CBSE Class 10 Mathematics Areas Related to Circles VBQs Set 04

Question. A toothed wheel of diameter 50 cm is attached to a smaller wheel of diameter 30 cm. How many revolutions will the smaller wheel make when the larger one makes 15 revolutions?
(a) 23
(b) 24
(c) 25
(d) 26
Answer: (c) 25

Question. The circumferences of two concentric circles forming a ring are 88 cm and 66 cm respectively. The width of the ring is
(a) 14 cm
(b) 7 cm
(c) \( \frac{7}{2} \) cm
(d) 21 cm
Answer: (c) \( \frac{7}{2} \) cm

Question. A wheel makes 1000 revolutions in covering a distance of 0.88 km. The radius of the wheel is
(a) 7 cm
(b) 14 cm
(c) 21 cm
(d) 28 cm
Answer: (b) 14 cm

Question. The radius of a circle is 50 cm. If the radius is decreased by 50%, its area will be decreased by
(a) 50%
(b) 75%
(c) 80%
(d) 25%
Answer: (b) 75%

Question. The area of a quadrant of a circle where the circumference of circle is 176 m, is
(a) 2464 m²
(b) 1232 m²
(c) 616 m²
(d) 308 m²
Answer: (c) 616 m²

Question. The diameter of a car wheel is 42 cm. The number of complete revolutions it will make in moving 132 km is
(a) 10⁴
(b) 10⁵
(c) 10⁶
(d) 10³
Answer: (b) 10⁵

Question. The number of rounds that a wheel of diameter \( \frac{7}{11} \) m will make in going 4 km is
(a) 800
(b) 8000
(c) 200
(d) 2000
Answer: (d) 2000

Question. If difference between the circumference and the radius of a circle is 37 cm, the circumference of the circle is [use \( \pi = \frac{22}{7} \)]
(a) 35 cm
(b) 39 cm
(c) 42 cm
(d) 44 cm
Answer: (d) 44 cm

Question. The circumference of two circles are in the ratio 4 : 9. The ratio of their areas is
(a) 9 : 4
(b) 4 : 9
(c) 16 : 81
(d) 21 : 81
Answer: (c) 16 : 81

Question. The circumference of the wheel of an engine of a train is \( 4 \frac{2}{7} \) m. If it makes seven revolutions in 4 seconds then the speed of the train is [use \( \pi = \frac{22}{7} \)]
(a) 5.5 m/s
(b) 7.5 m/s
(c) 8.5 m/s
(d) 15 m/s
Answer: (b) 7.5 m/s

Question. The perimeter of a semi-circular protractor is 108 cm. Then its diameter is
(a) 42 cm
(b) 41 cm
(c) 21 cm
(d) 14 cm
Answer: (a) 42 cm

Question. If the circumference of a circle and the perimeter of a square are equal, then
(a) Area of the circle = Area of the square
(b) Area of the circle > Area of the square
(c) Area of the circle < Area of the square
(d) nothing can be said about the relation between the areas of the circle and square
Answer: (b) Area of the circle > Area of the square

Question. Area of the largest triangle that can be inscribed in a semicircle of radius r is
(a) r²
(b) \( \frac{1}{2} r^2 \)
(c) 2r²
(d) \( \sqrt{2} r^2 \)
Answer: (a) r²

Question. The area of the sector of a circle of radius r and central angle \( \theta \) is
(a) \( \frac{1}{2} r \)
(b) \( \frac{2 \pi r^2 \theta}{720^\circ} \)
(c) \( \frac{\pi r^2 \theta}{360^\circ} \)
(d) \( \frac{\pi r \theta}{360^\circ} \)
Answer: (b) \( \frac{2 \pi r^2 \theta}{720^\circ} \)

Question. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameter 36 cm and 20 cm is
(a) 56 cm
(b) 42 cm
(c) 28 cm
(d) 16 cm
Answer: (c) 28 cm

Question. Area enclosed between two concentric circles is 770 cm². If the radius of the outer circle is 21 cm, then the radius of the inner circle is
(a) 12 cm
(b) 13 cm
(c) 14 cm
(d) 15 cm
Answer: (c) 14 cm

Question. An arc of a circle is of length 5\( \pi \) cm and the sector it bounds has an area of 20\( \pi \) cm². The radius of the circle is
(a) 5 cm
(b) 10 cm
(c) 7 cm
(d) 8 cm
Answer: (d) 8 cm

Question. The area of the circle that can be inscribed in a square of side 8 cm is
(a) 64\( \pi \) cm²
(b) 32\( \pi \) cm²
(c) 24\( \pi \) cm²
(d) 16\( \pi \) cm²
Answer: (d) 16\( \pi \) cm²

Question. A piece of wire 20 cm long is bent into the form of an arc of a circle subtended an angle of 60° at its centre. The radius of the circle is
(a) \( \frac{60}{\pi} \) cm
(b) \( \frac{30}{\pi} \) cm
(c) \( \frac{20}{\pi} \) cm
(d) \( \frac{50}{\pi} \) cm
Answer: (a) \( \frac{60}{\pi} \) cm

Very short Questions

Question. Show that if the circumferences of two circles are equal, then their areas are also equal.
Answer: Let radii of two circles be ‘r’ and ‘R’.
We are given that
\( 2\pi r = 2\pi R \)
\( \implies \) \( r = R \)
and hence \( \pi r^2 = \pi R^2 \)
\( \implies \) The areas are equal.

Question. The diameter of a cycle wheel is 21 cm. How many revolutions will it make in moving 66 m?
Answer: Since Circumference of the wheel = \( 2\pi R \)
\( = 2 \times \frac{22}{7} \times 10.5 = 66 \text{ cm} \)
Hence, to cover a distance of 66 m, the wheel will make \( = \frac{66 \times 100}{66} = 100 \text{ revolutions} \).

Question. Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why? 
Answer: No.
It is true only in case of minor segment. In case of major segment, area of segment is always greater than the area of its corresponding sector.

Question. Is it true to say that the area of a square inscribed in a circle of diameter p cm is p\(^2\) cm\(^2\)? Why? 
Answer: No.
When the square is inscribed in the circle, the diameter of the circle is equal to the diagonal of the square but not the side of the square.
Let side of square be a.
∴ Length of diagonal = \( \sqrt{a^2 + a^2} = \sqrt{2}a \)
\( \implies \) diameter of circle, \( p = \sqrt{2}a \)
\( \implies \) \( p^2 = 2a^2 \)
∴ Area of square \( a^2 = \frac{p^2}{2} \)

Question. Find the area of a sector of a circle of diameter 56 cm and central angle 45º. 
Answer: Given: diameter of the circle (d) = 56 cm
Then, the radius of circle (r) = 28 cm
Central angle, \( \theta = 45^{\circ} \)
The area of the sector = \( \frac{\theta}{360^{\circ}} \times \pi r^2 \)
\( = \frac{45}{360} \times \frac{22}{7} \times 28 \times 28 \)
\( = 308 \text{ cm}^2 \)
Hence, the area of the sector is 308 cm\(^2\).

Question. Find the area of a sector of a circle of radius 28 cm and central angle 45°.
Answer: Given: Radius of circle, r = 28 cm and central angle, \( \theta = 45^{\circ} \)
We know that,
If \( \theta \) is measured in degrees then
∴ Area of sector = \( \frac{\theta}{360^{\circ}} \times \pi r^2 \)
\( = \frac{45}{360} \times \frac{22}{7} \times 28 \times 28 \)
\( = 308 \text{ cm}^2 \)
Hence, the required area of the sector of the circle is 308 cm\(^2\).

Short Answer(SA-I) Type Questions

Question. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the area of the sector. 
Answer: Let \( \theta \) be the central angle of the sector. Then,
length of arc of circle = \( \frac{\theta}{360^{\circ}} \times 2\pi r \)
Perimeter = \( \frac{\theta}{360^{\circ}} \times 2\pi r + 2r \)
\( 16.4 = \frac{\theta}{360^{\circ}} \times 2\pi (5.2) + 2 \times 5.2 \)
[\( \because r = 5.2 \text{ cm} \)]
\( \implies \) \( 5.2 \pi \theta = 1080 \text{ (on simplification)} \)
Now, area of the sector = \( \frac{\theta}{360^{\circ}} \times \pi (5.2)^2 \)
\( = \frac{1080 \times 5.2}{360} \text{ sq cm.} \)
\( = 15.6 \text{ sq. cm.} \)

Question. The diameter of two circles with centre A and B are 16 cm and 30 cm respectively. If area of another circle with centre C is equal to the sum of areas of these two circles, then find the circumference of the circle with centre C.
Answer: Area of circle = \( \pi r^2 \)
Let the radius of circle with centre C = R
According to the question we have,
\( \pi(8)^2 + \pi(15)^2 = \pi R^2 \)
\( \implies \) \( 64\pi + 225\pi = \pi R^2 \)
\( \implies \) \( 289\pi = \pi R^2 \)
\( \implies \) \( R^2 = 289 \)
\( \implies \) \( R = 17 \text{ cm} \)
Circumference of circle:
\( 2\pi r = 2\pi \times 17 \)
\( = 34\pi \text{ cm} \)
So, the circumference of the circle with centre C is \( 34\pi \text{ cm} \).

Question. The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? Why? 
Answer: No. The given statement will be true for arcs of the same circle. But in different circles, it is not possible.
Area of 1st sector = \( \frac{1}{2} (r_1^2) \theta_1 \)
where, \( r_1 \) is the radius and \( \theta_1 \) is the angle.
Area of 2nd sector = \( \frac{1}{2} (r_2^2) \theta_2 \)
where, \( r_2 \) is the radius and \( \theta_2 \) is the angle subtended at the centre of the circle by the arc.
It is given that \( \frac{1}{2} r_1^2 \theta_1 = \frac{1}{2} r_2^2 \theta_2 \)
\( \implies \) \( r_1^2 \theta_1 = r_2^2 \theta_2 \)
Thus, it depends both on radius and angle subtended at the centre. But arc lengths depend only on radius.

Question. A circular road runs round a circular ground. If the difference between the circumference of the outer circle and the inner circle is 66 metres, then find the width of the road.
Answer: Let the radius of the outer circle be \( R \) and the inner circle be \( r \).
\( \implies \) Given, \( 2\pi R - 2\pi r = 66 \)
\( \implies \) \( 2\pi(R - r) = 66 \)
\( \implies \) \( 2 \times \frac{22}{7} \times (R - r) = 66 \)
\( \implies \) \( R - r = \frac{66 \times 7}{44} \)
\( \implies \) \( R - r = 1.5 \times 7 = 10.5 \) Therefore, the width of the road is 10.5 m.

Question. If the circumference of a circle exceeds the diameter by \( \pi \) units, then find the diameter of the circle.
Answer: Let the diameter of the circle be \( d \).
\( \implies \) Circumference \( (C) = \pi d \)
\( \implies \) According to the question, \( C - d = \pi \)
\( \implies \) \( \pi d - d = \pi \)
\( \implies \) \( d(\pi - 1) = \pi \)
\( \implies \) \( d = \frac{\pi}{\pi - 1} \) units. Therefore, the diameter of the circle is \( \frac{\pi}{\pi - 1} \) units.

Question. Find the number of revolutions taken by a wheel of diameter 84 cm to cover 792 m. \( [ \pi = \frac{22}{7} ] \)
Answer: Diameter of the wheel \( (d) = 84 \text{ cm} \).
\( \implies \) Radius \( (r) = \frac{84}{2} = 42 \text{ cm} \).
\( \implies \) Circumference of the wheel = \( 2\pi r = 2 \times \frac{22}{7} \times 42 = 264 \text{ cm} \).
\( \implies \) Total distance to be covered = 792 m = 79200 cm.
\( \implies \) Number of revolutions = \( \frac{\text{Total Distance}}{\text{Circumference}} = \frac{79200}{264} = 300 \). Therefore, the wheel takes 300 revolutions.

Question. The circumference of a circular plot is 220 m. A 15 m wide concrete track runs round outside the plot. Find the area of the track. [Use \( \pi = 22/7 \)]
Answer: Circumference of the plot = 220 m.
\( \implies \) \( 2 \times \frac{22}{7} \times r = 220 \)
\( \implies \) \( r = \frac{220 \times 7}{44} = 35 \text{ m} \).
\( \implies \) Radius of the outer circle \( (R) = r + 15 = 35 + 15 = 50 \text{ m} \).
\( \implies \) Area of the track = \( \pi R^2 - \pi r^2 = \pi(R^2 - r^2) \)
\( \implies \) \( = \frac{22}{7} (50^2 - 35^2) = \frac{22}{7} (2500 - 1225) \)
\( \implies \) \( = \frac{22}{7} \times 1275 = \frac{28050}{7} \approx 4007.14 \text{ m}^2 \).

Question. A human chain in the form of a circle of radius 56 m is formed at India Gate to protest against the rising prices of petrol. If each person is given two metres of space to stand find how many persons can be accommodated in the chain.
Answer: Radius of the circle \( (r) = 56 \text{ m} \).
\( \implies \) Circumference of the circle = \( 2\pi r = 2 \times \frac{22}{7} \times 56 = 352 \text{ m} \).
\( \implies \) Space per person = 2 m.
\( \implies \) Number of persons = \( \frac{352}{2} = 176 \).

Question. You are required to create a model of a circular wall clock and paste the numbers from 1 to 12 on its dial. What is the angle made at the centre between 3 and 7? Find the area of this region, if the length of the minute hand of the clock is 21 cm.
Answer: The clock dial is divided into 12 equal parts. Angle for 1 part = \( \frac{360^\circ}{12} = 30^\circ \).
\( \implies \) Difference between 3 and 7 is 4 units.
\( \implies \) Angle at the centre between 3 and 7 \( (\theta) = 4 \times 30^\circ = 120^\circ \).
\( \implies \) Length of the minute hand \( (r) = 21 \text{ cm} \).
\( \implies \) Area of the region = \( \frac{\theta}{360} \times \pi r^2 = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \)
\( \implies \) \( = \frac{1}{3} \times 22 \times 3 \times 21 = 22 \times 21 = 462 \text{ cm}^2 \).

Question. AB is one of the direct common tangent of two circles of radii 12 cm and 4 cm respectively touching each other. Find the area of the region enclosed by the circles and the tangent. [Hots]
Answer: Let radii be \( R = 12 \text{ cm} \) and \( r = 4 \text{ cm} \). Since circles touch each other, distance between centers \( d = R + r = 16 \text{ cm} \).
\( \implies \) Length of direct common tangent \( AB = \sqrt{d^2 - (R - r)^2} = \sqrt{16^2 - (12 - 4)^2} = \sqrt{256 - 64} = \sqrt{192} = 8\sqrt{3} \text{ cm} \).
\( \implies \) Let the centers be \( O_1 \) and \( O_2 \). \( O_1ABO_2 \) is a trapezium.
\( \implies \) Area of trapezium = \( \frac{1}{2} (R + r) \times AB = \frac{1}{2} (12 + 4) \times 8\sqrt{3} = 8 \times 8\sqrt{3} = 64\sqrt{3} \text{ cm}^2 \).
\( \implies \) To find area of region: Area of trapezium - (Area of sector in large circle + Area of sector in small circle).
\( \implies \) Let angle \( \angle AO_1O_2 = \theta \). Then \( \cos \theta = \frac{R-r}{d} = \frac{12-4}{16} = \frac{8}{16} = \frac{1}{2} \implies \theta = 60^\circ \).
\( \implies \) Angle for smaller circle sector = \( 180^\circ - 60^\circ = 120^\circ \).
\( \implies \) Area of region = \( 64\sqrt{3} - [\frac{60}{360} \pi (12)^2 + \frac{120}{360} \pi (4)^2] = 64\sqrt{3} - [24\pi + \frac{16}{3}\pi] = 64\sqrt{3} - \frac{88}{3}\pi \text{ cm}^2 \).

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct answer out of the following choices.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A): Area of the square inscribed in a circle of radius x is \( 2x^2 \) sq. units.
Reason (R): Area of the major segment of a circle = Area of the circle – Area of minor segment.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (b) Both A and R are true but R is not the correct explanation of A.

Question. Assertion (A): The diameter of a wheel is 4.2 m. It makes 75 revolutions in one minute. The speed of the wheel is 59.4 km/hr.
Reason (R): Distance travelled in one minute = circumference \( \times \) Number of revolutions in one minute.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation of A.