Chapter 11 — Areas Related to Circles

The circumference of a circle is the distance around it, its perimeter. In earlier classes you measured, by winding a thread round circular objects, that the circumference divided by the diameter is always the same number, a little more than 3. This constant is denoted by the Greek letter π (pi). So circumference = π × diameter = 2πr, where r is the radius.

The number π is not a fraction; it is an irrational number whose decimal expansion never ends and never repeats: π = 3.14159265… For calculations we use the approximations π ≈ 22/7 or π ≈ 3.14, and the examination tells you which to use. The great Indian mathematician Aryabhata (476 CE) gave the value 62832/20000 = 3.1416, correct to four decimal places, and remarked that it was approximate, showing that he knew π could not be captured exactly by a fraction.

The area of a circle of radius r is πr². A simple demonstration: cut the circle into a large number of thin sectors and lay them alternately point-up and point-down; they form an approximate rectangle of length half the circumference, πr, and breadth r, so the area is πr × r = πr².

Related formulas follow at once. The area of a semicircle is ½πr² and its perimeter is πr + 2r (the curved part plus the diameter; a common error is to forget the diameter). A quadrant has area ¼πr² and perimeter ½πr + 2r.

Worked example. The radii of two circles are 19 cm and 9 cm. Find the radius of the circle whose circumference equals the sum of their circumferences. 2πR = 2π(19) + 2π(9), so R = 28 cm.

Worked example. The radii of two circles are 8 cm and 6 cm. Find the radius of the circle having area equal to the sum of their areas. πR² = π(64 + 36) = 100π, so R = 10 cm.

Worked example. The cost of fencing a circular field at ₹24 per metre is ₹5280. The field is to be ploughed at ₹0.50 per m². Find the cost of ploughing (π = 22/7). Circumference = 5280/24 = 220 m, so 2 × (22/7) × r = 220 and r = 35 m. Area = (22/7) × 35 × 35 = 3850 m²; ploughing cost = 3850 × 0.50 = ₹1925.

Worked example. Find the area of a circle whose circumference is 44 cm. 2πr = 44 gives r = 7 cm (π = 22/7), so area = (22/7) × 49 = 154 cm².

Keep units straight: circumference in cm, area in cm². When π = 22/7 is used, radii that are multiples of 7 give whole-number answers, which is a useful check.

A circular region can be divided into parts in two natural ways.

Two radii OA and OB of a circle with centre O divide the circular region into two parts. Each part, bounded by the two radii and the arc between them, is a sector of the circle. The smaller part, bounded by the minor arc AB, is the minor sector, and the larger, bounded by the major arc, is the major sector. The angle ∠AOB between the two radii is the angle of the sector (or the central angle); the major sector has angle 360° − ∠AOB. When the two radii lie along a diameter, both sectors are semicircles.

A chord AB divides the circular region into two parts differently. Each part, bounded by the chord and the arc it cuts off, is a segment of the circle. The part containing the minor arc is the minor segment, and the part containing the centre is the major segment. A diameter gives two semicircular segments.

Observe the relation between them: the minor sector OAB is made up of the triangle OAB together with the minor segment. Hence

Area of minor segment = Area of minor sector − Area of triangle OAB.

And the major segment is the whole circle minus the minor segment; the major sector is the whole circle minus the minor sector.

Everyday examples: a slice of a round cake or pizza is a sector; the shaded face of a clock between the hands is a sector; the region of a circular garden on one side of a straight path is a segment; the wet region cleaned by a windscreen wiper of length r sweeping through angle θ is a sector of radius r and angle θ; a circular table top with one edge cut straight leaves a major segment.

Names to keep separate: arc is a curve, measured by length; sector and segment are regions, measured by area. The perimeter of a sector is the arc length plus two radii; the perimeter of a segment is the arc length plus the chord.

Worked example. A circle of radius 10 cm has a chord subtending 90° at the centre. Name and describe the parts: the two radii to the ends of the chord cut a minor sector (a quadrant) and a major sector of 270°; the chord cuts a minor segment, which is the quadrant minus the right isosceles triangle with legs 10 cm, and a major segment containing the centre.

Every problem in the chapter reduces to identifying which of these regions is being asked for, and then using the formulas of the next topics.

The whole circle corresponds to a central angle of 360° and its length is the circumference 2πr. A sector of angle θ is the fraction θ/360 of the whole circle, so its arc is the same fraction of the circumference. This proportional reasoning, called the unitary method, gives

Length of an arc of a sector of angle θ = (θ/360) × 2πr.

The reasoning is worth stating in the examination: when the degree measure of the angle at the centre is 360°, the arc length is 2πr; so when it is 1°, the arc length is 2πr/360; so when it is θ°, the arc length is (θ/360) × 2πr.

Special cases: a semicircular arc (θ = 180°) has length πr; a quadrant arc (θ = 90°) has length πr/2; an arc of 60° has length πr/3.

Since the sector is bounded by the arc and the two radii, perimeter of the sector = (θ/360) × 2πr + 2r. Do not forget the two radii; the arc alone is not the perimeter.

Worked example. Find the length of the arc of a sector of a circle of radius 21 cm and angle 60° (π = 22/7). Arc = (60/360) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm.

Worked example. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the perimeter of the sector. Perimeter = 22 + 2 × 21 = 64 cm.

Worked example. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the arc length and the area of the sector. Arc = 16.4 − 2 × 5.2 = 6 cm. Area of the sector = ½ × arc × r = ½ × 6 × 5.2 = 15.6 cm². (The formula area = ½ × l × r is explained in the next topic.)

Worked example. A pendulum swings through an angle of 30° and the tip describes an arc of length 8.8 cm. Find the length of the pendulum (π = 22/7). (30/360) × 2 × (22/7) × r = 8.8, so (1/12) × (44/7) r = 8.8, r = 8.8 × 12 × 7/44 = 16.8 cm.

Worked example. The minute hand of a clock is 14 cm long. How far does its tip move in 20 minutes? In 60 minutes the hand turns through 360°, so in 20 minutes through 120°. Distance = (120/360) × 2 × (22/7) × 14 = (1/3) × 88 = 29.33 cm.

Angles and time on a clock: the minute hand turns 6° per minute and the hour hand 0.5° per minute (30° per hour). Convert time to angle first, then apply the arc formula.

The same proportional reasoning gives the area of a sector. The whole circle, with central angle 360°, has area πr²; a sector of angle 1° has area πr²/360; so a sector of angle θ has area

Area of a sector of angle θ = (θ/360) × πr².

A second form is useful when the arc length l is known instead of the angle. Since l = (θ/360) × 2πr, we have (θ/360) × πr² = ½ × [(θ/360) × 2πr] × r = ½ lr. So

Area of a sector = ½ × (arc length) × (radius) = ½ lr.

This is like the area of a triangle with base l and height r, which is what the sector nearly is when it is thin.

Special cases: semicircle ½πr², quadrant ¼πr², sector of 60° is πr²/6, sector of 120° is πr²/3. The major sector has area (1 − θ/360) × πr² = πr² − minor sector.

Worked example. Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60° (π = 22/7). Area = (60/360) × (22/7) × 36 = (1/6) × 792/7 = 132/7 = 18.86 cm².

Worked example. Find the area of a quadrant of a circle whose circumference is 22 cm. 2πr = 22 gives r = 7/2 cm. Quadrant area = ¼ × (22/7) × (49/4) = 77/8 = 9.625 cm².

Worked example. The length of the minute hand of a clock is 14 cm. Find the area swept by it in 5 minutes. Angle in 5 minutes = 30°. Area = (30/360) × (22/7) × 196 = (1/12) × 616 = 154/3 = 51.33 cm².

Worked example. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep. One wiper cleans (115/360) × (22/7) × 625 = (23/72) × 13750/7 = 158125/252 cm²; two wipers clean 158125/126 = 1254.96 cm².

Worked example. A horse is tied to a peg at one corner of a square grass field of side 15 m by a rope 5 m long. Find the area it can graze (π = 3.14). The rope sweeps a quadrant of radius 5 m inside the field: area = ¼ × 3.14 × 25 = 19.625 m². If the rope were 10 m, the area would be ¼ × 3.14 × 100 = 78.5 m², an increase of 58.875 m².

Worked example. An umbrella has 8 ribs equally spaced; assuming it is a flat circle of radius 45 cm, the area between two consecutive ribs is a sector of 45°: (45/360) × (22/7) × 2025 = 22275/28 = 795.54 cm².

A chord AB of a circle with centre O and radius r subtends angle θ at the centre. The minor segment cut off by AB is the minor sector OAB with the triangle OAB removed. Hence

Area of minor segment = Area of sector OAB − Area of ΔOAB = (θ/360) × πr² − Area of ΔOAB.

The triangle OAB is isosceles with OA = OB = r and vertical angle θ. Its area can be found in three ways.

• For θ = 90°, the triangle is right-angled at O and its area is ½ r².
• For θ = 60°, the triangle is equilateral with side r and its area is (√3/4) r².
• For θ = 120°, drop the perpendicular OM from O to AB; then ∠AOM = 60°, OM = r cos 60° = r/2, AM = r sin 60° = (√3/2) r, so AB = √3 r and the area is ½ × √3 r × r/2 = (√3/4) r².

In general, with OM ⊥ AB, OM = r cos(θ/2), AB = 2r sin(θ/2), and area of ΔOAB = ½ × AB × OM = r² sin(θ/2) cos(θ/2). At this level you only need the three standard angles above.

The major segment has area πr² − (minor segment).

Worked example. Find the area of the segment of a circle of radius 12 cm whose corresponding arc subtends 120° at the centre (π = 3.14, √3 = 1.73). Sector = (120/360) × 3.14 × 144 = 150.72 cm². Triangle: OM = 6, AB = 12√3 = 20.76, area = ½ × 20.76 × 6 = 62.28 cm². Segment = 150.72 − 62.28 = 88.44 cm².

Worked example. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the areas of the minor and major segments (π = 3.14). Sector = ¼ × 3.14 × 100 = 78.5; triangle = ½ × 10 × 10 = 50; minor segment = 28.5 cm². Major segment = 314 − 28.5 = 285.5 cm².

Worked example. A chord of a circle of radius 15 cm subtends 60° at the centre. Find the areas of the minor and major segments (π = 3.14, √3 = 1.73). Sector = (1/6) × 3.14 × 225 = 117.75; equilateral triangle = (1.73/4) × 225 = 97.31; minor segment = 20.44 cm²; major segment = 706.5 − 20.44 = 686.06 cm².

Worked example. A round table cover has six equal designs, each a segment cut off by a side of the inscribed regular hexagon, in a circle of radius 28 cm. Each design is the segment of a 60° sector: sector = (1/6) × (22/7) × 784 = 410.67, triangle = (1.7/4) × 784 = 333.2 (with √3 = 1.7), segment = 77.47 cm²; six designs = 464.8 cm²; at ₹0.35 per cm² the cost is ₹162.68.

Always write the three lines: sector, triangle, difference.

The region between two concentric circles of radii R and r (R > r) is a ring or annulus. Its area is the difference of the two circular areas:

Area of a ring = πR² − πr² = π(R² − r²) = π(R + r)(R − r).

The factorised form is handy: R + r is the sum of the radii and R − r is the width of the ring.

Circular paths around circular lawns, washers, the rim of a wheel, the track of a circular race course and the border of a circular table cloth are all rings. A path of width w outside a circular lawn of radius r is the ring between r and r + w; a path inside a lawn of radius R is the ring between R − w and R.

Worked example. A circular park of radius 20 m is surrounded by a path 4 m wide. Find the area of the path (π = 3.14). Area = π(24² − 20²) = 3.14 × 44 × 4 = 552.64 m².

Worked example. The inner and outer radii of a circular race track are 105 m and 112 m. Find the area of the track (π = 22/7). Area = (22/7) × (112 + 105)(112 − 105) = (22/7) × 217 × 7 = 4774 m².

Worked example. The area of a ring is 1232 cm² and its outer radius is 21 cm. Find the inner radius (π = 22/7). (22/7)(441 − r²) = 1232, so 441 − r² = 392, r² = 49, r = 7 cm.

Worked example. A brooch is made with silver wire in the form of a circle of diameter 35 mm. The wire is also used to make 5 diameters dividing the circle into 10 equal sectors. Find the total length of wire and the area of each sector (π = 22/7). Circumference = (22/7) × 35 = 110 mm; five diameters = 175 mm; total wire = 285 mm. Each sector has angle 36°: area = (36/360) × (22/7) × (35/2)² = (1/10) × (22/7) × 1225/4 = 385/4 = 96.25 mm².

Worked example. Two concentric circles have radii 7 cm and 14 cm, and the angle AOC of the sector is 40°. Find the area of the region between the two arcs of the sector (a sector of the ring). Area = (40/360) × (22/7) × (196 − 49) = (1/9) × (22/7) × 147 = (1/9) × 462 = 51.33 cm².

Worked example. An archery target has three concentric regions of radii 10.5 cm, 21 cm and 31.5 cm (gold, red, blue). The gold region is a circle of area (22/7) × 110.25 = 346.5 cm²; the red ring is (22/7)(441 − 110.25) = 1039.5 cm²; the blue ring is (22/7)(992.25 − 441) = 1732.5 cm². The rings' areas are in the ratio 1 : 3 : 5, because (n + 1)² − n² = 2n + 1.

The sector-of-a-ring formula, (θ/360) × π(R² − r²), will be used repeatedly in the combination problems that follow.

Many examination figures combine a square with a circle inscribed in it, a circle circumscribing it, or quadrants at its corners. The key is the relation between the side of the square and the radius.

• Circle inscribed in a square of side a: the diameter equals the side, so r = a/2. Area of the corner region = a² − π(a/2)² = a²(1 − π/4).
• Square inscribed in a circle of radius r: the diagonal of the square is the diameter, so a√2 = 2r and a = r√2, a² = 2r². Area between the circle and the square = πr² − 2r² = r²(π − 2).
• Quadrants at the four corners of a square of side a with radius a/2: the four quadrants together make one full circle of radius a/2, so the region left in the middle is a² − π(a/2)², the same as for the inscribed circle.

Worked example. Find the area of the shaded region where a circle of radius 7 cm is inscribed in a square, the shaded region being the square outside the circle (π = 22/7). Side = 14 cm. Area = 196 − (22/7) × 49 = 196 − 154 = 42 cm².

Worked example. A square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region between the square and the quadrant (π = 3.14). The diagonal OB of the square is the radius: r = 20√2 cm, r² = 800. Quadrant area = ¼ × 3.14 × 800 = 628; square = 400; shaded = 228 cm².

Worked example. From each corner of a square of side 4 cm a quadrant of radius 1 cm is cut, and a circle of diameter 2 cm is cut from the centre. Find the remaining area (π = 22/7). Square = 16; four quadrants = one circle of radius 1 = 22/7; central circle radius 1 = 22/7; remaining = 16 − 44/7 = 68/7 = 9.71 cm².

Worked example. In a square ABCD of side 14 cm, semicircles are drawn with each side as diameter, and the four petals formed inside are shaded (the design where the four semicircles overlap). Method: the area of the four semicircles = 2 × π × 7² = 2 × 154 = 308 cm²; this counts the shaded petals twice and the unshaded parts once, and the square (196) counts everything once. So shaded = 308 − 196 = 112 cm². Alternatively, the unshaded area = 2(square − two semicircles on opposite sides) = 2(196 − 154) = 84, and shaded = 196 − 84 = 112 cm².

Worked example. Four circular cardboard pieces of radius 7 cm are placed so that each piece touches two others. Find the area of the region enclosed between them (π = 22/7). The centres form a square of side 14 cm. Enclosed region = square − four quadrants = 196 − (22/7) × 49 = 196 − 154 = 42 cm².

The pattern in every case: write the total figure, subtract or add the circular pieces, and note that four quadrants of equal radius make a whole circle.

When sectors are cut from the corners of a triangle or a polygon, the sectors' angles are the angles of the polygon, and their total is the angle sum of the polygon. This gives a quick way to find the total area removed.

Sectors at the vertices of a triangle. If arcs of radius r are drawn with each vertex of a triangle as centre, the three sectors have angles ∠A, ∠B, ∠C adding to 180°, so their total area is (180/360) × πr² = ½πr², half a circle, whatever the triangle. The area of the triangle left over is (area of triangle) − ½πr².

Worked example. In an equilateral triangle ABC of side 12 cm, sectors of radius 6 cm are drawn at each vertex. Find the area of the region of the triangle not included in the sectors (√3 = 1.73, π = 3.14). Triangle = (1.73/4) × 144 = 62.28; three sectors of 60° each with radius 6 = ½ × 3.14 × 36 = 56.52; remaining = 5.76 cm².

Sectors at the vertices of a quadrilateral. The angles add to 360°, so four sectors of equal radius r make a full circle of area πr².

Worked example. In a trapezium ABCD with AB ∥ DC, AB = 18 cm, DC = 32 cm, distance between them 14 cm, and arcs of radius 7 cm drawn at each vertex, find the area of the remaining region (π = 22/7). Trapezium = ½(18 + 32) × 14 = 350 cm²; four sectors = one circle = (22/7) × 49 = 154 cm²; remaining = 196 cm².

Circle circumscribing an equilateral triangle. For an equilateral triangle of side a, the circumradius is R = a/√3 and the inradius is r = a/(2√3), so R = 2r. Area between the circumcircle and the triangle = πa²/3 − (√3/4)a².

Worked example. An equilateral triangle of side 12 cm is inscribed in a circle. Find the area of the circle outside the triangle. R = 12/√3 = 4√3, so R² = 48: circle = 48π ≈ 150.72 (π = 3.14); triangle = (1.73/4) × 144 = 62.28; region = 88.44 cm².

Semicircles on the sides of a right triangle. In a right triangle ABC with the right angle at A and BC as diameter of a semicircle, with semicircles also drawn on AB and AC as diameters, the two shaded crescents (lunes) between the small semicircles and the big one have total area equal to the triangle. Reason: ½π(AB/2)² + ½π(AC/2)² = ½π(BC/2)² by the Pythagoras theorem, so small semicircles + triangle − big semicircle = triangle. For legs AB = 14 cm and AC = 48 cm the crescents' area is ½ × 14 × 48 = 336 cm².

Worked example. A right triangle ABC, right-angled at A, has AB = 6 cm and AC = 8 cm, and a semicircle is drawn on BC = 10 cm as diameter; find the area of the region inside the semicircle but outside the triangle (π = 3.14). Semicircle = ½ × 3.14 × 25 = 39.25; triangle = 24; region = 15.25 cm².

The long question from this chapter shows a figure with a shaded region made from circles, sectors, squares and triangles and asks for its area or perimeter. A fixed strategy handles every such question.

• Name the pieces. Decide which standard regions are involved: circle, semicircle, quadrant, sector of angle θ, segment, square, rectangle, triangle, ring.
• Find the dimensions. Use the relationships in the figure: a diameter equals a side, a diagonal equals a diameter, a radius equals half a side, two touching circles have centres at distance equal to the sum of radii.
• Write the area as a sum or difference. Shaded = (whole) − (unshaded) is the usual form. Sometimes it is easier to compute the unshaded region.
• Keep π symbolic until the end, then substitute 22/7 or 3.14 as instructed, and simplify.
• Check the answer is positive and reasonable compared with the whole figure.

Worked example. OACB is a quadrant of a circle of radius 3.5 cm with centre O and OD = 2 cm on OB. Find the area of the quadrant OACB and of the shaded region between the quadrant and triangle AOD (π = 22/7). Quadrant = ¼ × (22/7) × 12.25 = 9.625 cm². Triangle AOD = ½ × 3.5 × 2 = 3.5 cm². Shaded = 6.125 cm².

Worked example. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the design (√3 = 1.7, π = 22/7). The circle's radius 32 is the circumradius, so side = 32√3 and triangle area = (√3/4)(32√3)² = (√3/4) × 3072 = 768√3 ≈ 1305.6 (√3 = 1.7). Circle = (22/7) × 1024 = 3218.29. Design = 3218.29 − 1305.6 = 1912.69 cm².

Worked example. AB and CD are two diameters of a circle, perpendicular to each other, and OD is the diameter of a smaller circle. If OA = 7 cm, find the area of the shaded region, where the shaded region = (semicircle on AB with the triangle ABC removed) + (small circle on OD). Semicircle = ½ × (22/7) × 49 = 77; triangle ACB = ½ × 14 × 7 = 49; small circle of radius 3.5 = (22/7) × 12.25 = 38.5. Shaded = 77 − 49 + 38.5 = 66.5 cm².

Worked example. A square of side 28 cm has a circle inscribed in it; find the area of the four corner pieces and their perimeter. Area = 784 − (22/7) × 196 = 784 − 616 = 168 cm². Perimeter of the four corner pieces together = perimeter of the square + circumference = 112 + 88 = 200 cm.

Perimeter questions are the trap: the boundary of a shaded region includes every curved and straight edge that borders it, and only those. Trace the boundary with a finger before adding lengths.

A wheel rolling without slipping moves forward by exactly one circumference in one complete revolution. This connects the circumference formula with distance, speed and time, and is a favourite source of examination problems.

Distance covered = number of revolutions × circumference = n × 2πr. Conversely, the number of revolutions = distance ÷ circumference.

Worked example. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at 66 km/h? Distance in 10 minutes = 66 × 1000 × 10/60 = 11000 m = 1100000 cm. Circumference = (22/7) × 80 = 1760/7 cm. Revolutions = 1100000 ÷ (1760/7) = 1100000 × 7/1760 = 4375.

Worked example. A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel. Circumference = 11 × 1000 × 100/5000 = 220 cm. (22/7) × d = 220, d = 70 cm.

Worked example. The diameter of a wheel is 1.26 m. How far will it travel in 500 revolutions? Circumference = (22/7) × 1.26 = 3.96 m; distance = 500 × 3.96 = 1980 m = 1.98 km.

Worked example. A circular pond of diameter 17.5 m is surrounded by a path 2 m wide; find the cost of paving at ₹25 per m² (π = 3.14). Ring = 3.14 × (10.75² − 8.75²) = 3.14 × 19.5 × 2 = 122.46 m²; cost = ₹3061.50.

Tick-marks on a wheel. If the radius of a wheel is r and it turns through angle θ, a point on the rim travels the arc (θ/360) × 2πr while the wheel's centre moves the same distance forward. Thus a wheel of radius 35 cm turning through 90° carries the vehicle forward ¼ × 220 = 55 cm.

Speed of the tip of a hand. The tip of a 14 cm minute hand moves 2π × 14 = 88 cm in 60 minutes, that is, at 88/60 ≈ 1.47 cm per minute; the tip of the hour hand of the same length moves 88 cm in 12 hours.

Worked example. In one revolution a wheel of radius 21 cm covers 132 cm; to cover 792 m it makes 79200/132 = 600 revolutions.

Worked example. Which has a greater area, a circle of circumference c or a square of perimeter c? For the circle r = c/(2π) and area = c²/(4π); for the square the side is c/4 and area c²/16. Since 4π < 16, c²/(4π) > c²/16 and the circle has the greater area. Of all figures with a given perimeter, the circle encloses the most area; this is why circular tanks and pipes are economical.

Convert all quantities to the same unit before computing, and count only complete revolutions when the question says complete.

Perimeter questions test whether you can trace the boundary of a region correctly. Here are the standard boundaries and worked cases.

Sector of angle θ and radius r: perimeter = arc + two radii = (θ/360) × 2πr + 2r.

Segment cut by a chord subtending θ: perimeter = arc + chord = (θ/360) × 2πr + 2r sin(θ/2). For θ = 90°, chord = r√2; for 60°, chord = r; for 120°, chord = r√3.

Semicircle: πr + 2r. Quadrant: ½πr + 2r.

Ring: the boundary consists of two circles, so perimeter = 2πR + 2πr = 2π(R + r).

Race track with semicircular ends. A track consists of two straight sides of length L and two semicircular ends of radius r; its inner boundary has length 2L + 2πr. If the track has width w, the outer boundary is 2L + 2π(r + w), and the area of the track is 2Lw + π[(r + w)² − r²].

Worked example. A racing track has straight sides 106 m long and inner semicircular ends of diameter 60 m, and the track is 10 m wide (π = 22/7). Distance around the inner edge = 2 × 106 + (22/7) × 60 = 212 + 188.57 = 400.57 m. Outer edge = 212 + (22/7) × 80 = 212 + 251.43 = 463.43 m. Area of the track = 2 × 106 × 10 + (22/7)(40² − 30²) = 2120 + (22/7) × 700 = 2120 + 2200 = 4320 m².

Worked example. Find the perimeter of a quadrant of a circle of radius 14 cm. Perimeter = ½ × (22/7) × 14 + 2 × 14 = 22 + 28 = 50 cm.

Worked example. A piece of wire 22 cm long is bent into the form of an arc of a circle subtending 60° at the centre. Find the radius. (60/360) × 2 × (22/7) × r = 22, so (1/6) × (44/7) r = 22, r = 22 × 6 × 7/44 = 21 cm.

Worked example. A rectangle of length 20 cm and breadth 14 cm has semicircles drawn outward on its two breadths. Find the perimeter and area of the resulting figure (π = 22/7). The two semicircles make a circle of radius 7: perimeter = 2 × 20 + 2 × (22/7) × 7 = 40 + 44 = 84 cm; area = 20 × 14 + (22/7) × 49 = 280 + 154 = 434 cm². Note the breadths are inside the figure and are not part of the perimeter.

Worked example. Find the perimeter of the shaded region formed by the segment of a circle of radius 10 cm cut by a chord subtending 90°: arc = ¼ × 2 × 3.14 × 10 = 15.7; chord = 10√2 = 14.14; perimeter = 29.84 cm.

Distinguish: the perimeter of a sector includes the radii, the perimeter of a segment includes the chord, and the perimeter of a ring counts both circles.

This topic shows two complete model answers in examination form and lists the mistakes that cost marks year after year.

Model answer 1. Find the area of the segment AYB of a circle of radius 21 cm if ∠AOB = 120° (π = 22/7, √3 = 1.73).
Area of sector OAYB = (120/360) × (22/7) × 21 × 21 = (1/3) × 1386 = 462 cm². … (1)
Draw OM ⊥ AB. In ΔOMA, ∠AOM = 60°, so OM = OA cos 60° = 21/2 cm and AM = OA sin 60° = 21√3/2 cm; AB = 21√3 cm.
Area of ΔOAB = ½ × AB × OM = ½ × 21√3 × 21/2 = (441√3)/4 = 441 × 1.73/4 = 190.73 cm². … (2)
Area of segment AYB = (1) − (2) = 462 − 190.73 = 271.27 cm².

Model answer 2. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the design (√3 = 1.7).
Join O to A, B, C. Each of the angles AOB, BOC, COA is 120°. Draw OM ⊥ BC; OM = 32 cos 60° = 16 cm, BM = 32 sin 60° = 16√3 cm, BC = 32√3 cm.
Area of ΔABC = (√3/4) × (32√3)² = (√3/4) × 3072 = 768√3 = 768 × 1.7 = 1305.6 cm².
Area of the circle = (22/7) × 32 × 32 = 22528/7 = 3218.29 cm².
Area of the design = 3218.29 − 1305.6 = 1912.69 cm².

Common errors and how to avoid them.

• Using the diameter as the radius. Read the question twice: a circle of diameter 14 cm has r = 7 cm.
• Forgetting the two radii in a sector's perimeter, or the chord in a segment's perimeter.
• Using the wrong value of π. If the question says π = 22/7, using 3.14 gives a different decimal and loses the accuracy mark.
• Computing the major segment as sector minus triangle. The major segment is the circle minus the minor segment.
• Treating the triangle in a 120° segment as equilateral. It is isosceles with a 120° apex; only the 60° case is equilateral.
• Mixing units: metres with centimetres, or cm with cm².
• Rounding too early. Keep fractions until the last line.
• Omitting the unit or writing cm for an area.

Quick reference. Circle 2πr, πr². Sector (θ/360) × 2πr, (θ/360) × πr², ½lr. Segment sector − triangle. Ring π(R² − r²). Triangle areas at 60°, 90°, 120°: (√3/4)r², ½r², (√3/4)r². Four vertex sectors of a quadrilateral = one circle; three of a triangle = half a circle.

With these formulas and the strategy of naming pieces, finding dimensions and writing a sum or difference, every question from this chapter is within reach.