Chapter 18 Areas of Circle, Sector & Segment RS Aggarwal Solutions MCQ Class 10 Maths
Chapter Name  RS Aggarwal Chapter 18 Areas of Circle, Sector & Segment 
Book Name  RS Aggarwal Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Areas of Circle, Sector & Segment MCQ Solutions
1. The area of a circle is 38.5 cm^{2}. The circumference of the circle is
A. 6.2 cm
B. 12.1 cm
C. 11 cm
D. 22 cm
Solution
Let the radius if the circle be r
Given, Area of circle = 38.5 cm^{2}
Area of circle = Ï€r^{2}
⇒ Ï€r^{2 }= 38.5
Since Ï€ = 22/7
⇒ 22/7 × r^{2} = 38.5
⇒ r^{2} = 38.5 × (7/22)
⇒ r^{2} = 12.25
⇒ r = √12.25 = 3.5 cm
∴ Radius of circle = 3.5 cm
Circumference of circle = 2Ï€r
= 2 × 22/7 × 3.5 cm
= 22 cm
∴ Circumference of the circle is 22 cm
Let the radius if circle be r
Given, Area of circle = 38.5 cm^{2}
Area of circle = Ï€r^{2}
Ï€r^{2} = 38.5
Since = 22/7
∴ Ï€r^{2} = 38.5
⇒ 22/7 × r^{2} = 38.5
⇒ r^{2} = 12.25
⇒ r = √12.25 = 3.50 cm
∴ Radius of circle = 3.5 cm
Circumference of circle = 2Ï€r
= 2 × 22/7 × 3.5
= 22 cm
∴ Circumference of the circle is 22 cm.
2. The area of a circle is 49Ï€ cm^{2}. Its circumference is
A. 7Ï€ cm
B. 14Ï€ cm
C. 21Ï€ cm
D. 28Ï€ cm
Solution
Let the radius if circle be r
Given, Area of circle = 49Ï€ cm^{2}
Area of circle = Ï€r^{2}
Ï€r^{2} = 49Ï€
Since = 22/7
∴ Ï€r^{2} = 49Ï€
⇒ r^{2} = 49
⇒ r = √49 = 7 cm
∴ Radius of circle = 7 cm
Circumference of circle = 2Ï€r
= 2 × r × 7 cm
= 14Ï€ cm
∴ Circumference of the circle is 14Ï€ cm.
3. The difference between the circumference and radius of a circle is 37 cm. The area of the circle is
A. 111 cm^{2}
B. 84 cm^{2}
C. 154 cm^{2}
D. 259 cm^{2}
Solution
Let the radius if circle be r
Circumference of circle = 2Ï€r
Difference between the circumference and radius of a circle = 37 cm
⇒ 2Ï€r – r = 37 cm
⇒ 2 × 22/7 × r – r = 37 cm
⇒ 44/7 × r – r = 37 cm
⇒ (44/7 – 1) × r = 37 cm
⇒ 37/7 × r = 37 cm
⇒ r = 37 × 7/37
⇒ r = 7 cm
Area of circle = Ï€r^{2}
= 22/7 × 7 × 7 cm^{2}
= 22/7 × 49 cm^{2}
= 22 × 7 cm^{2}
= 154 cm^{2}
∴ Area of the circle is 154 cm^{2}
4. The perimeter of a circular field is 242 m. The area of the field is
A. 9317 m^{2}
B. 18634 m^{2}
C. 4658 m^{2}
D. none of these
Solution
Let the radius if circular field be r
Perimeter of circular field = 2Ï€r
Perimeter of circular field = 242 m
⇒ 2Ï€r = 242 m
⇒ 2 × 22/7 × r = 242 m
⇒ r = 242 × 1/2 × 7/22 = 38.5 m
∴ Radius of circular field = 38.5 m
Area of the filed = Ï€r^{2}
= 22/7 × 38.52 m^{2}
= 22/7 × 1482.5 m^{2} = 4658.5 m^{2}
∴ Area of the field = 4658.5 m^{2}
5. On increasing the diameter of a circle by 40%, its area will be increased by
A. 40%
B. 80%
C. 96%
D. 82%
Solution
Let the radius if circle be r
Area of circle = A = Ï€r^{2 }
Radius increases by 40%
So, new radius r’ = r + 40/100 × r = 1.4r
New Area of circle = A’ = Ï€r’^{2} = Ï€ × (1.4r)^{2}
= 1.96Ï€r^{2}
Percentage increase in area = (A’ – A)/A × 100
= (1.96Ï€ r^{2} – Ï€r^{2})/Ï€r^{2} × 100
= 0.96 × 100
= 96
∴ Increase in area = 96%
6. On decreasing the radius of a circle by 30%, its area is decreased by
A. 30%
B. 60%
C. 45%
D. none of these
Solution
Let the radius if circle be r
Area of circle = A = Ï€r^{2}
Radius decreases by 30%
So, New Radius r’ = r – 30/100 × r = 0.7 r
New Area of circle = A’ = Ï€r’^{2} = Ï€ × (0.7r)^{2}
= 0.49 Ï€r^{2}
Percentage decrease in area = (A – A’)/A × 100
= (Ï€r^{2}  0.49Ï€r^{2})/Ï€r^{2} × 100
= 0.51 × 100
= 51
∴ Decrease in area = 51%
7. The area of a square is the same as the area of a circle. Their perimeters are in the ratio
A. 1 : 1
B. 2 : Ï€
C. Ï€ : 2
D. √Ï€ : 2
Solution
Let the length of the side of the square be a
Let the radius if circle be r
Area of a square = a^{2}
Area of circle = Ï€r^{2}
Area of a square = Area of a circle
a^{2} = Ï€r^{2}
⇒ a = √Ï€ × r
Perimeter of circle = 2Ï€r
Perimeter of square = 4a
= 4√Ï€r
Perimeter of square)/(Perimeter of circle) = 4√Ï€r/2Ï€r
Perimeter of circle = √Ï€/2
Ratio of perimeter of circle and square = √Ï€: 2
8. The circumference of a circle is equal to the sum of the circumferences of two circles having diameters 36 cm and 20 cm. The radius of the new circle is
A. 16 cm
B. 28 cm
C. 42 cm
D. 56 cm
Solution
Let the bigger circle be C_{1} and other circles be C_{2 }and C_{3}
Radius of circle C_{1} = r_{1}
Diameter of circle C_{2} = 36 cm
Radius of circle C_{2} = r_{2} = 36/2 cm = 18 cm
Diameters of circle C_{3} = 20 cm
Radius of circle C_{3} = r_{3} = 20/2 cm = 10 cm
Circumference of circle C_{2} = 2Ï€r_{2}
= 2 × Ï€ × 18 cm
= 36Ï€ cm
Circumference of circle C3 = 2Ï€r_{3}
= 2 × Ï€ × 10 cm
= 20Ï€ cm
Circumference of circle C_{1} = Circumference of circle C_{2 }+ Circumference of circle C_{3}
⇒ 2Ï€r_{1 }= 2Ï€r_{2} + 2Ï€r_{3}
⇒ 2Ï€r_{1} = 36Ï€ + 20Ï€
⇒ 2Ï€r_{1} = 56Ï€
⇒ r_{1} = 28 cm
Radius of circle C_{1} = r_{1} = 28 cm
9. The area of a circle is equal to the sum of the areas of two circles of radii 24 cm and 7 cm. The diameter of the new circle is
A. 25 cm
B. 31 cm
C. 50 cm
D. 62 cm
Solution
Let the bigger circle be C_{1} and other circles be C_{2} and C_{3}
Radius of circles C_{1} = r_{1}
Radius of circle C_{2} = r_{2} = 24 cm
Radius of circle C_{3} = r_{3} = 7 cm
r_{1} = 25 cm
Diameter of new circle = 25 × 2 cm
= 50 cm
10. If the perimeter of square is equal to the circumference of a circle then the ratio of their areas is
A. 4 : Ï€
B. Ï€: 4
C. Ï€ : 7
D. 7 : Ï€
Solution
Let the length of the side of the square be a
Let the radius if circle be r
Perimeter of circle = 2Ï€r
Perimeter of square = 4a
Perimeter of circle = Perimeter of square
⇒ 2Ï€r = 4a
a = Ï€ × r/2
Area of square = a^{2}
Area of circle = Ï€r^{2}
Area of square)/(Area of circle) = a^{2}/Ï€r^{2}
(Area of square)/(Area of circle) = (Ï€r/4)^{2}/Ï€r^{2}
(Area of square)(Area of circle) = Ï€^{2}r^{2}/4Ï€r^{2}
Area of square)/Area of circle) = Ï€/4
Ratio of area of square to circle = Ï€ : 4
11. If the sum of the areas of two circles with radii R_{1} and R_{2} is equal to the area of a circle of radius R then
A. R_{1} + R_{2} = R
B. R_{1} + R_{2} < R
SolutionLet the three circles be C_{1}, C_{2} and C
Area of circle C = Area of circle C_{1} + Area of circle C_{2}
12. If the sum of the circumferences of two circles with radii R_{1} and R_{2} is equal to the circumference of a circle of radius R then
A. R_{1} + R_{2} = R
B. R_{1} + R_{2} > R
C. R_{1} + R_{2 }< R
D. None of these
Solution
Let three circles be C_{1}, C_{2} and C
Circumference of circle C = Circumference of circle C_{1} + Circumference of circle C_{2}
⇒ 2Ï€R = 2Ï€R_{1} + 2Ï€R_{2}
R = R_{1} + R_{2}
13. 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. None of these
Solution
Let the length of the side of the square be a
Let the radius if circle be r
Perimeter of circle = 2Ï€r
Perimeter of square = 4a
Perimeter of circle = Perimeter of square
2Ï€r = 4a
⇒ a = Ï€ × r/2
Area of square = a^{2}
= (Ï€ × r/2)^{2}
= Ï€/4 × Ï€r^{2}
Area of circle = Ï€r^{2 }
Seeing the coefficient of Ï€r^{2}
1 > Ï€/4
∴ Ï€r^{2} > Ï€/4 × Ï€r^{2}
So, (area of the circle) > (area of the square)
14. The radii of two concentric circles are 19 cm and 16 cm respectively. The area of the ring enclosed by these circles is
A. 320 cm^{2}
B. 330 cm^{2}
C. 332 cm^{2}
D. 340 cm^{2}
Solution
Radius of circle 1 = r_{1} = 19 cmRadius of circle 2 = r_{2} = 16 cm
15. The areas of two concentric circles are 1386 cm^{2} and 962.5 cm^{2}. The width of the ring is
A. 2.8 cm
B. 3.5 cm
C. 4.2 cm
D. 3.8 cm
Solution
Let the radius of circle 1 & 2 be R_{1} and R_{2} respectively
Area of circle 1 = 1386 cm^{2}
R_{2} = 17.5 cm
Width of the ring = R_{1} – R_{2}
= 21 – 17.5
= 3.5 cm
16. The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is
A. 3 : 4
B. 4 : 3
C. 9 : 16
D. 16 : 9
Solution
Circumference of circle C_{1} = 2Ï€r_{1}
Circumference of circle C2 = 2Ï€r_{2}
(Circumference of circle C_{1})/(Circumference of circle C_{2}) = 2Ï€r_{1}/2Ï€r_{2} = 3/4
r_{1}/r_{2} = 3/4
∴ Ratio of two circles = 9 : 16
17. The areas of two circles are in the ratio 9 : 4. The ratio of their circumferences is
A. 3 : 2
B. 4 : 9
C. 2 : 3
D. 81 : 16
Solution
r_{1}/r_{2} = 3/2
(Circumference of circle C_{1})/(Circumference of circle C_{2}) = 2Ï€r_{1}/2Ï€r_{2} = r_{1}/r_{2} = 3/2
Ratio of their circumferences = 3 : 2
18. The radius of a wheel is 0.25 m. How many revolutions will it making in covering 11 km?
A. 2800
B. 4000
C. 5500
D. 7000
Solution
Radius of wheel = r = 0.25 m
Distance the wheel travels = 11 km = 11000 m
In 1 revolution wheel travels 2Ï€r distance
No. of revolutions a wheel makes = (distance travelled by the wheel)/2Ï€r
= 11000/(2Ï€ × 0.25)
= 11000/(2 × 22/7 × 0.25)
= (11000 × 7)/(2 × 22 × 0.25
= 7000 revolutions
19. The diameter of a wheel is 40 cm. How many revolutions will it make in covering 176 cm?
A. 140
B. 150
C. 160
D. 166
Solution
Diameter of wheel = 40 cm
Radius of wheel = r = 40/2 cm = 20 cm
Distance the wheel travels = 176 m = 17600 cm
In 1 revolution wheel travels 2Ï€r distance
No. of revolutions a wheel makes = (distance travelled by the wheel)/2Ï€r
= 17600/(2Ï€ × 20)
= 17600/(2 × 22/7 × 20)
= 17600 × 7)/(2 × 22 × 20)
= 140 revolutions
20. In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is
A. 14 m
B. 24 m
C. 28 m
D. 40 m
Solution
Distance the wheels travels = 88 km = 88000 m
In 1 revolution wheel travels 2Ï€r distance
No. of revolutions a wheel makes = (distance travelled by the wheel)/2Ï€r
No. of revolutions a wheel makes = 1000
r = (distance travelled by the wheel)/(2Ï€ × No. of revolutions a wheel makes) = 88000/(2 × 22/7 × 1000)
= (88000 × 7)/(2 × 22 × 1000)
r = 14 m
Radius of wheel = 14 m
Diameter of wheel = 2 × 14 m = 28 m
21. The area of a sector of angle Î¸° of a circle with radius R is
A. 2Ï€R^{2}Î¸/180
B. 2Ï€R^{2}Î¸/360
C. Ï€R^{2}Î¸/180
D. Ï€R^{2}Î¸/360
Solution
Area of a sector of angle Î¸° of a circle with radius R = area of circle × Î¸/360
= Ï€R^{2}Î¸/360
22. The length of an arc of a sector of angle Î¸° of a circle with radius R is
A. 2Ï€RÎ¸ /180
B. 2Ï€RÎ¸/360
C. Ï€R^{2}Î¸/180
D. Ï€R^{2}Î¸/360
Solution
Length of an arc of a sector of angle Î¸° of a circle with radius R
= Circumference of circle × Î¸/360
= 2Ï€RÎ¸/360
23. The length of the minute hand of a clock is 21 cm. The area swept by the minute hand in 10 minutes is
A. 231 cm^{2}
B.210 cm^{2}
C .126 cm^{2}
D. 252 cm^{2}
Solution
Length of the minute hand of a clock = 21 cm
∴ Radius = R = 21 cm
In 1 minute, minute hand sweeps 6°
So, in 10 minutes, minute hand will sweep 10 × 6° = 60°
Area swept by minute hand in 10 minutes = Area of a sector of angle of a circle with radius R = Ï€R^{2}Î¸/360
= 22/7 × 21 × 21 × 60/360
= 231 cm^{2}
24. A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segments (given, Ï€ = 3.14) is
A. 32.5 cm^{2}
B. 34.5 cm^{2}
C. 28.5 cm^{2}
D. 30.5 cm^{2}
Solution
Radius of circle = R = 10 cm
Area of minor segment = Area of sector subtending 90°  Area of triangle ABC
Area of sector subtending 90° = Ï€R^{2}Î¸/ 360
= 3.14 × 10 × 10 × 90/360 cm^{2}
= 78.5 cm^{2}
Area of triangle ABC = 1/2 × AC × BC
= 1/2 × 10 × 10 cm^{2}
= 50 cm^{2}
Area of Minor segment = 78.5 cm^{2} – 50 cm^{2}
= 28.5 cm^{2}
25. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the arc is
A. 21 cm
B. 22 cm
C. 18.16 cm
D. 23.5 cm
Solution
Radius of circle = R = 21 cm
Angle subtended by the arc = 60°
Length of an arc of a sector of angle Î¸° of a circle with radius R = 2Ï€RÎ¸/360
Length of arc = 2 × 22/7 × 21 × 60/360 cm
= 22 cm
26. In a circle of radius 14 cm, an arc subtends an angle of 120° at the centre. If √3 = 1.73 then the area of the segment of the circle is
A. 120.56 cm^{2}
B. 124.63 cm^{2}
C. 118.24 cm^{2}
D. 130.57 cm^{2}
Solution
Radius of Circle = R = 14 cmAngle subtended by the arc = Î¸ = 120°
Area of sector subtending 120° = Ï€R^{2}Î¸/360
= 22/7 × 14 × 14 × 120/360 cm^{2}
= 205.33 cm^{2}
In triangle ABC
AC = BC = 14 cm = R
Area of triangle ABC = 1/2 × base × height
= 2 × 1/2 × R sin Î¸/2 × R × cos Î¸/2
= 2 × 1/2 × 14 × 14 × sin 60° × cos 60°
= 84.77 cm^{2}
Area of segment = Area of sector subtending 120°  Area of triangle ABC
= 205.33 – 84.77 cm^{2}
= 120.56 cm^{2}
Areas of Circle, Sector & Segment
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 Ï€ = 3.14]
A. 214 cm^{2}B. 228 cm^{2}
C. 242 cm^{2}
D. 248 cm^{2}
Solution
Length of side of square = OA = 20 cm
Radius of Quadrant = Ï€R^{2} × Î¸/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}
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
Solution
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 2Ï€r distance
No. of revolutions a wheel makes = (distance travelled by the wheel)/2Ï€r
= 79200/(2Ï€ × 42)
= 79200/(2 × 22/7 × 42)
= (79200 × 7)/(2 × 22 × 42)
= 300 revolutions
3. The area of a sector of a circle with radius r, making an angle of x° at the centre is x
A. x/360 × 2Ï€r
B. x/180 × Ï€r^{2}
C. x/360 × 2Ï€r
D. x/360 × Ï€r^{2}
Solution
Area of a sector of angle Î¸° of a circle with radius R = area of circle = Î¸/360
= Ï€r^{2} × x°/360°
4. In the given figure, ABCD is a rectangle inscribed in a circle having length 8 cm and breadth 6 cm. If Ï€ = 3.14 then the area of the shaded region is
Solution
Given
Length of rectangle = 8 cm
Breadth of rectangle = 6 cm
Area of rectangle = length × breadth
= 8 × 6
= 48 cm^{2}
Consider △ABC,
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 = 5 cm
Let the radius of circle be r = 5 cm
Then, Area of circle = Ï€r^{2}
= 22/7 × 5 × 5
= (22 × 25)/7
= 550/7
= 78.57 cm^{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}.
[None of the option is correct].
5. The circumference of a circle is 22 cm. Find its area. [Take Ï€ = 22/7]
Solution
Let the radius if circle be r
Circumference of circle = 22 cm
2Ï€r = 22 cm
⇒ 2 × 22/7 × r = 22 cm
⇒ r = 22 × 1/2 × 7/22 cm
⇒ r = 3.5 cm
Area of Circle = Ï€r^{2}
= 22/7 × 3.5 × 3.5 cm^{2}
= 38.5 cm^{2}
∴ Area of Circle = 38.5 cm^{2}
6. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.
Solution
Radius of circle = R = 21 cm
Angle subtended by arc = 60°
Length of an arc of a sector of angle Î¸° of a circle with radius R
= Circumference of circle × Î¸/360
= 2Ï€RÎ¸/360
Length of arc = 2 × 22/7 × 21 × Î¸/360 cm
= 22 cm
Length of arc = 22 cm
7. The minute hand of a clock is 12 cm long. Find the area swept by it in 35 minutes.
Solution
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 Î¸° of a circle with radius R = Ï€R^{2}Î¸/360
= 22/7 × 12 × 12 × 60°/360°
= 264 cm^{2}
Area swept by minute hand in 35 minutes = 264 cm^{2}
8. The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector.
Solution
Radius of circle = 5.6 cm
Perimeter of a sector of a circle = 2R + Circumference of circle × Î¸/360
= 2R + 2Ï€RÎ¸/360
Perimeter of a sector of a circle = 2 × 5.6 + 2 × 22/7 × 5.6 × Î¸/360 cm
= 27.2 cm
⇒ 2 × 22/7 × 5.6 × Î¸/360 = 27.2 – 11.2 cm
= 16 cm
⇒ Î¸ = 16 × 1/2 × 1/5.6 × 7/22
⇒ Î¸ = 163.63°
Area of sector = Ï€r^{2 }× Î¸/360 = 22/7 × 5.6 × 5.6 × 163.63/360
= 44.8 cm^{2 }
∴ Area of sector = 44.8 cm^{2}
9. A chord of a circle of radius 14 cm makes right angle at the centre. Find the area of the sector.
Solution
Chord AB subtends an angle of 90° at the centre of the circleRadius of circle = R = 14 cm
Area of sector of circle of radius R = Ï€R^{2}Î¸/360
= 22/7 × 14 × 14 × 90/360 cm^{2}
= 154 cm^{2}
10. In the given figure, the sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region.
SolutionGiven,
Radius of smaller circle = R_{1} = 3.5 cm
Radius of bigger circle = R_{2} = 7 cm
Angle subtended = 30°
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 form of a circle, what will be the area of the circle? [Take Ï€ = 22/7]
Solution
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}
⇒ √3/4a^{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 = 2Ï€r
⇒ 2Ï€r = 66 cm
⇒ 2 × 22/7 × r = 66 cm
⇒ r = 66 × 1/2 × 7/22 cm
⇒ r = 10.5 cm
Area of circle = Ï€r^{2} = 22/7 × 22/7 × 10.5 × 10.5 cm^{2}
= 346.5 cm^{2}
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 Ï€ = 22/7]
Solution
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 = 2Ï€R
Speed of the wheel = 5 × 2Ï€R 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
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 Ï€ = 22/7]
SolutionRadius 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 = Ï€R^{2}Î¸/360
= 22/7 × 3.5 × 3.5 × 90°/360° cm^{2}
= 9.625 cm^{2}
(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} – 3.5 cm^{2}
= 6.125 cm^{2}
14. In the given figure, ABCD is a square each of whose side measures 28 cm. Find the area of the shaded region. [Take Ï€ = 22/7]
SolutionLength 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
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 = Ï€R^{2}Î¸/360
= 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}
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 Ï€ = 3.14 and √3 = 1.73]
SolutionRadius of circle = R = 4 cmOD perpendicular to AB is drawn
△ABC is equilateral triangle,
∠A = ∠B = ∠C = 60°
∠OAD = 30°
OD/AO = sin 30°
AO = 4 cm
OD/AO = 1/2
⇒ 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 = Ï€R^{2}
= 3.14 × 4 × 4 cm^{2}
= 50.24 cm^{2}
Area of shaded region = 29.53 cm^{2}
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.
Solution
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° = 360°
Area described by minute hand = Ï€R^{2}Î¸/360°
= 22/7 × 7.5 × 7.5 × 336°/360° cm^{2}
= 165 cm^{2}
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.
Solution
Let the inner radius be R_{1} and outer radius be R_{2}
Inner circumference = 2Ï€R_{1} = 352 m
⇒ 2 × 22/7 × R_{1} = 3652 m
⇒ R_{1} = 352 × 1/2 × 7/22
⇒ R_{1} = 56 m
Outer Circumference = 2Ï€R_{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 – 56m = 7 m
Solution
∠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°
∴ △ABC is a equilateral triangle
Length of side of an equilateral triangle = radius of circle = 30 cm
Area of equilateral triangle = √3/4 × side^{2}
= 1.732 × 30 × 30 cm^{2}
= 389.7 cm^{2}
Area of sector ACB = Ï€R^{2}Î¸/360
= 3.14 × 30 × 30 × 60°
= 471.45 cm^{2}
Area of minor segment = Area of sector ACB – Area of △ABC
= 471.45 cm^{2} – 389.7 cm^{2}
= 81.75 cm^{2}
Area of circle = Ï€R^{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}
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 Ï€ = 3.14]
Solution
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 = 1/4Ï€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}
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 Ï€ = 3.14]
Solution
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 Ï€R^{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