Why it happens: the sector formula rests on the same symmetry argument as the arc formula. Rotating the disc about its centre leaves it unchanged, so 360 one-degree sectors are all congruent and each has 1/360 of the area. A 60° sector is 60 of them — one sixth of the disc.
Q2.
Find the area of a quadrant of a circle whose circumference is 44 cm.
Answer
Area = 38.5 cm².
2πr = 44 2 × 22⁄7 × r = 44 ⟹ r = 7 cm
area of circle = πr² = 22⁄7 × 49 = 154 cm²
a quadrant is θ = 90°, i.e. 90⁄360 = ¼ of the circle area of quadrant = ¼ × 154 = 38.5 cm²
Tip: a "quadrant" is a quarter disc — the region — while a "quarter circle" usually means just the arc. Read which one the question wants.
Q3.
The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
Answer
Area swept = 77⁄3 ≈ 25.67 cm².
In 60 minutes the minute hand turns through 360°. In 10 minutes it turns through 360° × 10⁄60 = 60°
The region swept is a sector of radius 7 cm and angle 60°. area = πr² × 60⁄360 = 22⁄7 × 49 × 1⁄6 = 154 × 1⁄6 = 25.67 cm² (2 d.p.)
Why it happens: the hand is a segment pinned at the centre. As it turns, the region it sweeps out is bounded by the two positions of the hand and by the arc traced by its tip — that is exactly a sector.
Tip: the Hindi edition asks for the distance covered by the hand in 10 minutes instead. That is the arc traced by the tip: 2πr × 60⁄360 = 44 × 1⁄6 = 22⁄3 ≈ 7.33 cm.
Q4.
A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (i) minor sector (that subtends 90° at the centre), and (ii) major sector (that subtends 270° at the centre). (Use π ≈ 3.14.)
Answer
(i) 78.5 cm² (ii) 235.5 cm².
area of the whole circle = πr² = 3.14 × 10² = 314 cm²
(ii) major sector, θ = 270° = 314 × 270⁄360 = 314 × ¾ = 235.5 cm²
Check it yourself: 78.5 + 235.5 = 314 = the whole circle ✓. The two sectors between them use up every degree, so their areas must add to πr².
Why it happens: the chord splits the disc into two segments, but the two radii split it into two sectors. Here we were asked for the sectors, so no triangle has to be subtracted. Question 5 asks for the segments, and there the triangle does have to come off.
Q5.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and √3 ≈ 1.73.)
Answer
Minor segment ≈ 20.44 cm²; major segment ≈ 686.06 cm².
A segment is a sector with the triangle cut off.
area of sector (60°) = πr² × 60⁄360 = 3.14 × 225 × 1⁄6 = 706.5⁄6 = 117.75 cm²
The triangle formed by the chord and the two radii has two sides of 15 cm with 60° between them ⟹ it is equilateral with side 15 cm. area = √3⁄4 × 15² = 1.73⁄4 × 225 = 0.4325 × 225 = 97.3125 cm²
major segment = whole circle − minor segment = 3.14 × 225 − 20.4375 = 706.5 − 20.4375 = 686.06 cm² (2 d.p.)
Sector (blue + red) minus the equilateral triangle (blue) leaves the minor segment (red).
Why it happens: a chord that subtends 60° at the centre is special — the isosceles triangle with apex angle 60° has base angles 60° too, so it is equilateral and the chord equals the radius. That is why the whole calculation stays exact until the numerical approximations are put in. The minor segment turns out to be tiny (about 3% of the disc) because a 60° sector is already thin and the triangle fills most of it.
Tip: the Hindi edition does not state values for π and √3 in this question; the answers above use the same π ≈ 3.14 and √3 ≈ 1.73 that the English edition specifies.
Q6.
A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.
Answer
Total area cleaned = 4928⁄3 ≈ 1642.67 cm².
Each blade sweeps a sector of radius 28 cm and angle 120°.
area of one sector = πr² × θ°⁄360° = 22⁄7 × 28² × 120⁄360 = 22⁄7 × 784 × 1⁄3 = 2464 × 1⁄3 = 2464⁄3 ≈ 821.33 cm²
two wipers, no overlap: total = 2 × 2464⁄3 = 4928⁄3 = 1642.67 cm² (2 d.p.)
Why it happens: the phrase "which do not overlap" is the mathematical content of the question. If the two swept sectors overlapped, adding them would count the common part twice and the answer would be too big. Because they are disjoint, the areas simply add.
Tip: a real wiper blade does not reach the pivot, so the region it cleans is an annular sector — the difference of two sectors. The question idealises the blade as reaching all the way in.
Q7.
*A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to πr²(1/6 − √3/4).
Answer
Take the sector and subtract the triangle.
area of the 60° sector = πr² × 60⁄360 = πr²⁄6
The chord and the two radii make a triangle with two sides r and 60° between them. Its base angles are each ½(180° − 60°) = 60°, so it is equilateral with side r. area of triangle = √3⁄4 r²
Tip — a misprint to be aware of: the English edition prints the answer as πr²(1/6 − √3/4). Multiplying out gives πr²⁄6 − π√3 r²⁄4, which is not the segment area (it is in fact negative). The correct expression is r²(π/6 − √3/4) — with π inside the bracket — and that is exactly what the Hindi edition prints. Numerically, r²(0.5236 − 0.4330) = 0.0906 r², a small positive area, as a thin segment should be.
Why it happens: a segment is always "sector minus triangle", and the 60° case is the one where the triangle is easiest, since the chord equals the radius. Substituting r = 15, π ≈ 3.14 and √3 ≈ 1.73 gives 225(0.52333 − 0.4325) = 225 × 0.09083 = 20.44 cm² — exactly the answer to Question 5.
Q8.
*An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is equal to 3√3/(4π) ≈ 0.413.
Answer
First find the side of the triangle in terms of r.
Let the triangle be ABC with centre O of the circle, OA = OB = OC = r. By symmetry the three central angles ∠AOB, ∠BOC, ∠COA are equal, so each is 360° ÷ 3 = 120°.
Drop a perpendicular OM from O to the chord AB. It bisects both AB and ∠AOB. ∴ ∠AOM = 60°, and in right triangle OAM, AM = OA sin 60° = r·√3⁄2 ∴ side AB = 2AM = √3 r
area of the equilateral triangle = √3⁄4 (side)² = √3⁄4 × 3r² = 3√3⁄4 r²
Why it happens: the r² cancels, so the ratio is a pure number — it does not depend on how big the circle is. That is the same scaling principle as for C/D: any two "equilateral triangle inscribed in a circle" figures are similar, so every ratio of areas within them is fixed. The number 0.413 says the triangle wastes nearly 60% of the disc — three big segments are left over.
Tip: a useful fact to carry away — for an equilateral triangle, side = √3 × circumradius, and circumradius = 2 × inradius.
Q9.
*A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to 2/π ≈ 0.637.
Answer
The diagonal of an inscribed square is a diameter.
Each vertex of the square lies on the circle, and opposite vertices subtend 180° at the centre, so the diagonal passes through O. diagonal = 2r
If the side is a, then by the Baudhāyana–Pythagoras theorem a² + a² = (2r)² 2a² = 4r² a² = 2r² ⟹ area of the square = 2r²
area of circle = πr²
ratio = 2r²⁄πr² = 2⁄π = 2⁄3.14159… ≈ 0.637 ∎
Why it happens: notice that the side of the square never had to be found — only its square, which is what the area needs. The inscribed square covers about 64% of the disc, more than the inscribed equilateral triangle's 41%. That is no accident: the more sides an inscribed regular polygon has, the closer it hugs the circle, and the closer the ratio creeps to 1. Question 10 does the hexagon (82.7%), and Archimedes pushed this all the way to 96 sides.
Q10.
*A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to 3√3/(2π) ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?
Answer
A regular hexagon inscribed in a circle is six equilateral triangles glued at the centre.
Each central angle = 360° ÷ 6 = 60°. The two radii and the side make an isosceles triangle with a 60° apex, so it is equilateral with side r.
area of one such triangle = √3⁄4 r² area of hexagon = 6 × √3⁄4 r² = 3√3⁄2 r²
ratio = (3√3/2) r²⁄πr² = 3√3⁄2π ≈ 5.196⁄6.283 ≈ 0.827
Why exactly twice the answer to Question 8.
hexagon = 3√3⁄2 r² inscribed equilateral triangle = 3√3⁄4 r² hexagon = 2 × triangle, so the ratios are in the same 2 : 1 relation.
Geometrically: label the hexagon's vertices A, B, C, D, E, F. Joining the alternate vertices A, C, E gives exactly the inscribed equilateral triangle of Question 8. It leaves three corner triangles ABC, CDE, EFA. Each of these has two sides equal to a side of the hexagon and the 120° interior angle between them, and all three are congruent. Now ΔACE is made of three equilateral triangles of side r (join O to A, C, E), so
So the three corner triangles together equal the middle triangle: the hexagon is twice the inscribed equilateral triangle. ∎
The inscribed triangle ACE (blue) and the three leftover corner triangles (pink) are equal in total area — so the hexagon is twice the triangle.
Why it happens: six equilateral triangles round a point close up exactly, because 6 × 60° = 360°. This one fact makes the regular hexagon the easiest polygon to inscribe in a circle — the side equals the radius, so a single compass setting steps round the circle in six moves. It is also the reason honeycombs and floor tilings are hexagonal.