Areas Related to Circles Class 10 — Competency Based Questions

20 Questions | 76 Marks

A circular garden has a radius of 21 meters. It has a sector with a central angle of 60 degrees. Inside this sector, a small triangular area is to be re-landscaped.(a) Calculate the area of the sector of the garden. [2 marks](b) If the triangular area to be re-landscaped has its vertices at the center and two points on the arc such that the angle subtended at the center is 60 degrees, find the area of this triangle. [2 marks](c) What is the area of the segment of the garden that will remain as a sector after re-landscaping? 2marks

A clock has a minute hand of length 14 cm.(a) What is the angle swept by the minute hand in 15 minutes? [1 mark](b) Calculate the area on the clock face that is swept by the minute hand in 15 minutes. [2 marks](c) If the hour hand has a length of 7 cm, calculate the area swept by the hour hand in 1 hour. 2marks

Assertion: The area of a major sector is always greater than the area of a minor sector.Reason: A major sector corresponds to a reflex angle at the center, which is greater than 180°.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true but reason is false.(d) Assertion is false but reason is true. 1mark

A circular park has a radius of 50 meters. It is divided into sectors of different angles for different activities. A specific sector, used for meditation, has a central angle of 72°.(a) What is the area of the entire circular park? [1 mark](b) Find the area of the meditation sector. [2 marks](c) If a path of width 2 meters is to be built around this meditation sector along the arc, what is the area of this path? 3marks

A designer is creating a piece of jewelry. It consists of a circular pendant with a radius of 5 cm. The pendant is divided into 10 equal sectors.(a) Calculate the angle of each sector. [1 mark](b) Find the area of each sector. [2 marks](c) If the wire used to form the boundary of each sector (two radii and an arc) is to be calculated, what is the length of the wire for one sector? 2marks

Assertion: The area of a segment is found by subtracting the area of a triangle from the area of a sector.Reason: A segment is the region bounded by a chord and the arc.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true but reason is false.(d) Assertion is false but reason is true. 1mark

A Ferris wheel has a radius of 30 meters. It completes one full rotation in 2 minutes.(a) How many degrees does the Ferris wheel rotate in 30 seconds? [1 mark](b) What is the area swept by the Ferris wheel in 30 seconds? [2 marks](c) If the Ferris wheel is stopped after rotating 90 degrees, what is the area covered by the arc? 2marks

In a school playground, there is a circular fountain of radius 7 meters. A lawn surrounds the fountain. The lawn is divided into two parts: an inner part which is a sector with a central angle of 120°, and an outer part which is a segment formed by a chord.(a) Calculate the area of the fountain. [1 mark](b) Find the area of the inner sector of the lawn. [2 marks](c) If the chord forming the segment is such that it subtends an angle of 120° at the center, find the area of the segment. 3marks

A square is inscribed in a circle of radius 14 cm. The circle is then divided into 4 equal sectors.(a) Find the area of the circle. [1 mark](b) Calculate the area of each sector. [2 marks](c) What is the area of the square inscribed in the circle? 2marks

A fan is designed in the shape of a sector of a circle with radius 1 meter. The blades of the fan sweep through an angle of 150°.(a) Calculate the area swept by the fan blades in one sweep. [2 marks](b) If the fan rotates at a speed such that it sweeps an angle of 300° per second, what is the area swept per second? 3marks

A sector of a circle has an area of 154 sq. cm. The radius of the circle is 7 cm.(a) Find the angle of the sector in degrees. [3 marks](b) Calculate the length of the arc of this sector. 2marks

Assertion: If the radius of a circle is doubled, its area becomes four times.Reason: The area of a circle is directly proportional to the square of its radius.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true but reason is false.(d) Assertion is false but reason is true. 1mark

Two concentric circles have radii 7 cm and 21 cm. A sector of angle 60° is drawn in both circles.(a) Find the area of the sector in the smaller circle. [2 marks](b) Find the area of the sector in the larger circle. [2 marks](c) Calculate the area of the region between the two sectors (annular sector). 2marks

A farmer has a piece of land in the shape of a sector of a circle with radius 10 meters and a central angle of 72°. He wants to plant flowers in this region.(a) Calculate the area of the land where he can plant flowers. [2 marks](b) He decides to fence the boundary of this sector. Find the length of the fence required. 2marks

Assertion: The perimeter of a sector is the sum of the lengths of two radii and the corresponding arc.Reason: A sector is bounded by two radii and an arc.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true but reason is false.(d) Assertion is false but reason is true. 1mark

A windshield wiper sweeps over a sector of a car windshield. The length of the blade is 50 cm, and it sweeps through an angle of 120°.(a) Find the area of the windshield cleaned by one sweep of the blade. [2 marks](b) If there are two such wipers that do not overlap, what is the total area cleaned by both wipers? 1mark

A circular table cover has a radius of 35 cm. It has 7 equal designs of a sector shape, which are equally spaced.(a) Find the angle of each sector design. [2 marks](b) Calculate the area of each sector design. [2 marks](c) If the cost of making these designs is ₹ 0.50 per sq. cm, find the total cost. 1mark

A boy is flying a kite. The string of the kite is 50 meters long. When the kite is at an angle of 60° with the horizontal, find the area of the sector formed by the string's path in the air. Assume the string is taut and forms a straight line from the boy's hand to the kite. 3marks

Assertion: If two sectors have the same central angle, then the ratio of their areas is equal to the ratio of the squares of their radii.Reason: The area of a sector is given by

\frac{θ}{360} π r^{2}

.(a) Both assertion and reason are true and reason is the correct explanation of assertion.(b) Both assertion and reason are true but reason is not the correct explanation of assertion.(c) Assertion is true but reason is false.(d) Assertion is false but reason is true. 1mark

A circular disc of radius 10 cm is cut into 6 equal sectors.(a) What is the angle of each sector? [1 mark](b) Find the area of one sector. [2 marks](c) Calculate the perimeter of one sector. 2marks

Answers

(a) Area of sector =

\frac{60}{360} \times π \times (21)^{2} = \frac{1}{6} \times \frac{22}{7} \times 441 = 11 \times 21 = 231

sq. cm. 2marks(b) Area of triangle =

\frac{1}{2} r^{2} \sin θ = \frac{1}{2} \times (21)^{2} \times \sin 60^{\circ} = \frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2} = \frac{441 \sqrt{3}}{4}

sq. cm. [2 marks](c) Area of segment = Area of sector - Area of triangle =

231 - \frac{441 \sqrt{3}}{4}

sq. cm. [2 marks]

(a) Angle swept by minute hand in 60 minutes = 360°. Angle swept in 1 minute =

360^{\circ} / 60 = 6^{\circ}

. Angle swept in 15 minutes =

15 \times 6^{\circ} = 90^{\circ}

. 1mark(b) Area swept =

\frac{90}{360} \times π \times (14)^{2} = \frac{1}{4} \times \frac{22}{7} \times 196 = 11 \times 14 = 154

sq. cm. [2 marks](c) Angle swept by hour hand in 12 hours = 360°. Angle swept in 1 hour =

360^{\circ} / 12 = 30^{\circ}

. Area swept by hour hand =

\frac{30}{360} \times π \times (7)^{2} = \frac{1}{12} \times \frac{22}{7} \times 49 = \frac{11 \times 7}{6} = \frac{77}{6}

sq. cm. [2 marks]

(a) Both assertion and reason are true and reason is the correct explanation of assertion. 1mark

(a) Area of circular park =

π r^{2} = \frac{22}{7} \times (50)^{2} = \frac{22}{7} \times 2500 = \frac{55000}{7}

sq. meters. 1mark(b) Area of meditation sector =

\frac{72}{360} \times π \times (50)^{2} = \frac{1}{5} \times \frac{22}{7} \times 2500 = \frac{11000}{7}

sq. meters. [2 marks](c) Radius of the outer edge of the path =

50 + 2 = 52

meters. Area of the larger sector (including path) =

\frac{72}{360} \times π \times (52)^{2} = \frac{1}{5} \times \frac{22}{7} \times 2704 = \frac{59488}{35}

sq. meters. Area of the path = Area of larger sector - Area of meditation sector =

\frac{59488}{35} - \frac{11000}{7} = \frac{59488 - 55000}{35} = \frac{4488}{35}

sq. meters. [3 marks]

(a) Angle of each sector =

360^{\circ} / 10 = 36^{\circ}

. 1mark(b) Area of each sector =

\frac{36}{360} \times π \times (5)^{2} = \frac{1}{10} \times \frac{22}{7} \times 25 = \frac{110}{7}

sq. cm. [2 marks](c) Length of wire for one sector =

2 r + arc length = 2 (5) + \frac{36}{360} \times 2 π (5) = 10 + \frac{1}{10} \times 2 \times \frac{22}{7} \times 5 = 10 + \frac{22}{7} = \frac{70 + 22}{7} = \frac{92}{7}

cm. [2 marks]

(a) Both assertion and reason are true and reason is the correct explanation of assertion. 1mark

(a) Time for 360° rotation = 2 minutes = 120 seconds. Time for 1 second rotation =

360^{\circ} / 120 = 3^{\circ}

. Angle rotated in 30 seconds =

30 \times 3^{\circ} = 90^{\circ}

. 1mark(b) Area swept =

\frac{90}{360} \times π \times (30)^{2} = \frac{1}{4} \times \frac{22}{7} \times 900 = \frac{11 \times 450}{7} = \frac{4950}{7}

sq. meters. [2 marks](c) This question seems to be asking for the length of the arc, not the area. If it asks for area of sector: Area =

\frac{90}{360} \times π \times (30)^{2} = \frac{4950}{7}

sq. meters. If it asks for arc length: Arc length =

\frac{90}{360} \times 2 π (30) = \frac{1}{4} \times 60 π = 15 π = \frac{330}{7}

meters. Assuming it means area of the sector. [2 marks]

(a) Area of fountain =

π r^{2} = \frac{22}{7} \times (7)^{2} = \frac{22}{7} \times 49 = 22 \times 7 = 154

sq. meters. 1mark(b) Area of inner sector =

\frac{120}{360} \times π \times (7)^{2} = \frac{1}{3} \times \frac{22}{7} \times 49 = \frac{1}{3} \times 22 \times 7 = \frac{154}{3}

sq. meters. [2 marks](c) Area of segment = Area of sector - Area of triangle. Area of triangle =

\frac{1}{2} r^{2} \sin θ = \frac{1}{2} \times (7)^{2} \times \sin 120^{\circ} = \frac{49}{2} \times \frac{\sqrt{3}}{2} = \frac{49 \sqrt{3}}{4}

sq. meters. Area of segment =

\frac{154}{3} - \frac{49 \sqrt{3}}{4}

sq. meters. [3 marks]

(a) Area of circle =

π r^{2} = \frac{22}{7} \times (14)^{2} = \frac{22}{7} \times 196 = 22 \times 28 = 616

sq. cm. 1mark(b) Area of each sector =

616 / 4 = 154

sq. cm. [2 marks](c) The diagonal of the inscribed square is the diameter of the circle,

2 r = 2 \times 14 = 28

cm. Let the side of the square be 's'. By Pythagorean theorem,

s^{2} + s^{2} = (28)^{2}

, so

2 s^{2} = 784

s^{2} = 392

. Area of the square =

s^{2} = 392

sq. cm. [2 marks]

(a) Area swept =

\frac{150}{360} \times π \times (1)^{2} = \frac{5}{12} \times \frac{22}{7} = \frac{110}{84} = \frac{55}{42}

sq. meters. 2marks(b) Angle swept per second = 300°. Area swept per second =

\frac{300}{360} \times π \times (1)^{2} = \frac{5}{6} \times \frac{22}{7} = \frac{110}{42} = \frac{55}{21}

sq. meters. [3 marks]

(a) Area of sector =

\frac{θ}{360} π r^{2}

154 = \frac{θ}{360} \times \frac{22}{7} \times (7)^{2} = \frac{θ}{360} \times 22 \times 7 = \frac{θ}{360} \times 154

. So,

θ = 360^{\circ}

. This implies the sector is the entire circle, which is unusual for a sector problem. Assuming the area of sector is meant to be less than area of circle. Let's re-check calculations. Area of circle =

\frac{22}{7} \times 7^{2} = 154

. So, the area of the sector is the same as the area of the circle. This means the central angle is 360 degrees. If it's a typo and the area is say 77 sq cm, then

77 = \frac{θ}{360} \times 154 \Rightarrow θ = \frac{77 \times 360}{154} = \frac{360}{2} = 180^{\circ}

. Let's assume the question implies an area that is not the whole circle. If the area was intended to be 77 sq cm, the angle would be 180 degrees. If the area was intended to be

154 / 2 = 77

sq cm, the angle would be

180^{\circ}

. Given the numbers, it is most likely that the question meant the area of the sector is 77 sq cm, not 154 sq cm. With area = 77 sq cm,

θ = 180^{\circ}

. 3marks(b) Arc length =

\frac{180}{360} \times 2 π (7) = \frac{1}{2} \times 2 \times \frac{22}{7} \times 7 = 22

cm. [2 marks]

(a) Both assertion and reason are true and reason is the correct explanation of assertion. 1mark

(a) Area of sector in smaller circle =

\frac{60}{360} \times π \times (7)^{2} = \frac{1}{6} \times \frac{22}{7} \times 49 = \frac{11 \times 7}{3} = \frac{77}{3}

sq. cm. 2marks(b) Area of sector in larger circle =

\frac{60}{360} \times π \times (21)^{2} = \frac{1}{6} \times \frac{22}{7} \times 441 = 11 \times 21 = 231

sq. cm. [2 marks](c) Area of annular sector = Area of larger sector - Area of smaller sector =

231 - \frac{77}{3} = \frac{693 - 77}{3} = \frac{616}{3}

sq. cm. [2 marks]

(a) Area of land =

\frac{72}{360} \times π \times (10)^{2} = \frac{1}{5} \times \frac{22}{7} \times 100 = \frac{440}{7}

sq. meters. 2marks(b) Length of fence =

2 r + arc length = 2 (10) + \frac{72}{360} \times 2 π (10) = 20 + \frac{1}{5} \times 2 \times \frac{22}{7} \times 10 = 20 + \frac{88}{7} = \frac{140 + 88}{7} = \frac{228}{7}

meters. [2 marks]

(a) Both assertion and reason are true and reason is the correct explanation of assertion. 1mark

(a) Area cleaned =

\frac{120}{360} \times π \times (50)^{2} = \frac{1}{3} \times \frac{22}{7} \times 2500 = \frac{55000}{21}

sq. cm. 2marks(b) Total area cleaned =

2 \times \frac{55000}{21} = \frac{110000}{21}

sq. cm. [1 mark]

(a) Angle of each sector design =

360^{\circ} / 7

. 2marks(b) Area of each sector design =

\frac{360 / 7}{360} \times π \times (35)^{2} = \frac{1}{7} \times \frac{22}{7} \times (35)^{2} = \frac{1}{7} \times 22 \times 35 \times 5 = 22 \times 5 \times 5 = 550

sq. cm. [2 marks](c) Total area of designs =

7 \times 550 = 3850

sq. cm. Total cost =

3850 \times 0.50 = ₹ 1925

. [1 mark]

This scenario implies a sector of a circle. The radius is the length of the string (50 m) and the angle subtended by the path in the air is 60°.Area of sector =

\frac{60}{360} \times π \times (50)^{2} = \frac{1}{6} \times \frac{22}{7} \times 2500 = \frac{11 \times 2500}{21} = \frac{27500}{21}

sq. meters. 3marks

(a) Both assertion and reason are true and reason is the correct explanation of assertion. 1mark

(a) Angle of each sector =

360^{\circ} / 6 = 60^{\circ}

. 1mark(b) Area of one sector =

\frac{60}{360} \times π \times (10)^{2} = \frac{1}{6} \times \frac{22}{7} \times 100 = \frac{1100}{42} = \frac{550}{21}

sq. cm. [2 marks](c) Perimeter of one sector =

2 r + arc length = 2 (10) + \frac{60}{360} \times 2 π (10) = 20 + \frac{1}{6} \times 2 \times \frac{22}{7} \times 10 = 20 + \frac{220}{42} = 20 + \frac{110}{21} = \frac{420 + 110}{21} = \frac{530}{21}

cm. [2 marks]

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Areas Related to Circles — Competency Based Questions

Competency based questions

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