $OAB$ that has the same two endpoints, and the difference between them is exactly the triangle $OAB$ formed by the two radii and the chord:

where: $\triangle OAB$ is isosceles because $OA = OB = r$, and its included angle at $O$ is the same $\theta$ as the sector's angle. This one line is the entire new content of the chapter — everything else is either the sector-area formula you already have, or a way of computing the area of that isosceles triangle. When $\theta = 90°$ the triangle is right-angled and its area is simply $\tfrac{1}{2} r \times r = \tfrac{1}{2} r^2$; for a general $\theta$, dropping a perpendicular from $O$ to the chord (as in Example 2 below) or using the trigonometric area formula $\tfrac{1}{2} r^2 \sin\theta$ both work.

Minor and major: the complement relations

Because the minor and major pieces of each kind always fill up the whole circle between them, you never need to compute a major sector or major segment from scratch — subtract the minor one from the whole circle:

where: both relations follow because minor and major pieces of the same kind are non-overlapping and together fill the full circular region of area $\pi r^2$. The major sector's angle is likewise $360° - \theta$, so it can equally be computed directly from formula $(1)$ with $360° - \theta$ in place of $\theta$ — both routes must agree, and checking that they do is a good way to catch an arithmetic slip.

Board level - area of a sector and its major complement

Step 1. A sector $OAPB$ is cut from a circle of radius $r = 4$ cm with central angle $\theta = 30°$, and $\pi = 3.14$ is to be used throughout — always fix the value of $\pi$ a question specifies before starting, since the final answer depends on it.

Step 2. Apply formula $(1)$ directly: Area of sector $= \dfrac{\theta}{360}\pi r^2 = \dfrac{30}{360} \times 3.14 \times 4 \times 4$ cm$^2$.

Step 3. Simplify the fraction first to keep the arithmetic clean: $\dfrac{30}{360} = \dfrac{1}{12}$, so the area is $\dfrac{1}{12} \times 3.14 \times 16 = \dfrac{50.24}{12} \approx 4.19$ cm$^2$.

Step 4. For the major sector, use the complement relation $(4)$: Area of major sector $= \pi r^2 -$ Area of minor sector $= 3.14 \times 16 - 4.19 = 50.24 - 4.19 = 46.05 \approx 46.1$ cm$^2$. As a cross-check, the major sector's own angle is $360° - 30° = 330°$, and $\dfrac{330}{360} \times 3.14 \times 16 \approx 46.05$ cm$^2$ — the two routes agree. The area of the minor sector is approximately 4.19 cm², and the area of the major sector is approximately 46.1 cm².

Board level - arc length and area swept by a clock's minute hand

Step 1. The minute hand of a clock has length $14$ cm, and a minute hand behaves exactly like a rotating radius, so this is a sector problem with $r = 14$ cm. In $60$ minutes the hand sweeps the full $360°$, so in $5$ minutes it sweeps an angle $\theta = \dfrac{5}{60} \times 360° = 30°$.

Step 2. The area swept in those $5$ minutes is the area of the sector of angle $30°$: using formula $(1)$ with $\pi = \dfrac{22}{7}$, Area $= \dfrac{30}{360} \times \dfrac{22}{7} \times 14 \times 14 = \dfrac{1}{12} \times \dfrac{22}{7} \times 196$ cm$^2$.

Step 3. Simplify: $\dfrac{22}{7} \times 196 = 22 \times 28 = 616$, so Area $= \dfrac{616}{12} = \dfrac{154}{3} \approx 51.33$ cm$^2$.

Step 4. As a companion quantity, the length of arc traced by the tip of the hand in those $5$ minutes is Length $= \dfrac{30}{360} \times 2 \times \dfrac{22}{7} \times 14 = \dfrac{1}{12} \times 88 = \dfrac{22}{3} \approx 7.33$ cm — note this is a length, in cm, not to be confused with the area just found, in cm$^2$. The area swept by the minute hand in $5$ minutes is 154/3 cm², approximately 51.33 cm².

Board HOTS / foundation JEE-NEET - area of a segment using trigonometry

Step 1. A chord $AB$ of a circle of radius $r = 21$ cm subtends $\angle AOB = 120°$ at the centre $O$, and the required region is the minor segment $AYB$ cut off by that chord, with $\pi = \dfrac{22}{7}$.

Step 2. Find the area of sector $OAYB$ using formula $(1)$: Area of sector $= \dfrac{120}{360} \times \dfrac{22}{7} \times 21 \times 21 = \dfrac{1}{3} \times \dfrac{22}{7} \times 441$ cm$^2$. Since $441 = 7 \times 63$, this is $\dfrac{1}{3} \times 22 \times 63 = \dfrac{1386}{3} = 462$ cm$^2$.

Step 3. Find the area of $\triangle OAB$ by dropping a perpendicular $OM$ from $O$ to chord $AB$. Since $OA = OB = r$, triangle $OAM$ is congruent to triangle $OBM$ by the RHS rule, so $M$ bisects $AB$ and bisects the angle, giving $\angle AOM = 60°$. In right triangle $OMA$: $\cos 60° = OM/OA$ gives $OM = 21 \times \tfrac{1}{2} = \tfrac{21}{2}$ cm, and $\sin 60° = AM/OA$ gives $AM = 21 \times \tfrac{\sqrt3}{2} = \tfrac{21\sqrt3}{2}$ cm, so the full chord is $AB = 2\,AM = 21\sqrt3$ cm.

Step 4. Compute the triangle's area as one half base times height: Area of $\triangle OAB = \tfrac{1}{2} \times AB \times OM = \tfrac{1}{2} \times 21\sqrt3 \times \tfrac{21}{2} = \dfrac{441\sqrt3}{4}$ cm$^2$. (The same number falls out instantly from the trigonometric shortcut of the third derivation: $\tfrac{1}{2} r^2 \sin\theta = \tfrac{1}{2} \times 441 \times \sin120° = \tfrac{441}{2} \times \tfrac{\sqrt3}{2} = \dfrac{441\sqrt3}{4}$, the same value, confirming the perpendicular method and the formula agree.)

Step 5. Subtract using equation $(3)$: Area of segment $AYB = 462 - \dfrac{441\sqrt3}{4} = \dfrac{21}{4}\left(88 - 21\sqrt3\right)$ cm$^2$, taking $441 = 21 \times 21$ common with the $462 = \tfrac{21}{4}\times 88$ rewriting to combine the two terms over the denominator $4$. The area of the segment is (21/4)(88 minus 21 root 3) cm², approximately 199.4 cm² once $\sqrt3 \approx 1.732$ is substituted. Timing note: the perpendicular-and-RHS-congruence method above is the safe, fully-justified board-exam route and is worth writing out in full for proof-style marks; if you already trust the $\tfrac12 r^2\sin\theta$ shortcut from the third derivation, you can jump from Step 2 straight to the triangle's area in one line and save two to three minutes on a timed paper — but only after you have practised the full derivation enough times that the shortcut is a genuine recall, not a guess.

Mixing up sector and segment is the single most frequent slip: a sector is bounded by two radii and an arc, while a segment is bounded by a chord and an arc — draw the figure before choosing a formula. A close second is computing a segment's area as if it were just the sector, forgetting the "minus the triangle" step in equation $(3)$; the sector is always the larger of the two (for $\theta < 180°$), so if your "segment" answer comes out bigger than the sector you started from, you have made a sign error. Students also frequently forget that the major sector's angle is $360° - \theta$, not $\theta$ itself, and then apply formula $(1)$ with the wrong angle instead of using the simpler complement relation in $(4)$. Watch units carefully: arc length is a length (cm, m), sector and segment area are areas (cm$^2$, m$^2$) — writing "cm" for an area answer or "cm$^2$" for a length answer costs marks even when the number is right. Finally, always use the value of $\pi$ the question specifies ($\dfrac{22}{7}$ or $3.14$) consistently through every step of that problem; switching midway, or rounding too early, produces an answer that will not match the expected final figure.

• State precisely what a sector and a segment are, and distinguish minor from major for each.
• Derive, not just quote, the sector-area formula $\dfrac{\theta}{360}\pi r^2$ and the arc-length formula $\dfrac{\theta}{360} \times 2\pi r$ using the unitary method.
• Compute the area of a segment as sector area minus triangle area, choosing the right-angle, equilateral, or trigonometric method depending on $\theta$.
• Use the complement relations to get a major sector's or major segment's area from the corresponding minor one without recomputing from scratch.
• Keep the correct value of $\pi$ fixed throughout a problem and keep length and area units clearly separate.
• Solve real-world "area swept" problems: rotating hands, wiper blades, tethered grazing animals, and equally spaced decorative sectors (umbrella ribs, brooch sectors, table-cover designs).
• Recognise and correctly compute the perimeter of a sector as arc length plus two radii, when asked.
• Cross-check an answer for a major region by computing it two independent ways: by subtraction from $\pi r^2$, and directly with angle $360° - \theta$.

Ask the tutor: Want to work through the umbrella-ribs or tethered-horse style problem step by step, or see the segment formula derived for a few more standard angles like 150 degrees?