Q.1. In the given figure, a square OABC has been inscribed in the quadrant OPBQ. If OA = 20 cm then the area of the shaded region is [take π = 3.14]
(a) 214 cm^{2}
(b) 228 cm^{2}
(c) 242 cm^{2}
(d) 248 cm^{2}
Length of side of square = OA = 20 cm
Radius of Quadrant = OB =
Area of Quadrant = πR^{2} × θ/360 = 3.14 × 20√2 × 20√2 × 90/360 = 628 cm^{2}
Area of Square = a^{2} = 20^{2} cm^{2} = 400 cm^{2}
Area of Shaded region = Area of Quadrant  Area of Square
= 628 cm^{2} – 400 cm^{2}
= 228 cm^{2}
Q.2. The diameter of a wheel is 84 cm. How many revolutions will it make to cover 792 m?
(a) 200
(b) 250
(c) 300
(d) 350
Diameter of wheel = 84 cm
Radius of wheel = r = 84/2 cm = 42 cm
Distance the wheel travels = 792 m = 79200 cm
In 1 revolution wheel travels 2πr distance
No. of revolutions a wheel makes =
= 300 revolutions
Q.3. The area of a sector of a circle with radius r, making an angle of x° at the centre is x
(a)
(b)
(c)
(d)
Area of a sector of angle θº of a circle with radius R = area of circle × θ/360
Q.4. In the given figure, ABCD is a rectangle inscribed in a circle having length 8 cm and breadth 6 cm. If π = 3.14 then the area of the shaded region is
(a) 264 cm^{2}
(b) 266 cm^{2}
(c) 272 cm^{2}
(d) 254 cm^{2}
Given:
Length of rectangle = 8 cm
Breadth of rectangle = 6 cm
Area of rectangle = length × breadth
= 8 × 6 = 48 cm^{2}
Consider ΔABC,
By Pythagoras theorem,
AC^{2} = AB^{2} + BC^{2}
= 8^{2} + 6^{2} = 64 + 36 = 100
AC = √100 = 10 cm
⇒ Diameter of circle = 10 cm
Thus, radius of circle = 10/2 = 5cm
Let the radius of circle be r = 5 cm
Then, Area of circle = πr^{2}
Area of shaded region = Area of circle – Area of rectangle
= 78.57  48
= 30.57 cm^{2}
Hence, the area of shaded region is 30.57 cm^{2}.
Q.5. The circumference of a circle is 22 cm. Find its area. [Take π = 22/7]
Let the radius if circle be r
Circumference of circle = 22 cm
2πr = 22 cm
2 × 22/7 × r = 22 cm
r = 22 × 1/2 ×7/22 cm
r = 3.5 cm
Area of Circle = πr^{2}
= 22/7 × 3.5 × 3.5 cm^{2}
= 38.5 cm^{2}
∴ Area of Circle = 38.5 cm^{2}
Q.6. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.
Radius of circle = R = 21 cm
Angle subtended by arc = 60°
Length of an arc of a sector of angle θº of a circle with radius R
= Circumference of circle × θ/360
Length of arc = 2 × 22/7 × 21 × θ/360 cm = 22 cm
Length of arc = 22 cm
Q.7. The minute hand of a clock is 12 cm long. Find the area swept by it in 35 minutes.
Length of the minute hand of a clock = 12 cm
∴ Radius = R = 12 cm
In 1 minute, minute hand sweeps 6°
So, in 35 minutes, minute hand will sweep 35 × 6° = 210°
Area swept by minute hand in 35 minutes = Area of a sector of angle θº of a circle with radius R = = 22/7 × 12 × 12 × 60°/360° = 264 cm^{2}
Area swept by minute hand in 35 minutes = 264 cm^{2}
Q.8. The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector.
Radius of circle = 5.6 cm
Perimeter of a sector of a circle = 2R + Circumference of circle × θ/360
Perimeter of a sector of a circle = 2 × 5.6 + 2 × 22/7 × 5.6 × θ/360 cm
= 27.2 cm
⇒ 2 × 22/7 × 5.6 × θ/360 = 27.2 – 11.2 cm = 16 cm
⇒ θ = 16 × 1/2 × 1/5.6 × 7/22
⇒ θ = 163.63°
Area of Sector = πr^{2} × θ\360 = 22/7 × 5.6 × 5.6 × 163.63/360 = 44.8cm^{2}
∴ Area of Sector = 44.8 cm^{2}
Q.9. A chord of a circle of radius 14 cm makes a right angle at the centre. Find the area of the sector.
Chord AB subtends an angle of 90° at the centre of the circle
Radius of Circle = R = 14 cm
Area of sector of circle of radius R =
= 22/7 × 14 × 14 × 90/360 cm^{2} = 154 cm^{2}
Q.10. In the give figure, the sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region.
Given,
Radius of smaller circle = R_{1} = 3.5 cm
Radius of bigger circle = R_{2} = 7 cm
Angle subtended = 30°
Area of Shaded region =
= 22/7 × (7^{2 }– 3.5^{2}) × 30/360 cm^{2}
= 22/7 × (49 – 12.22) × θ/360 cm^{2}
= 9.625 cm^{2}
∴ Area of shaded region = 9.625 cm^{2}
Q.11. A wire when bent in the form of an equilateral triangle encloses an area of 121√3 cm^{2}. If the same wire is bent into the form of a circle, what will be the area of the circle? [Take π = 22/7]
Let the sides of equilateral triangle be a cm
Area of equilateral triangle = 121√3 cm^{2}
Area of equilateral triangle = √3/4 × a^{2}
⇒ √3/4 a^{2 }= 121√3
⇒ a^{2 }= 121√3 × 4/√3 = 121 × 4 cm^{2}
⇒ a^{2 }= 484 cm^{2}
⇒ a = 22 cm
Perimeter of equilateral triangle = 3a
= 3 × 22 cm = 66 cm
Perimeter of equilateral triangle = Circumference of circle
Circumference of circle = 66 cm
Let the radius of circle be r
Circumference of circle = 2πr
⇒ 2πr = 66 cm
⇒ 2 × 22/7 × r = 66 cm
⇒ r = 66 × 1/2 × 7/22 cm
⇒ r = 10.5 cm
Area of circle = πr^{2} = 22/7 × 22/7 × 10.5 × 10.5 cm^{2}
= 346.5 cm^{2}
Q.12. The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km per hour. [Take π = 22/7]
Diameter of the wheel = 84 cm
Let the radius of the wheel be R cm
Radius of the wheel = 84/2 cm = 42 cm
No. of revolutions wheel makes = 5 rev/sec
Since, 1 revolution = 2πR
Speed of the wheel = 5 × 2πR rev/sec
= 5 × 2 × 22/7 × 42 = 1320 cm/sec
= 13.20 m/sec
= 13.20 × 3600/1000 km/h
= 47.52 km/h
Since, 1 m/sec = 3600/1000 km/h
Q.13. OACB is a quadrant of a circle with centre O and its radius is 3.5 cm. If OD = 2 cm, find the area of (i) the quadrant OACB (ii) the shaded region. [Take π = 22/7]
Radius of circle = R = 3.5 cm
OD = 2 cm
OA = OB = R = 3.5 cm
Since, OACB is a quadrant of a circle ∴ angle subtended by it at the centre = 90°
(i) Area of quadrant =
= 22/7 × 3.5 × 3.5 × 90°/360° cm2
= 9.625 cm2
(ii) Area of shaded region = Area of quadrant – Area of triangle OAD
Area of triangle OAD = 1/2 × base × height
= 1/2 × OA × OD
= 1/2 × 3.5 × 2 cm^{2}
= 3.5 cm^{2}
Area of shaded region = 9.625 cm^{2} – 3.5cm^{2}
= 6.125 cm^{2}
Q.14. In the given figure, ABCD is a square each of whose sides measures 28 cm. Find the area of the shaded region. [Take π = 22/7]
Length of the sides of square = 28 cm
Area of square = a^{2} = 282 cm^{2}
= 784 cm^{2}
Since, all the circles are identical so, they have same radius
Let the radius of circle be R cm
From the figure 2R = 28 cm
R = 28/2 cm
R = 14 cm
Quadrant of a circle subtends 90° at the centre.
Area of quadrant of circle =
= 22/7 × 14 × 14 × 90°/360° cm^{2} = 154 cm^{2}
Area of 4 quadrants of circle = 154 × 4 cm^{2} = 616 cm^{2}
Area of shaded region = Area of square – Area of 4 quadrants of circle
= 784 cm^{2} – 616 cm^{2}
= 168 cm^{2}
Q.15. In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm. Find the area of the shaded region. [Take π = 3.14 and √3 = 1.73]
Radius of circle = R = 4 cm
OD perpendicular to AB is drawn
ΔABC is equilateral triangle,
∠A = ∠B = ∠C = 60°
∠OAD = 30°
OD/AO = sin 30°
AO = 4 cm
OD = 1/2 × 4 cm
OD = 2 cm
AD^{2} = OA^{2} – OD^{2}
= 4^{2 }– 2^{2 }= 16 – 4 = 12 cm^{2}
AD = 2√3 cm
AB = 2 × AD
= 2 × 2√3 cm = 4√3 cm
Area of triangle ABC = √3/4 × AB^{2}
= √3/4 × 4√3 × 4√3
= 20.71 cm^{2}
Area of circle = πR^{2}
= 3.14 × 4 × 4 cm^{2}
= 50.24 cm^{2}
Area of shaded region = 29.53 cm^{2}
Q.16. The minute hand of a clock is 7.5 cm long. Find the area of the face of the clock described by the minute hand in 56 minutes.
Length of minute hand = 7.5 cm
In a clock, length of minute hand = radius
Radius = R = 7.5 cm
In 1 minute, minute hand moves 6°
So, in 56 minutes, minute hand moves 56 × 6° = 336°
Area described by minute hand =
= 22/7 × 7.5 × 7.5 × 336°/360° cm^{2}
= 165 cm^{2}
Q.17. A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track.
Let the inner radius be R_{1} and outer radius be R_{2}
Inner circumference = 2πR_{1} = 352 m
⇒ 2 × 22/7 × R_{1 }= 352 m
⇒ R_{1 }= 352 × 1/2 × 7/22
⇒ R_{1} = 56 m
Outer Circumference = 2πR_{2} = 396 m
⇒ 2 × 22/7 × R_{2} = 396 m
⇒ R_{2} = 396 × 1/2 × 7/22 m
⇒ R_{2} = 63 m
Width of the track = R_{2} – R_{1} = 63 m – 56 m = 7 m
Area of track = = 22/7 × (63^{2} 56^{2})
= 22/7 × (3969 – 3136) m^{2}
= 22/7 × 833 m^{2} = 2618 m^{2}
Q.18. A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the areas of the minor and major segments. [Take π = 22/7 and √3 = 1.732.]
∠ACB = 60°
Chord AB subtends an angle of 60° at the centre
Radius = 30 cm
Let Radius be R
In triangle ABC, AC = BC
So, ∠CAB = ∠CBA
∠ ACB + ∠ CAB + ∠ CBA = 180°
60° + 2∠CAB = 180°
2∠CAB = 180°  60° = 120°
∠CAB = 120°/2 = 60°
∠CAB = ∠CBA = 60°
∴ ΔABC is a equilateral triangle
Length of side of an equilateral triangle = radius of circle = 30 cm
Area of equilateral triangle = √3/4 × side2 = 1.732/4 × 30 × 30 cm^{2}
= 389.7 cm^{2}
Area of sector ACB = = 3.14 × 30 × 30 × 60°/360° = 471.45 cm^{2}
Area of minor Segment = Area of sector ACB – Area of ΔABC
= 471.45 cm^{2} – 389.7 cm^{2} = 81.75 cm^{2}
Area of circle = πR^{2} = 3.14 × 30 × 30 cm^{2} = 2828.57 cm^{2}
Area of major segment = Area of circle – Area of minor segment
= 2826 cm^{2 }– 81.75 cm^{2}
= 2744.25 cm^{2}
Q.19. Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshared area. What area will be left ungrazed? [Take π = 3.14]
From the figure we see that cows are tethered at the corners of the square so while grazing they form four quadrants as shown in the figure
Length of side of square = 50 m
Length of side of square = 2 × Radius of quadrant
Radius of quadrant = R = 50/2 m
= 25 m
Area of square = side^{2}
= 50^{2} m^{2} = 2500 m^{2}
Area of quadrant = 1/4 π R^{2} = 1/4 × 3.14 × 25 × 25 m^{2}
= 490.625 m^{2}
Area of 4 quadrants = 4 × 490.625 m^{2}
= 1962.5 m^{2}
Area left ungrazed = Area of shaded part
= Area of square – Area of 4 quadrants
= 2500 m^{2} – 1962.5 m^{2}
= 537.5 m^{2}
Q.20. A square tank has an area of 1600 m^{2}. There are four semicircular plots around it. Find the cost of turfing the plots at Rs. 12.50 per m^{2}. [Take π = 3.14]
Let the length of side of the square tank be a
Area of square tank = a^{2} = 1600 m^{2}
⇒ a = √1600 m = 40 m
Let the radius of semicircle be R
From the figure we can see that
Length of the side of the square = Diameter of semicircle
40 m = 2 × R
R = 40/2 m
R = 20 m
Area of semicircle = 1/2 πR^{2} = 1/2 × 3.14 × 20 × 20 m^{2}
= 628 m^{2}
Area of 4 semicircles = 4 × 628 m^{2}
= 2512 m^{2}
Cost of turfing the plots = Rs. 12.50 per m^{2}
Cost of Turfing = Cost of turfing per m^{2} × Area of 4 semicircle
= Rs. 12.50 × 2512
= Rs. 31400
53 docs15 tests

53 docs15 tests


Explore Courses for Class 10 exam
