Question. The perimeter of a circle having radius 5cm is equal to:
(a) 30 cm
(b) 3.14 cm
(c) 31.4 cm
(d) 40 cm
Answer: (c) 31.4 cm
Explanation: The perimeter of the circle is equal to the circumference of the circle.
Circumference = 2πr
= 2 x 3.14 x 5
= 31.4 cm
Question. Area of the circle with radius 5cm is equal to:
(a) 60 sq.cm
(b) 75.5 sq.cm
(c) 78.5 sq.cm
(d) 10.5 sq.cm
Answer: (c) 78.5 sq.cm
Explanation: Radius = 5cm
Area = πr2 = 3.14 x 5 x 5 = 78.5 sq.cm
(a) r2
(b) 1/2r2
(c) 2r2
(d) √2r2
Answer: (a) r2
Explanation: The height of the largest triangle inscribed will be equal to the radius of the semi-circle and base will be equal to the diameter of the semi-circle.
Area of triangle = ½ x base x height
= ½ x 2r x r
= r2
Question. If the perimeter of the circle and square are equal, then the ratio of their areas will be equal to:
(a) 14:11
(b) 22:7
(c) 7:22
(c) 11:14
Answer: (a) 14:11
Explanation: Given,
The perimeter of circle = perimeter of the square
2πr = 4a
a=πr/2
Area of square = a2 = (πr/2)2
Acircle/Asquare = πr2/(πr/2)2
= 14/11
Question. The area of the circle that can be inscribed in a square of side 8 cm is
(a) 36 π cm2
(b) 16 π cm2
(c) 12 π cm2
(d) 9 π cm2
Answer: (b) 16 π cm2
Explanation: Given,
Side of square = 8 cm
Diameter of a circle = side of square = 8 cm
Therefore, Radius of circle = 4 cm
Area of circle
= π(4)2
= π (4)2
= 16π cm2
Question. The area of the square that can be inscribed in a circle of radius 8 cm is
(a) 256 cm2
(b) 128 cm2
(c) 642 cm2
(d) 64 cm2
Answer: (b) 128 cm2
Explanation: Radius of circle = 8 cm
Diameter of circle = 16 cm = diagonal of the square
Let “a” be the triangle side, and the hypotenuse is 16 cm
Using Pythagoras theorem, we can write
162= a2+a2
256 = 2a2
a2= 256/2
a2= 128 = area of a square.
Question. The area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.
(a) 142/7
(b) 152/7
(c) 132/7
(d) 122/7
Answer: (c) 132/7
Explanation: Angle of the sector is 60°
Area of sector = (θ/360°) × π r2
∴ Area of the sector with angle 60° = (60°/360°) × π r2 cm2
= (36/6) π cm2
= 6 × (22/7) cm2
= 132/7 cm2
Question. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the arc is;
(a) 20cm
(b) 21cm
(c) 22cm
(d) 25cm
Answer: (c) 22cm
Explanation: Length of an arc = (θ/360°) × (2πr)
∴ Length of an arc AB = (60°/360°) × 2 × 22/7 × 21
= (1/6) × 2 × (22/7) × 21
Or Arc AB Length = 22cm
Question. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The area of the sector formed by the arc is:
(a) 200 cm2
(b) 220 cm2
(c) 231 cm2
(d) 250 cm2
Answer: (c) 231 cm2
Explanation: The angle subtended by the arc = 60°
So, area of the sector = (60°/360°) × π r2 cm2
= (441/6) × (22/7) cm2
= 231 cm2
Question. Area of a sector of angle p (in degrees) of a circle with radius R is
(a) p/180 × 2πR
(b) p/180 × π R2
(c) p/360 × 2πR
(d) p/720 × 2πR2
Answer: (d) p/720 × 2πR2
Explanation: The area of a sector = (θ/360°) × π r2
Given, θ = p
So, area of sector = p/360 × π R2
Multiplying and dividing by 2 simultaneously,
= [(p/360)/(π R2)]×[2/2]
= (p/720) × 2πR2
Question. If the area of a circle is 154 cm2, then its perimeter is
(a) 11 cm
(b) 22 cm
(c) 44 cm
(d) 55 cm
Answer: (c) 44 cm
Explanation:
Given,
Area of a circle = 154 cm2
πr2 = 154
(22/7) × r2 = 154
r2 = (154 × 7)/22
r2 = 7 × 7
r = 7 cm
Perimeter of circle = 2πr = 2 × (22/7) × 7 = 44 cm
Question. If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then
(a) R1 + R2 = R
(b) R12 + R22 = R2
(c) R1 + R2 < R
(d) R12 + R22 < R2
Answer: (b) R12 + R22 = R2
Explanation:
According to the given,
πR12 + πR22 = πR2
π(R12 + R22) = πR2
R12 + R22 = R2
Question. If θ is the angle (in degrees) of a sector of a circle of radius r, then the length of arc is
(a) (πr2θ)/360
(b) (πr2θ)/180
(c) (2πrθ)/360
(d) (2πrθ)/180
Answer: (a) (2πrθ)/360
If θ is the angle (in degrees) of a sector of a circle of radius r, then the area of the sector is (2πrθ)/360.
Question. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be
(a) 10 m
(b) 15 m
(c) 20 m
(d) 24 m
Answer: (a) 10 m
Explanation:
Radii of two circular parks will be:
R1 = 16/2 = 8 m
R2 = 12/2 = 6 m
Let R be the radius of the new circular park.
If the areas of two circles with radii R1 and R2 is equal to the area of circle with radius R, then
R2 = R12 + R22
= (8)2 + (6)2
= 64 + 36
= 100
R = 10 m
Question. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is
(a) 56 cm
(b) 42 cm
(c) 28 cm
(d) 16 cm
Answer: (c) 28 cm
Explanation:
If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then R1 + R2 = R.
Here,
R1 = 36/2 = 18 cm
R2 = 20/2 = 10 cm
R = R1 + R2 = 18 + 10 = 28 cm
Therefore, the radius of the required circle is 28 cm
Question. Find the area of a sector of circle of radius 21 cm and central angle 120°.
(a) 441 cm2
(b) 462 cm2
(c) 386 cm2
(d) 512 cm2
Answer: (b) 462 cm2
Explanation:
Given, radius (r) = 21 cm
Central angle = θ = 120
Area of sector = (πr2θ)/360
= (22/7) × (21 × 21) × (120/360)
= 22 × 21
= 462 cm2
Question. The wheel of a motorcycle is of radius 35 cm. The number of revolutions per minute must the wheel make so as to keep a speed of 66 km/hr will be
(a) 50
(b) 100
(c) 500
(d) 1000
Answer: (c) 500
Explanation:
Circumference of the wheel = 2πr = 2 × (22/7) × 35 = 220 cm
Speed of the wheel = 66 km/hr
= (66 × 1000)/60 m/min
= 1100 × 100 cm/min
= 110000 cm/min
Number of revolutions in 1 min = 110000/220 = 500
Question. If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(a) 2 units
(b) π units
(c) 4 units
(d) 7 units
Answer: (a) 2 units
Explanation:
According to the given,
Perimeter of circle = Area of circle
2πr = πr2
⇒ r = 2
Therefore, radius = 2 units
Question. The area of a quadrant of a circle with circumference of 22 cm is
(a) 77 cm2
(b) 77/8 cm2
(b) 35.5 cm2
(c) 77/2 cm2
Answer: (b) 77/8 cm2
Explanation:
Given, circumference = 22 cm
2πr = 22
2 × (22/7) × r = 22
r = 7/2 cm
Area of quadrant of a circle = (1/4)πr2
= (1/4) × (22/7) × (7/2) × (7/2)
= 77/8 cm2
Question. . In a circle of radius 14 cm, an arc subtends an angle of 30° at the centre, the length of the arc is
(a) 44 cm
(b) 28 cm
(c) 11 cm
(d) 22/3 cm
Answer: (d) 22/3 cm
Explanation:
Given, radius = r = 14 cm
Length of arc = (2πrθ)/360
= 2 × (22/7) × 14 × (30/360)
= 22/3 cm
Question. Perimeter of a sector of a circle whose central angle is 90° and radius 7 cm is
(a) 35 cm
(b) 25 cm
(c) 77 cm
(d) 7 cm
Answer: (b) 25 cm
Question. The area of a circle that can be inscribed in a square of side 10 cm is
(a) 40π cm²
(b) 30π cm²
(c) 100π cm²
(d) 25π cm²
Answer: (d) 25π cm²
Question. The perimeter of a square circumscribing a circle of radius a units is
(a) 2 units
(b) 4α units
(c) 8α units
(d) 16α units
Answer: (c) 8α units
Question. The perimeter of the sector with radius 10.5 cm and sector angle 60° is
(a) 32 cm
(b) 23 cm
(c) 41 cm
(d) 11 cm
Answer: (a) 32 cm
Question. In a circle of diameter 42 cm, if an arc subtends an angle of 60° at the centre, where π = \(\frac{22}{7}\) then length of arc is
(a) 11 cm
(b) 22/7cm
(c) 22 cm
(d) 44 cm
Answer: (c) 22 cm
Question. The perimeter of a sector of radius 5.2 cm is 16.4 cm, the area of the sector is
(a) 31.2 cm²
(b) 15 cm²
(c) 15.6 cm²
(d) 16.6 cm²
Answer: (c) 15.6 cm²
Question. If the perimeter of a semicircular protractor is 72 cm where π = \(\frac{22}{7}\), then the diameter of protractor is
(a) 14 cm
(b) 33 cm
(c) 28 cm
(d) 42 cm
Answer: (c) 28 cm
Question. If the radius of a circle is doubled, its area becomes
(a) 2 times
(b) 4 times
(c) 8 times
(d) 16 times
Answer: (b) 4 times
Question. If the sum of the circumferences of two circles with radii R1 and R2 is equal to circumference of a circle of radius R, then
(a) R1 + R2 = R
(b) R1 + R2 > R
(c) R1 + R2 < R
(d) Can’t say;
Answer: (a) R1 + R2 = R
Question. The perimeter of a circular and square fields are equal. If the area of the square field is 484 m² then the diameter of the circular field is
(a) 14 m
(b) 21 m
(c) 28 m
(d) 7 m
Answer: (c) 28 m
Question. The radius of sphere is r cm. It is divided into . two equal parts. The whole surface area of two parts will be
(a) 8πr² cm²
(b) 6πr² cm²
(c) 4πr² cm²
(d) 3πr² cm²
Answer: (b) 6πr² cm²
Question. if the diameter of a semicircular protractor is 14 cm, then its perimeter is .
(a) 27 cm
(b) 36 cm
(c) 18 cm
(d) 9 cm
Answer: (b) 36 cm
Question. A race track is in the form of a circular ring whose outer and inner circumferences are 396 m and 352 m respectively. The width of the track is
(a) 63 m
(b) 56 m
(c) 7 m
(d) 3.5 m
Answer: (c) 7 m
Question. The area of the largest square that can be inscribed in a circle of radius 12 cm is
(a) 24 cm²
(b) 249 cm²
(c) 288 cm²
(d) 196√2 cm²
Answer: (c) 288 cm²
Question. The area of the largest triangle that can be inscribed in a semicircle of radius r is
(a) r²
(b) 2r²
(c) r³
(d) 2r³
Answer: (a) r²
Question.The area (in cm²) of the circle that can be inscribed in a square of side 8 cm is
(a) 64 π
(b) 16 π
(c) 8 π
(d) 32 π
Answer: (b) 16 π
Question. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 22 : 7
(b) 14 : 11
(c) 7 : 22
(d) 11 : 14
Answer: (b) 14 : 11
Question. The circumference of two concentric circles forming a ring are 88 cm and 66 cm. Taking π = \(\frac{22}{7}\), the width of the ring is
(a) 14 cm
(b) 7 cm
(c) 7/2cm
(d) 21 cm
Answer: (c) 7/2cm
Question. A steel wire when bent in the form of a square encloses an area of 121 cm². If the same wire is bent in the form of a circle, then the circumference of the circle is
(a) 88 cm
(b) 44 cm
(c) 22 cm
(d) 11 cm
Answer: (b) 44 cm
Question. The diameter of a circle whose area is equal to sum of the areas of the two circles of radii 40 cm and 9 cm is
(a) 41 cm
(b) 49 cm
(c) 82 cm
(d) 62 cm
Answer: (c) 82 cm
Question. If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then
(a) R1 + R2 = R
(b) R12 + R22 = R2
(c) R1 + R2 < R
(d) R12 + R22 < R2
Answer: (b) R12 + R22 = R2
Explanation: According to given condition,
Area of circle = Area of first circle + Area of second circle
πR2 = πR12 + πR22
R2 = R12 + R22
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 definite can be said about the relation between the areas of the circle and square.
Answer: (b) Area of the circle > Area of the square
Explanation: According to given condition
Circumference of a circle = Perimeter of square
Hence Area of the circle > Area of the square
Question. Area of the largest triangle that can be inscribed in a semi-circle of radius r units, in square units is:
(a) r2
(b) 1/2r2
(c) 2 r2
(d) √2r2
Answer: (a) r2
Explanation: The triangle inscribed in a semi-circle will be the largest when the perpendicular height of the triangle is the same size as the radius of the semi-circle.
Consider the following figure:
Question. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is:
(A) 22:7
(B) 14:11
(C) 7:22
(D) 11:14
Answer: (b) 14:11
Explanation: Perimeter of circle = Perimeter of square
Question. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be
(a) 10 m
(b) 15 m
(c) 20 m
(d) 24 m
Answer: (a) 10 m
Explanation: Area of first circular park, whose diameter is 16m
= πr2 = π (16/2)2 = 64π m2
Area of second circular park, whose diameter is 12m
= πr2 = π (12/2)2 = 36π m2
According to question,
Area of new circular park =
πR2 = (64π + 36π) m2
πR2 = 100π m2
R = 10m
Question. The area of the circle that can be inscribed in a square of side 6 cm is
(a) 36 π cm2
(b) 18 π cm2
(c) 12 π cm2
(d) 9 π cm2
Answer: (d) 9 π cm2
Explanation: Given,
Side of square = 6 cm
Diameter of a circle = side of square = 6cm
Therefore, Radius of circle = 3cm
Area of circle
= πr2
= π (3)2
= 9π cm2
Question. The area of the square that can be inscribed in a circle of radius 8 cm is
(a) 256 cm2
(b) 128 cm2
(c) 642 cm2
(d) 64 cm2
Answer: (b) 128 cm2
Explanation: Radius of circle = 8 cm
Diameter of circle = 16 cm = diagonal of the square
Therefore side of square = diagonal/√2
= 16/√2
Therefore, are of square is = (side)2 = (16/√2)2
= 256/2
= 128 cm2
Question. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is
(a) 56 cm
(b) 42 cm
(c) 28 cm
(d) 16 cm
Answer: (c) 28 cm
Explanation: According to question,
Circumference of circle = Circumference of first circle + Circumference of second circle
πD = πd1 + πd2
D = 36 + 20
D = 56cm
So, Radius = 56/2 = 28cm
Question. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm respectively, is
(a) 31 cm
(b) 25 cm
(c) 62 cm
(d) 50 cm
Answer: (d) 50 cm
Explanation: According to question
Therefore diameter = 2 × 25 = 50cm
Question. If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then
(a) the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.
(b) the angle of the corresponding sector of the first circle is equal the angle of the corresponding sector of the other circle.
(c) the angle of the corresponding sector of the first circle is half the angle of the corresponding sector of the other circle.
(d) the angle of the corresponding sector of the first circle is 4 times the angle of the corresponding sector of the other circle.
Answer: (a) the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.
Explanation: According to Question,
Question. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/h?
(A) 300
(B) 400
(C) 450
(D) 500
Answer: (D)
Explanation:
Question. A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m, then the area of the field in which the cow can graze is:
(a) 154 m2
(b) 156 m2
(c) 158 m2
(d) 160 m2
Answer: (a) 154 m2
Explanation: Figure according to question is:
Area of the field in which cow can graze= Area of a sector AFEG
= (θ/360) X πr2
= (90/360) X π (14)2
= (1/4) X (22/7) X 196
= 154 m2
Question. The area of the shaded region in Fig., where arcs drawn with centres P, Q, Rand S intersect in pairs at mid-points A, B, C and D of the sides PQ, QR, RS and SP, respectively of a square PQRS, is:
(a) 25.25 cm2
(b) 27.45 cm2
(c) 29.65 cm2
(d) 30.96 cm2
Answer: (d) 30.96 cm2
Explanation:
Question. Area of a sector of central angle 120° of a circle is 3π cm2. Then the length of the corresponding arc of this sector is:
(a) 5.8cm
(b) 6.1cm
(c) 6.3cm
(d) 6.8cm
Answer: (c) 6.3cm
Explanation:
Given that
Area of a sector of central angle 120° of a circle is 3π cm2
Question. A round table cover has six equal designs as shown in the figure. If the radius of thecover is 28 cm, then the cost of making the design at the rate of Rs. 0.35 per cm2 is:
(a) Rs.146.50
(b) Rs.148.75
(c) Rs.152.25
(d) Rs.154.75
Answer: (B) Rs.148.75
Explanation: The area of the hexagon will be equal to six equilateral triangles with each side equal to the radius of circle.
Area of given hexagon = Area of 6 equilateral triangles.
= 6 X (√3/4) X (side)2
= 6 X (√3/4) X (28)2
= 1999.2 cm2 (Taking √3 = 1.7)
Area of circle = πr2
= π × 282
= 2464 cm2
So, area of designed portion = 2464 – 1999.2 = 464.8 cm2
Cost of making design = 464.8 × 0.35
= Rs. 162.68
Question. ABC is an equilateral triangle. The area of the shaded region if the radius of each of the circle is 1 cm, is
Answer: c
Explanation:
Question. ABCDEF is any hexagon with different vertices A, B, C, D, E and F as the centres of circles with same radius r are drawn. The area of the shaded portion is
(a) πr²
(b) 2πr²
(c) 3πr²
(d) 4πr²
Answer: b 2πr²
Explanation:
Question. In the figure, PQRS is a square and O is centre of the circle. If RS = 10 √2, then area of shaded region is
(a) 90 π – 90
(b) 80 π – 80
(c) 50 π-100
(d) 100 π – 100
Answer: c 50 π-100
Explanation:
(c) Diagonal of square = √2 × (10√2) = 20 units
∴ Diameter of circle = 20 units
Area of circle = π × (10)² = 100π sq.units
Area of square = (10√2)² = 200 sq. units
Area of circle not included in the square = (100π – 200) sq.units
∴ Area of shaded portion = 1/2 (100π – 200)
= 50π – 100.
Question. The diameter of the wheel of a bus is 1.4 m. The wheel makes 10 revolutions in 5 seconds. The speed of the vehicle (in kmph) is ______ .
Answer:
Explanation:
Question. The area of a quadrant of a circle whose circumference is 44 cm is ______ .
Answer:
Explanation:
Question.. If the wheel of an engine of a train is m in circumference makes seven revolutions in 4 seconds, then the speed of the train is _____ km/h.
Answer:
Explaination:
Question. The area of the largest possible square inscribed in a circle of unit radius (in sq. units) is ______.
Explanation:
2 units
Hints: Diameter of circle = 2 units
∴ Diagonal of the square = 2 units
Side of the square √2 units
∴ Area of the square = (√2)² = 2 sq. units36.
Explaination:
Length of AEDˆ = πr
= π x 7 cm = 7π cm
AB + BC + DC = 20 + 14 + 20 = 54 cm
∴ Perimeter of figure = (7π + 54) cm38. A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel. (use π = 22/7)Answer:
Explaination:
Explanation:
Explaination:
Explanation:
Perimeter = AD + BC + length of DPC + length of APB
= 14 + 14 + πr + πr
= 28 + 2 × 22/7 × 14/2= 72 cm.
(a) 77
(b) 154
(c) 44
(d) 22
Explanation:
(60° + 80° + 40°) = 180∘/360∘×22/7×7×7
= 77 cm²26. The ratio of the areas of the incircle and circumcircle of a square is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 1 : √2
(a) Let side of square = x units
∴ Diagonal of the square = √2 x units
Diameter of the incircle = x units
Diameter of the circumcircle = √2 x units
(b) 4 : 9
(c) 2 : 3
(d) 16 : 81Answer: d 16 : 81
Explanation:
