Areas Related to Circles

Step 1 — what is given. Radius \(r = 6\) cm and angle \(\theta = 60^{\circ}\). No value of \(\pi\) is stated, so the exercise's own instruction applies and \(\pi = \tfrac{22}{7}\).

Step 2 — the formula.

Step 3 — substitute.

Step 4 — simplify. Here \(\tfrac{60}{360} = \tfrac{1}{6}\), and \(36 \div 6 = 6\):

So the area of the sector is \(\tfrac{132}{7}\) cm², which is \(18\tfrac{6}{7}\) cm² or about 18.86 cm².

Step 1 — what a quadrant is. A quadrant is one quarter of the disc, so it is the sector of angle \(90^{\circ}\). The radius is not given directly, so it has to be found from the circumference first.

Step 2 — find the radius. The circumference of a circle of radius \(r\) is \(2\pi r\), and here that equals 22 cm:

Step 3 — solve for \(r\). Multiply both sides by 7 and divide by 44:

Step 4 — apply the sector formula with \(\theta = 90^{\circ}\).

Step 5 — simplify. Since \(r^{2} = \left(\tfrac{7}{2}\right)^{2} = \tfrac{49}{4}\),

So the area of the quadrant is \(\tfrac{77}{8}\) cm² = 9.625 cm².

Step 1 — see the region. As the minute hand turns, its tip traces an arc and the hand sweeps out a sector of a circle whose radius is the length of the hand, \(r = 14\) cm. The only thing left to find is the angle of that sector.

Step 2 — find the angle. The minute hand takes 60 minutes to go once round, which is \(360^{\circ}\). So in 5 minutes it turns through \(\tfrac{5}{60}\) of a full turn:

Step 3 — apply the sector formula.

Step 4 — simplify. Here \(\tfrac{30}{360} = \tfrac{1}{12}\), and \(\tfrac{22}{7}\times 196 = 22 \times 28 = 616\):

So the area swept in 5 minutes is \(\tfrac{154}{3}\) cm², which is \(51\tfrac{1}{3}\) cm² or about 51.33 cm².

Step 1 — what is given. The radius is \(r = 10\) cm and the chord AB subtends a right angle at the centre, so \(\theta = 90^{\circ}\). Take \(\pi = 3.14\) as instructed.

Step 2 — the minor sector. Both parts need this, so find it first:

Step 3 — the triangle OAB. This is the one place where a right angle at the centre makes life easy: OA and OB are the two perpendicular sides, so they are the base and the height of the triangle and no trigonometry is needed at all.

Step 4 — part (i), the minor segment. The minor segment is what is left when the triangle is cut away from the minor sector:

Step 5 — part (ii), the major sector. The minor and major sectors together make the whole disc, so

Answers. (i) minor segment = 28.5 cm². (ii) major sector = 235.5 cm².

Worth noticing. Part (ii) asks for the major SECTOR, not the major segment — so the triangle plays no part in it. The major segment would be a different answer again: \(314 - 28.5 = 285.5\) cm². Read which of the four regions is being asked for before starting.

Step 1 — what is given. Radius \(r = 21\) cm and angle \(\theta = 60^{\circ}\). No value of \(\pi\) is stated, so \(\pi = \tfrac{22}{7}\).

Step 2 — part (i), the length of the arc. The arc is the fraction \(\theta/360\) of the whole circumference \(2\pi r\):

The whole circumference is 132 cm, and a sixth of it is 22 cm — a good check that the fraction was taken correctly.

Step 3 — part (ii), the area of the sector.

Step 4 — part (iii), set up the triangle. The segment is the sector minus the triangle OAB, so △OAB is needed next. Here something special happens: OA and OB are both radii, so the triangle is isosceles, and its apex angle is \(60^{\circ}\). The other two angles must therefore share the remaining \(120^{\circ}\) equally, giving \(60^{\circ}\) each. All three angles are \(60^{\circ}\), so △OAB is equilateral and every side equals the radius, 21 cm.

Step 5 — the area of the triangle. For an equilateral triangle of side \(a\), the area is \(\tfrac{\sqrt{3}}{4}a^{2}\):

Step 6 — part (iii), the segment.

Answers. (i) arc = 22 cm. (ii) sector = 231 cm². (iii) segment = \(\left(231 - \tfrac{441\sqrt{3}}{4}\right)\) cm².

A note on leaving \(\sqrt{3}\) in the answer. Unlike Q6 and Q7, this question gives no value for \(\sqrt{3}\), so the exact form above is the answer. If you do want a decimal, \(\sqrt{3}\approx 1.732\) gives \(\tfrac{441 \times 1.732}{4}\approx 190.96\), so the segment is about 40.04 cm².

Step 1 — what is given. Radius \(r = 15\) cm, angle \(\theta = 60^{\circ}\), and this time both \(\pi = 3.14\) and \(\sqrt{3} = 1.73\) are supplied, so the answers come out as decimals.

Step 2 — the minor sector.

Step 3 — the triangle. As in Q5, a \(60^{\circ}\) angle at the centre of a circle makes △OAB equilateral, so its side is the radius, 15 cm:

Step 4 — the minor segment.

Step 5 — the major segment. The minor and major segments together fill the whole disc, so subtract from \(\pi r^{2}\) rather than starting again:

Answers. minor segment = 20.4375 cm² (about 20.44 cm²); major segment = 686.0625 cm² (about 686.06 cm²).

A check worth doing. The two answers should add back up to the whole circle: \(20.4375 + 686.0625 = 706.5\) cm², which is exactly \(3.14 \times 225\). If your two segments do not add to \(\pi r^{2}\), one of them is wrong.

Step 1 — what is given. Radius \(r = 12\) cm and angle \(\theta = 120^{\circ}\), with \(\pi = 3.14\) and \(\sqrt{3} = 1.73\).

Step 2 — the sector.

Step 3 — why the triangle is harder here than in Q5 and Q6. At \(60^{\circ}\) the triangle came out equilateral and its area followed from a standard formula. At \(120^{\circ}\) it does not, so drop the perpendicular OM from the centre to the chord AB, exactly as in Example 2. Since OA = OB, that perpendicular bisects both the chord and the angle at O:

Step 4 — find the height OM. In the right triangle OMA, with hypotenuse OA = 12 cm,

Step 5 — find the base AB.

Step 6 — the area of the triangle.

Step 7 — the segment.

So the area of the corresponding segment is 88.44 cm².

Step 1 — see the shape of the grazing region. The horse can reach every point within 5 m of the peg, which would be a full circle if it were standing in open ground. But the peg is at a corner of the field, and the two sides of the field meeting at that corner are at right angles — so only the quarter of that circle lying inside the field is grass the horse can actually reach. The grazing region is a quadrant, the sector of angle \(90^{\circ}\).

Step 2 — check the quadrant really fits inside the field. This is worth a moment, because if the rope were longer than the side of the field the quarter circle would run off the far edge and the answer would not be a plain quadrant at all. Here the rope is 5 m and later 10 m, and both are less than the 15 m side, so in each case the quarter circle lies wholly inside the field.

Step 3 — part (i), the grazing area with a 5 m rope.

Step 4 — the grazing area with a 10 m rope. The same quadrant formula, with \(r = 10\):

Step 5 — part (ii), the increase. The question asks for the increase, not the new area, so subtract:

Answers. (i) 19.625 m². (ii) an increase of 58.875 m².

Worth noticing. Doubling the rope did not double the grazing area — it multiplied it by four, because area depends on \(r^{2}\). That is why the increase (58.875 m²) is three times the original area (19.625 m²), not one times it.

Step 1 — what is given. The diameter is \(d = 35\) mm, so the radius is \(r = \tfrac{35}{2}\) mm. No value of \(\pi\) is stated, so \(\pi = \tfrac{22}{7}\).

Step 2 — part (i), the wire round the rim. The wire runs once round the circle, so that part of it is the circumference. Using \(\pi d\) rather than \(2\pi r\) saves a step here:

Step 3 — the wire in the diameters. There are 5 diameters and each is 35 mm long:

Step 4 — the total.

Step 5 — part (ii), the angle of each sector. The 5 diameters cut the circle into 10 equal sectors, and the angles at the centre must share \(360^{\circ}\) equally:

Step 6 — the area of one sector. Rather than substituting \(36^{\circ}\) into the formula, note that one of ten equal sectors is simply one tenth of the disc:

Answers. (i) total wire = 285 mm. (ii) each sector = \(\tfrac{385}{4}\) mm² = 96.25 mm².

Step 1 — see the region. Flattened out, the umbrella is a circle of radius \(r = 45\) cm, and the 8 ribs are 8 equally spaced radii running from the centre to the rim. The region between two consecutive ribs is therefore a sector, and the 8 ribs cut the disc into 8 such sectors, all equal.

Step 2 — the angle between consecutive ribs.

Step 3 — the area of one sector. One of eight equal sectors is one eighth of the disc:

Step 4 — simplify. Here \(45^{2} = 2025\), and \(2025\) is divisible by 7 no further, so keep the fraction:

So the area between two consecutive ribs is \(\tfrac{22275}{28}\) cm², which is about 795.54 cm².

Step 1 — see the region for ONE wiper. A wiper blade of length 25 cm turning through \(115^{\circ}\) sweeps a sector of radius \(r = 25\) cm and angle \(\theta = 115^{\circ}\).

Step 2 — why “do not overlap” matters. The question says the two wipers do not overlap. That is the permission to simply add the two areas: if they overlapped, the shared strip would be counted twice and the total would be too big.

Step 3 — the area swept by one wiper.

Step 4 — simplify. Cancel 5 into \(115\) and \(360\), giving \(\tfrac{23}{72}\), and note \(25^{2} = 625\):

Step 5 — the total for both wipers. The blades are identical, so double it:

So the total area cleaned at each sweep is \(\tfrac{158125}{126}\) cm², which is about 1254.96 cm².

Step 1 — see the region. The light reaches 16.5 km in every direction within an \(80^{\circ}\) spread, so the warned region is a sector with \(r = 16.5\) km and \(\theta = 80^{\circ}\). Take \(\pi = 3.14\).

Step 2 — the formula.

Step 3 — substitute.

Step 4 — simplify. First \(\tfrac{80}{360} = \tfrac{2}{9}\), and \(16.5 \times 16.5 = 272.25\):

So the area of sea over which the ships are warned is 189.97 km².

Worth noticing. This one comes out to an exact two-decimal answer rather than a rounded one — \(\tfrac{2}{9}\times 854.865\) is exactly \(189.97\), because \(854.865 \times 2 = 1709.73\) is exactly divisible by 9.

Step 1 — identify what one “design” actually is. The six designs are the six pieces left between the sides of the regular hexagon and the rim of the cover. Each of those pieces is bounded by a chord (a side of the hexagon) and its arc — so each design is a segment of the circle.

Step 2 — find the angle each side subtends. The six sides of a regular hexagon are equal, so the six angles they subtend at the centre are equal and share \(360^{\circ}\) between them:

Step 3 — the sector. With \(r = 28\) cm and \(\theta = 60^{\circ}\), and \(\pi = \tfrac{22}{7}\) since no other value is given:

Step 4 — the triangle. Once again a \(60^{\circ}\) angle at the centre makes the triangle equilateral, so its side equals the radius, 28 cm. (This is also why a regular hexagon inscribed in a circle always has its side equal to the radius.)

Step 5 — the area of ONE design.

Step 6 — the area of all six.

Step 7 — the cost. At ₹ 0.35 for every square centimetre,

So the cost of making the designs is ₹ 162.68.

Worth noticing. The question asks for a COST, not an area, so the work is only finished at Step 7. It is also worth checking the size is sensible: the six designs come to 464.8 cm² out of a total cover area of \(\tfrac{22}{7}\times 784 = 2464\) cm², so the designs cover a little under a fifth of the cover — which matches the picture.

Step 1 — write down what the area actually is. A sector of angle \(p\) is the fraction \(\tfrac{p}{360}\) of the whole disc, whose area is \(\pi R^{2}\). So the area of the sector is

None of the four options is written in that form, so the task is to find which one is equal to it.

Step 2 — a shortcut before any algebra. Area is measured in square units, so the correct option must contain \(R^{2}\), not \(R\). Options (A) and (C) both contain only \(R\) — they are lengths, not areas — so they can be ruled out at a glance. That leaves (B) and (D).

Step 3 — test option (D). Simplify it by cancelling 2 into 720:

This is exactly the expression from Step 1, so (D) is correct.

Step 4 — confirm option (B) is wrong.

That is twice the correct area, because \(180\) has been used where \(360\) belongs. So (B) is wrong.

Step 5 — what (A) and (C) really are. It is worth naming them rather than just discarding them. Option (C), \(\tfrac{p}{360}\times 2\pi R\), is the length of the arc of the same sector — a genuine formula from this chapter, just answering a different question. Option (A) is twice that.

So the correct option is (D).