Step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 6 Exercise Set 6.3 (Questions 1–10). Covers sector areas, car wipers, segments, and ratio proofs for inscribed equilateral triangles, squares, and regular hexagons.
A sector is a portion of a circle enclosed by two radii and the intercepted arc. Its area is proportional to its central angle θ out of the full 360° circle.
The angle subtended by the arc of a sector at the centre of the circle. It determines what fraction (θ / 360°) of the full circle's area is enclosed by the sector.
A quadrant is one-fourth of a complete circle, formed by two perpendicular radii meeting at a right angle of 90° at the centre.
When the circumference C of a circle is known, the radius r is determined using C = 2πr before calculating the area.
A minute hand completes a full revolution of 360° in 60 minutes (6° per minute). The angle swept in m minutes is (m / 60) × 360°.
A pair of radii divides a circle into two sectors. The sector with central angle θ < 180° is the minor sector, and the remaining region with angle (360° − θ) is the major sector. Their sum equals the total area of the circle.
A chord divides the circular region into two parts called segments. The minor segment is bounded by the chord and the minor arc, while the major segment is bounded by the chord and the major arc.
When a chord subtends 60° at the centre of a circle, the triangle formed with two radii is equilateral (all three angles 60° and sides equal to radius r).
When an equilateral triangle ABC is inscribed in a circle of radius r, each side subtends a central angle of 120°. Dropping a perpendicular from the centre forms a 30°–60°–90° triangle giving side length a = √3 r.
When a square is inscribed in a circle of radius r, its diagonal equals the diameter of the circle (2r). By Pythagoras' theorem, the side length is a = √2 r and its area is 2r².
A regular hexagon inscribed in a circle of radius r consists of 6 congruent equilateral triangles of side r (since 360° / 6 = 60°). Its area is exactly twice that of an inscribed equilateral triangle.
Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Radius of the circle, r = 7 cm
Central angle of the sector, θ = 60°
Using the textbook default approximation π = 22/7:
Find the area of a quadrant of a circle whose circumference is 44 cm.
A quadrant represents one-fourth of a circle with central angle 90°. From circumference 44 cm, radius is 7 cm and quadrant area is 77/2 cm² (38.50 cm²).
Using π = 22/7:
A quadrant is one-fourth of a circle (central angle θ = 90°):
The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
In 10 minutes, the minute hand sweeps through (10/60) × 360° = 60°. The sector swept has radius 7 cm and area 77/3 cm² (approx. 25.67 cm²).
The minute hand completes one full revolution of 360° in 60 minutes.
Using π = 22/7:
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.)
A chord subtends 90° at the centre of a 10 cm circle. The minor sector subtends 90° (area = 78.5 cm²) and the major sector subtends 270° (area = 235.5 cm²).
Radius of the circle, r = 10 cm
Central angle for minor sector, θ = 90°
We know that the area of a sector of a circle with radius r and central angle θ is given by:
The major sector subtends an angle of 360° − 90° = 270° at the centre.
Alternative verification for major sector:
Both methods give the exact same result.
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.)
A chord subtends 60° at the centre of a 15 cm circle, forming equilateral △OAB. Minor segment area = minor sector (117.75 cm²) − △OAB (97.31 cm²) ≈ 20.44 cm²; major segment area ≈ 686.06 cm².
Radius of the circle, r = 15 cm
Central angle subtended by chord AB, θ = 60°
Approximations specified: π ≈ 3.14 and √3 ≈ 1.73
Area of the minor sector OAB:
Area of triangle OAB:
In △OAB, OA = OB = 15 cm (radii of the circle).
Since OA = OB, the angles opposite to these sides are equal:
In △OAB:
2∠OAB = 120° ⟹ ∠OAB = ∠OBA = 60°
Since each angle is 60°, △OAB is an equilateral triangle with side a = 15 cm.
We know that the area of an equilateral triangle of side a is:
Area of the minor segment:
Rounding to two decimal places: ≈ 20.44 cm²
Area of the major segment:
The major segment comprises the entire circle except the minor segment:
Rounding to two decimal places: ≈ 686.06 cm²
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.
Each wiper sweeps out a 120° sector with radius 28 cm. Since the two swept sectors do not overlap, the total area cleaned is twice the area of one sector.
Sweeping angle of each wiper, θ = 120°
Number of wipers = 2 (which do not overlap)
Each wiper sweeps out a sector of a circle with radius r = 28 cm and central angle θ = 120°.
Using π = 22/7:
Total area cleaned by two wipers:
Since the two wipers do not overlap, the total area cleaned is twice the area of one sector:
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).
Chord AB subtends 60° at centre O, forming equilateral △AOB of side r. The minor segment area is obtained by subtracting the area of equilateral △AOB from the area of minor sector AOB.
The mathematically correct result to be proved is:
r²(π/6 − √3/4)
This correction is necessary because the area of the minor segment is:
Area of minor sector − Area of triangle.
Let the chord be AB, and O be the centre of the circle.
Radius of the circle, OA = OB = r
Central angle subtended by chord AB, ∠AOB = 60°
In △AOB:
OA = OB = r (radii of the same circle)
Since OA = OB, the angles opposite to these sides are equal:
In △AOB, sum of interior angles is 180°:
2∠OAB = 120° ⟹ ∠OAB = ∠OBA = 60°
Since all three interior angles are 60°, △AOB is an equilateral triangle with side AB = r.
Area of the minor sector:
The minor sector subtends 60° at the centre.
Area of △AOB:
Since △AOB is an equilateral triangle with each side equal to r:
Area of the minor segment:
The minor segment is the region bounded by chord AB and the minor arc AB.
Taking r² common from both terms:
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.
Equilateral triangle ABC is inscribed in a circle with centre O and radius r. Perpendicular OM from centre O bisects central angle AOB (120°) and side AB, yielding 30°–60°–90° triangle OMA with side AB = √3 r.
Let ABC be an equilateral triangle inscribed in a circle with centre O and radius r.
Since △ABC is equilateral, all three sides are equal chords of the circle:
Equal chords of a circle subtend equal angles at the centre. Therefore, the three central angles are equal:
∠AOB = ∠BOC = ∠COA = 360° / 3 = 120°
Draw perpendicular OM from centre O to side AB (with M on AB).
In △OAB, OA = OB = r (radii of the circle).
Since △OAB is an isosceles triangle with OA = OB, the perpendicular OM from vertex O bisects both the central angle ∠AOB and the chord AB:
∠AOM = ∠BOM = 120° / 2 = 60°
and M is the midpoint of AB (AM = MB).
In right-angled triangle OMA (where ∠OMA = 90°):
∠OAM = 180° − (90° + 60°) = 30°
Thus, △OMA is a 30°–60°–90° right triangle.
Using trigonometry in right triangle OMA:
AM = r × (√3 / 2) = (√3 / 2) r
Since M is the midpoint of AB:
Area of the equilateral triangle:
We know that the area of an equilateral triangle with side a is:
Area of the circle:
Ratio of the area of the triangle to the area of the circle:
Ratio = Area of triangle / Area of circle
Cancelling r² from the numerator and denominator:
Using π ≈ 3.14159 and √3 ≈ 1.73205:
(3 × 1.73205) / (4 × 3.14159) = 5.19615 / 12.56637 ≈ 0.41348 ≈ 0.413 (or 0.41)
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.
Square ABCD is inscribed in a circle with centre O and radius r. The diagonal AC passes through centre O and equals diameter 2r. By Pythagoras' theorem, side a = √2 r and area of the square = 2r².
Let ABCD be a square inscribed in a circle with centre O and radius r.
Since ABCD is a square inscribed in the circle, each interior angle is 90° (for instance, ∠ABC = 90°).
By the angle in a semicircle theorem, the diagonal AC subtending 90° must be a diameter of the circle:
Let the side of the square be a.
In right-angled triangle ABC (with ∠B = 90° and AB = BC = a):
Area of the square:
Area of the circle:
Ratio of the area of the square to the area of the circle:
Ratio = Area of square / Area of circle
Cancelling r² from the numerator and denominator:
Using π ≈ 3.14159:
2 / 3.14159 ≈ 0.63662... ≈ 0.637 (or 0.64)
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?
A regular hexagon inscribed in a circle of radius r is partitioned into six congruent equilateral triangles of side r. Its area is 6 × (√3/4)r² = (3√3/2)r², exactly twice the area of an inscribed equilateral triangle.
As intended by the question and textbook context, the inscribed hexagon is a regular hexagon.
Let ABCDEF be a regular hexagon inscribed in a circle with centre O and radius r.
Partitioning into six central triangles:
Join the centre O to each of the six vertices A, B, C, D, E, F.
This divides the regular hexagon into 6 congruent triangles:
△OAB, △OBC, △OCD, △ODE, △OEF, △OFA
The six central angles around centre O are equal and sum to 360°:
In △OAB:
OA = OB = r (radii of the circumcircle)
Since OA = OB, the base angles are equal:
∠OAB = ∠OBA = (180° − 60°) / 2 = 60°
Since each angle is 60°, △OAB is an equilateral triangle with side length equal to r.
Area of one equilateral triangle:
Area of the regular hexagon:
Since the regular hexagon consists of 6 congruent equilateral triangles:
Area of the circle:
Ratio of the area of the hexagon to the area of the circle:
Ratio = Area of hexagon / Area of circle
Cancelling r² from numerator and denominator:
Using π ≈ 3.14159 and √3 ≈ 1.73205:
(3 × 1.73205) / (2 × 3.14159) = 5.19615 / 6.28318 ≈ 0.8270... ≈ 0.827 (or 0.83)
Why the answer is exactly twice the answer to Question 8:
From Question 8, the ratio for an equilateral triangle inscribed in a circle of radius r is:
(Area of triangle) / (Area of circle) = (3√3) / (4π)
From Question 10, the ratio for a regular hexagon inscribed in the same circle is:
(Area of hexagon) / (Area of circle) = (3√3) / (2π)
Notice that:
(3√3) / (2π) = 2 × ((3√3) / (4π))
Geometric explanation:
An inscribed regular hexagon has twice the area of an inscribed equilateral triangle in the same circle (Area of hexagon = (3√3/2)r² = 2 × (3√3/4)r²). In fact, joining alternate vertices of a regular hexagon creates an inscribed equilateral triangle covering exactly half the hexagon's area. Therefore, the hexagon ratio is exactly twice the triangle ratio.
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