Chapter 18 Areas of Circle, Sector & Segment RS Aggarwal Solutions Exercise 18A Class 10 Maths
Chapter Name  RS Aggarwal Chapter 18 Areas of Circle, Sector & Segment 
Book Name  RS Aggarwal Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Areas of Circle, Sector & Segment Exercise 18A Solutions
1. The difference between the circumference and radius of a circle is 37 cm. using Ï€ = 22/7, find the circumference of the circle.
Solution
Given,
Difference between the circumference and the radius of the circle = 37 cm
Let the radius of the circle be ‘r’.
Circumference of the circle = 2Ï€r
So, Difference between the circumference and the radius of the circle = 2Ï€r – r = 37
2Ï€r – r = 37
⇒ 2 × 22/7 × r – r = 37
⇒ 44/7 × r – r = 37
⇒ r(44/7 – 1) = 37
⇒ 37/7 × r = 37
⇒ r = 37 × 7/37
⇒ r = 7 cm
∴ Circumference of the circle = 2 × 22/7 × 7
= 2 × 22
= 44 cm
Hence, the circumference of the circle is 44 cm.
2. The circumference of a circle is 22 cm. Find the area of its quadrant.
Solution
Given,
Circumference of circle = 22 cm
Let the radius of the circle be ‘r’.
∵ Circumference of circle = 2Ï€r
∴ 22 = 2 × Ï€ × r
⇒ 22 = 2 × 22/7 × r
⇒ 22 × 7/22 × 1/2 = r or 7/2 = r
or r = 7/2
∵ Area of circle = Ï€r^{2}
∴ Area of its quadrant = 1/4Ï€r^{2}
= 1/4 × 22/7 × 7/2 × 7/2
= 77/8
Hence, the area of the quadrant of the circle is 77/8 cm.
3. What is the diameter of a circle whose area is equal to the sum of the areas of two circles of diameter 10 cm and 24 cm?
Solution
Let the two circles be C_{1} and C_{2} with diameters 10 cm and 24 cm respectively.
Area of circle, C = Area of C_{1} + Area of C_{2} ...(i)
∵ Diameter = 2 × radius
∴ Radius o C_{1}, r_{1} = 10/2 = 5 cm
and Radius of C_{2}, r_{2} = 24/2 = 12 cm
∵ Area of circle = Ï€r^{2} …(ii)
⇒ Diameter = 2 × r
= 2 × 13
= 26 cm
Hence, the diameter of the circle is 26 cm.
4. If the area of a circle is numerically equal to twice its circumference, then what is the diameter of the circle?
Solution
Given,
Area of circle = 2 × Circumference of circle …(i)
Let the radius of the circle be ‘r’.
Then, the area of the circle = Ï€r^{2}
and the circumference of the circle = 2Ï€r
Using (i), we have
Ï€r^{2 }= 2 × 2Ï€r
⇒ Ï€r^{2} = 4 Ï€r
⇒ r = 4 cm
∵ Diameter = 2 × radius
∴ Diameter = 2 × 4
= 8 cm
Hence, the diameter of the circle is 8 cm.
5. What is the perimeter of a square which circumscribes a circle of radius a cm?
Solution
Given,
Perimeter of square circumscribes a circle of radius ‘a’.
Side of square = Diameter of circle
Diameter of circle = 2 × radius
= 2a
So, side of square = 2a
∵ Perimeter of square = 4 × side
∴ Perimeter of square = 4 × 2a
= 8a
Hence, the perimeter of the square is 8a.
6. Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60° at the centre.
Solution
Given,
Diameter of circle = 42 cm
⇒ Radius of circle = 42/2 cm = 21 cm
Angle subtend at the centre = 60°
∵ Length of arc = Î¸/360° × 2Ï€r
= 60/360 × 2 × 22/7 × 21
= 22 cm
Hence, the length of the arc is 22 cm.
7. Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii 4 cm and 3 cm.
Solution
Let the two circles with radii 4 cm and 3 cm be C_{1} and C_{2} respectively.
⇒ r_{1} = 4 cm and r_{2 }= 3 cm
Area of circle, C = Area of C_{1} + Area of C_{2} …(i)
∵ Area of circle = Ï€r^{2} ...(ii)
Area of C = 352/7 + 198/7 = 550/7 cm^{2 }
Now, using (ii), we have
Ï€^{r2} = 550/7
⇒ 22/7 × r^{2 }= 550/7
⇒ r^{2} = 550/7 × 7/22 = 25
⇒ r = √25 = 5
⇒ r = 5 cm
∵ Diameter = 2 × radius
∴ Diameter = 2 × 5 = 10 cm
Hence, diameter of the circle with area equal to the sum of two circles of radii 4 cm and 3 cm is 10 cm.
8. Find the area of a circle whose circumference is 8Ï€.
Solution
Given,
Circumference of circle = 8Ï€.
∵ Circumference of a circle = 2Ï€r
∴ 8Ï€ = 2Ï€r
⇒ r = 4
∵ Area of circle = Ï€r^{2}
∴ Area of circle = Ï€ × 4 × 4
= 16Ï€
Hence, the area of the circle is 16Ï€.
9. Find the perimeter of a semicircular protractor whose diameter is 14 cm.
Solution
Given,
Diameter of the semicircular protractor = 14 cm
Radius of the protractor = 14/2 cm = 7 cm
∵ Perimeter of semicircle = Ï€r + d
∴ Perimeter of semicircular protractor = 22/7 × 7 + 14
= 22 + 14
= 36 cm
Hence, the perimeter of the semicircular protractor is 36 cm.
10. Find the radius of a circle whose perimeter and area are numerically equal.
Solution
Given,
Perimeter of a circle = Area of circle ...(i)
∵ Perimeter of circle = 2Ï€r and Area of circle = Ï€r^{2}
∴ Using (i), we have
2Ï€r = Ï€r^{2}
⇒ 2 = Ï€r^{2}/2Ï€r
⇒ 2 = r or r = 2
Hence, the radius of the circle is 2 cm.
11. The radii of two circles are 19 cm and 9 cm. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution
Given,
Radius of one of the circles, C_{1 }= 19 cm = r_{1}
Radius of the other circle, C_{2} = 9 cm = r_{2 }
Let the other circle be C with radius ‘r’.
Circumference of C = Circumference of C_{1} + circumference of C_{2} …(i)
∵ Circumference of circle = 2Ï€r
∴ Circumference of C_{1} = 2Ï€r_{1 }= 2 × 22/7 × 19 = 836/7
and circumference of C_{2} = 2 Ï€r_{2} = 2 × 22/7 × 9 = 396/7
Using (i), we have
2 Ï€r = 836/7 + 396/7 = 1232/7
⇒ 2 × 22/7 × r = 1232/7
⇒ r = 1232/7 × 7/22 × 1/2 = 28
⇒ r = 28 cm
Hence, the radius of the circle is 28 cm.
12. The radii of two circles are 8 cm and 6 cm. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution
Given,
Radius of one of the circles, C_{1 }= 8 cm = r_{1}
Radius of the other circle, C_{2} = 6 cm = r_{2 }
Let the other circle be C with radius ‘r’.
Area of C = Area of C_{1 }+ Area of C_{2} …(i)
∵ Area of circle = Ï€r^{2}
Hence, the radius of the circle is 10 cm.
13. Find the area of the sector of a circle having radius 6 cm and of angle 30°.
Solution
Given,
Radius of circle = 6 cm
Angle of the sector = 30°
∵ Area of sector = Î¸/360 × Ï€r^{2}
= 30/360 × 3.14 × 6 × 6
= 3 × 3.14
= 9.42 cm^{2}
Hence, the area of the sector is 9.42 cm^{2}.
14. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.
Solution
Given,
Radius of circle = 21 cm
Angle subtended by the arc = 60°
∵ Length of arc = Î¸/360 × 2Ï€r
= 60/360 × 2 × 22/7 × 21
= 22 cm
Hence, the length of the arc is 22 cm.
15. The circumference of two circles are in the ratio 2 : 3. What is the ratio between their areas?
Solution
Given,
Ratio of circumferences of two circles = 2 : 3
Let the two circles be C_{1} and C_{2} with radii ‘r_{1}’ and 'r_{2}’.
∵ Circumference of circle = 2Ï€r
∴ Circumference of C_{1} = 2Ï€r_{1}
and circumference of C_{2} = 2Ï€r_{2}
⇒ 2Ï€r_{1}/2Ï€r_{2} = 2/3
⇒ r_{1}/r_{2 }= 2/3
Squaring both sides, we get
Hence, the ratio between the areas of C_{1} and C_{2} is 4 : 9.
16. The areas of two circles are in the ratio 4 : 9. What is the ratio between their circumferences ?
Solution
Given,
Ratio of areas of two circles = 2 : 3
Let the two circles be C_{1} and C_{2} with radii ‘r_{1}’ and ‘r_{2}’.
∵ Area of circle = Ï€r^{2}
⇒ r_{1}/r_{2} = 2/3
Multiplying and dividing L.H.S. by ‘Ï€’, we get
⇒ Ï€r_{1}/Ï€r_{2 }= 2/3
Multiplying and dividing L.H.S. by ‘2’, we get
⇒ 2Ï€r_{1}/2Ï€r_{2} = 2/3
As Circumference of circle = 2Ï€r
⇒ Circumference of C_{1}/Circumference of C_{2} = 2/3
Hence, the ratio between the circumference of C_{1 }and C_{2} is 2 : 3.
17. A square is inscribed in a circle. Find the ration of the areas of the circle and the square.
Solution
Given,
A square is inscribed in a circle.
Let the radius of circle be ’r’ and the side of the square be ‘x’.
⇒ the length of the diagonal = 2r
∵ Length of side of square = Length of diagonal/√2
∴ Length of side of square = 2r/√2 = √2r
Area of square = side × side = x × r = √2r × √2r = 2r^{2}
Area of circle = Ï€r^{2}
Ratio of areas of circle and square = (Area of circle)/(Area of square) = Ï€r^{2}/2r^{2 }= Ï€/2
Hence, the ratio of areas of circle and square is Ï€ : 2.
18. The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72°.
Solution
Given:
Circumference of circle = 8 cm
Central angle = 72°
∵ Circumference of a circle = 2Ï€r
∴ 2Ï€r = 8
⇒ 2 × 22/7 × r = 8
⇒ r = 8 × 7/22 × 1/2
⇒ r = 14/11 cm
∴ Area of sector = Î¸/360 × Ï€r^{2}
= 72/360 × Ï€ × 14/11 × 14/11
= 1.02 cm^{2}
19. A pendulum swings through an angle of 30° and describes an arc 8.8 cm in length. Find the length of the pendulum.
Solution
Given,
Angle made by the pendulum = 30°
Length of the arc made by the pendulum = 8.8 cm
Then the length of the pendulum is equal to the radius of the sector made by the pendulum.
Let the length of the pendulum be ‘r’.
∵ Length of arc = Î¸/360 × 2Ï€r
∴ We have,
Î¸/360 × 2Ï€r = 8.8
⇒ 30/360 × 2 × 3.14 × r = 8.8
⇒ r = 8.8 × 360/30 × 1/2 × 1/3.14
⇒ r = 16.8 cm
Hence, the length of the pendulum is 16.8 cm.
20. The minute hand of a clock is 15 cm long. Calculate the area swept by it in 20 minutes.
Solution
Given,
Length of minute hand = 15 cm
Here, the length of the minute hand is equal to the radius of the sector formed by the minute hand.
Angle made by the minute hand in 1 minute = 360/60 = 6°
Angle made by the minute hand in 20 minutes = 20 × 6 = 120°
Here, the area swept by the minute hand is equal to the area of the corresponding sector made.
∵ Area of sector = Î¸/360 × Ï€r^{2 }
= 120/360 × 3.14 × 15 × 15
= 235.5 cm^{2 }
Hence, the area swept by it in 20 minutes is 235.5 cm^{2}.
21. A sector of 56°, cut from a circle, contains 17.6 cm^{2}. Find the radius of the circle.
Solution
Given:
Angle of the sector = 56°
Area of the sector = 17.6 cm^{2}
Let the radius of the circle be ‘r’.
∵ Area of sector = Î¸/360 × Ï€r^{2 }
∴ 17.6 = 56/360 × 22/7 × r^{2 }
r^{2} = 360/56 × 7/22 × 17.6
⇒ r^{2} = 36
⇒ r = √36
⇒ r = 6 cm
Hence, the radius of the circle is 6 cm.
22. The area of the sector of a circle of radius 10.5 cm is 69.3 cm^{2}. Find the central angle of the sector.
Solution
Given,
Radius of the circle = 10.5 cm
Area of the sector = 69.3 cm^{2 }
∵ Area of the sector = Î¸/360 × Ï€r^{2 }
∴ 69.3 = Î¸/360× 22/7 × 10.5 × 10.5
⇒ Î¸ = 69.3 × 360 × 7/22 × 1/10.5 × 1/10.5
⇒ Î¸ = 72°
Hence, the central angle is 72°.
23. The perimeter of a certain sector of a circle of radius 6.5 cm is 31 cm. Find the area of sector.
Solution
Given,
Radius of circle = 6.5 cm
Perimeter of sector = 31 cm
Now, Perimeter of sector = 2 × radius + Length of arc
∵ Length of arc = Î¸/360 × 2r × 2Ï€r
∴ Perimeter of sector = 2 × r + Î¸/360 × 2r × Ï€
= 21 × [1 + Î¸/360 × Ï€]
31 = 2 × 6.5 × [1 + Î¸/360 × 22/7]
⇒ 31 = 13 × [1 + Î¸/360 × 22/7]
⇒ 31/13 = 1 + Î¸/360 × 22/7
⇒ 31/13 – 1 = Î¸/360 × 22/7
⇒ 18/13 = Î¸/360 × 22/7
⇒ Î¸ = 18/13 × 360 × 7/22 …(i)
∵ Area of sector = Î¸/360 × Ï€r^{2}
∴ Using (i), we have
Area = 18/13 × 360 × 7/22 × 1/360 × 22/7 × 6.5 × 6.5
= 18 × 3.25
= 58.5 cm^{2}
Hence, the area of the sector is 58.5 cm^{2}.
24. The radius of a circle is 17.5 cm. Find the area of the sector enclosed by two radii and an arc 44 cm in length.
Solution
Given,
Radius of circle = 17.5 cm
Length of arc = 44 cm
∵ Length of arc = Î¸/360 × 2Ï€r
∴ 44 = Î¸/360 × 2 × 22/7 × 17.5
Î¸ = 44 × 360 × 1/2 × 7/22 × 10/17.5
⇒ Î¸ = 2520/17.5 = 144°
Now, Area of sector = Î¸/360 × Ï€r^{2}
= 144/360 × 22/7 × 17.5 × 17.5 = 385 cm^{2 }
Hence, the area of the sector is 385 cm^{2}.
25. Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular cardboard of dimensions 14 cm × 7 cm. Find the area of remaining cardboard.
Solution
Given:
Length of the rectangular cardboard = 14 cm
Breadth of the rectangular cardboard = 7 cm
∵ Area of rectangle = length × breadth
∴ Area of cardboard = 14 × 7 = 98 cm^{2}
Let the two circles with equal radii and maximum area have a radius of ‘r’ cm. each.
Then, 2r = 7
r = 7/2 cm
∵ Area of circle = Ï€r^{2}
∴ Area of two circular cut outs = 2 × Ï€r^{2}
= 2 × 22/7 × 7/2 × 7/2
= 11 × 7
= 77 cm^{2}
Thus, the area of remaining cardboard = 98 – 77
= 21 cm^{2}
Hence, the area of remaining cardboard is 21 cm^{2}.
26. In the given figure, ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is also drawn. Find the area of the shaded region.
SolutionGiven,
Side of the square = 4 cm
Radius of the quadrants at the corners = 1 cm
Radius of the circle in the centre = 1 cm
∵ 4 quadrants = 1 circle
∴ There are 2 circles of radius 1 cm
Area of square = side × side
= 4 × 4
= 16 cm^{2}
Area of 2 circles = 2 × Ï€r^{2}
= 2 × 22/7 × 1 × 1
= 44 cm^{2}
∵ Area of shaded region = Area of square – Area of 2 circles
= 16  44/7
= (112 – 44)/7
= 68/7 cm^{2}
= 9.7 cm^{2}
Hence, the area of shaded region is 9.72 cm^{2}.
27. From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, semicircular portion with BC as diameter is cut off. Find the area of the remaining paper.
Solution
Given,
Length of rectangular sheet of paper = 40 cm
Breadth of rectangular sheet of paper = 28 cm
Radius of the semicircular cut out = 14 cm
∵ Area of rectangle = length × breadth
∴ Area of rectangular sheet of paper = 40 × 28
= 1120 cm^{2}
∵ Area of semicircle = 1/2Ï€r^{2}
∴ Area of semicircular cut out = 1/2 × 22/7 × 14 × 14
= 22 × 14
= 308 cm^{2}
Thus, the area of remaining sheet of paper = Area of rectangular sheet of paper – Area of semicircular cut out
= 1120 – 308
= 812 cm^{2}
Hence, the area of remaining sheet of paper is 812 cm^{2}.
28. In the given figure, OABC is a square of side 7 cm. If COPB is a quadrant of a circle with centre C find the area of the shaded region.
SolutionGiven:
Side of square = 7 cm
Radius of the quadrant = 7 cm
Area of square = side × side
= 7 × 7
= 49 cm2
∵ Area of circle = Ï€r^{2}
∴ Area of a quadrant = 1/4 Ï€r^{2}
= 1/4 × 22/7 × 7 × 7
= 77/2
= 38.5 cm^{2}
Thus, the area of shaded region = Area of square – Area of quadrant
= 49 – 38.5
= 10.5 cm^{2}
Hence, the area of the shaded region is 10.5 cm^{2}.
29. In the given figure, three sectors of a circle of radius 7 cm, making angle of 60°, 80° and 40° at the centre are shaded. Find the area of the shaded region.
SolutionGiven,
Radius of circle = 7 cm
Let the sectors with central angles 80°, 60° and 40° be S_{1}, S_{2}, and S_{3} respectively.
Then, the area of shaded region = Area of S_{1 }+ Area of S_{2} + Area of S_{3} …(i)
∵ Area of sector = Î¸/360 × Ï€r^{2}
∴ Area of S_{1} = 80/360 × 22/7 × 7 × 7
= 308 cm^{2}
Similarly, Area of S_{2} = 60/360 × 22/7 × 7 × 7
= 154/6 cm^{2}
and area of S_{3} = 60/360 × 22/7 × 7 × 7
= 154 cm^{2}
Thus, using (i), we have
Area of shaded region = 308/9 + 154/6 + 154/9
= (616 + 462 + 308)/2
= 1386/18
= 77 cm^{2}
Hence, the area of shaded region is 77 cm^{2}.
30. In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm with centre O. If ∠POQ = 30°, find the area of the shaded region.
Given,
Radius of inner circle = 3.5 cm
Radius of outer circle = 7 cm
∠POQ = 30°
Let the sector made by the arcs PQ and AB be S_{1} and S_{2} respectively.
Then, Area of shaded region = Area of S_{1} – Area of S_{2} …(i)
∵ Area of sector = Î¸/360 × Ï€r^{2}
∴ Area of S_{1} = 30/360 × 22/7 × 7 × 7
= 71/6 cm^{2}
Similarly, Area of S_{2} = 30/360 × 22/7 × 3.5 × 3.5
= 77/24 cm^{2 }
Thus, using (i), we have
Area of shaded region = 77/6 – 77/24
= (308 – 77)/24
= 231/24
= 77/8 cm^{2}
Hence, the area of shaded region is 77/8 cm^{2}.
31. In the given figure, find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircle.
SolutionGiven
Side of square = 14 cm
Diameter of each semicircle = 14 cm
Radius of each semicircle = 14/2 = 7 cm
∵ Both are semicircles have same radius.
∴ We consider one circle of radius 7 cm.
Area of shaded region = Area of square – Area of circle …(i)
Area of square = side × side
= 14 × 14
= 196 cm^{2}
Area of circle = Ï€r^{2 }
= 22/7 × 7 × 7
= 22 × 7
= 154 cm^{2}.
Thus, using (i), we have
Are of shaded region = 196 – 154
= 42 cm^{2 }
Hence, the area of shaded region is 42 cm^{2}.
32. In the given figure, the shape of the top of a table is that of a sector of circle with centre O and ∠AOB = 90°. IF AO = OB = 42 cm, then find the perimeter of the top of the table.
SolutionGiven:
Radius of the circle = 42 cm
Central angle of the sector = ∠AOB = 90°
Perimeter of the top of the table = Length of the major arc AB + 2 × radius ...(i)
Length of major arc AB = (360 – Î¸)/360 × 2Ï€r
= (360 – 90)/360 × 2 × 22/7 × 42
= 270/360 × 2 × 22 × 6
= 3/4 × 264
= 3 × 66
= 198 cm
Thus, using (i), we have
Perimeter of the top of the table = 198 + 2 × 42
= 198 + 84
= 282 cm
Hence, the perimeter of the top of the table is 282 cm.
33. In the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants of circles each of the radius 7 cm. Find the area of shaded region.
SolutionGiven:
Side of square = 7 cm
Radius of each quadrant = 7 cm
Area of square = side × side
= 7 × 7
= 49 cm^{2}
∵ Area of quadrant = 1/4Ï€r^{2}
∴ Area of 2 quadrants = 2 × 1/4 × Ï€r^{2}
= 1/2 × 22/7 × 7 × 7
= 77 cm^{2}
Area of shaded region = Area of 2 quadrants – Area of square
= 77 – 49
= 28 cm^{2}
Hence, the area of shaded region is 28 cm^{2}.
34. In the given figure, OABC is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region.
SolutionGiven,
Radius of Circle = 3.5 cm
OD = 2 cm
∵ Area of Quadrant = 1/4Ï€r^{2}
∴ Area of Quadratic OABC = 1/4 × 22/7 × 3.5 × 3.5
= 9.625 cm^{2}
∵ Area of Triangle = 1/2 × Base × Height
∴ Area of △COD = 1/2 × 3.5 × 2
= 3.5 cm^{2 }
Area of Shaded Region = Area of Quadratic OABC – Area of △COD
= 38.5 – 3.5
= 35 cm^{2}
Hence, the area of shaded region is 35 cm^{2}.
35. Find the perimeter of shaded region in the figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles.
SolutionGiven,
Side of square = 14 cm
Diameter of semi –circle = 14 cm
⇒ Radius of semicircle = 14/2 = 7 cm
∵ There are 2 semi circles of same radius.
∴ We consider it as one circle with radius 7 cm.
So,
Perimeter of 2 semicircles = Perimeter of circle = 2Ï€r
= 2 × 22/7 × 7
= 2 × 22
= 44 cm
Perimeter of shaded region = Perimeter of 2 semicircles + 2×Side of square
= 44 + 2 × 14
= 44 + 28
= 72 cm
Hence, the area of the shaded region is 72 cm.
36. In a circle of radius 7 cm, a square ABCD is inscribed. Find the area of the circle which is outside the square.
Solution
Given,
Radius of the circle = 7 cm
Diameter of the circle = 14 cm
Here, diagonal of square = 14 cm
∵ Side of a square = diagonal/√2
⇒ Side = 14/√2
= 7√2 cm
⇒ Area of square = side × side
= 7√2 × 7√2
= 49 × 2
= 98 cm^{2 }
Area of circle = Ï€r^{2}
= 22/7 × 7 × 7
= 22 × 7
= 154 cm^{2 }
Thus, the area of the circle outside the square = Area of circle – Area of square
= 154 – 98
= 56 cm^{2 }
Hence, the area of the required region is 56 cm^{2}.
37. In the given figure, APB and COD are semicircle of diameter 7 cm each, while ARC and BSD are semicircles of diameter 14 cm each. Find the (i) perimeter, (ii) area of the shaded region.
Solution(i) Given,
Diameter of semicircles APB and CQD = 7 cm
⇒ Radius of semicircles ARC and CQD = 7/2 cm = r_{1}
Diameter of semicircles ARC and BSD = 14 cm
Radius of semicircles ARC and BSD = 14/2 cm = 7 cm = r_{2 }
Perimeter of APB = Perimeter of CQD
Area of APB = Area of CQD …(i)
Perimeter of ARC = Perimeter of BSD
Area of ARC = Area of BSD …(ii)
∵ Perimeter of semicircle = Ï€r …(iii)
∴ Perimeter of APB = Ï€r_{1 }
= 22/7 × 7/2
= 11 cm
Then, using (i), we have
Perimeter of CQD = 11 cm
Now, using (iii), we have
Perimeter of ARC = Ï€r_{2}
= 22/7 × 7
= 22 cm
Then, using (ii), we have
Perimeter of BSD = 22 cm
Perimeter of shaded region = (Perimeter of ARC + Perimeter of APB) + (Perimeter of BSD + Perimeter of CQD)
= (22 + 11) + (22 + 11)
= 33 + 33
= 66 cm
Hence, the perimeter of the shaded region is 66 cm.
(ii) Now,
∵ Area of semicircle = 1/2Ï€r^{2} …(iv)
= 1/2 × 22/7 × 7 × 7
= 11 × 7
= 77 cm^{2}
Then, by using (ii), we have
Area of BSD = 77 cm^{2}
Area of shaded region = (Area of ARC – Area of APB) + (Area f BSD  Area of CQD)
= (77 – 77/4) + (77 – 77/4)
= (308 – 77)/4 + (308 – 77)/4
= 231/4 + 231/4
= 462/4
= 115.5 cm^{2 }
Hence, the area of the shaded region is 115.5 cm^{2}.
38. In the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively. Find the perimeter of shaded region.
Solution
Given,
Diameter of semicircle PSR = 10 cm
⇒ Radius of semicircle PSR = 10/2 = 5 cm = r_{1}
Diameter of semicircle RTQ = 3 cm
⇒ Radius of semicircle RTQ = 3/2 = 1.5 cm = r_{2}
Diameter of semicircle PAQ = 7 cm
⇒ Radius of semicircle PAQ = 7/2 = 3.5 cm = r_{3}
∵ Perimeter of semicircle = Ï€r
∴ Perimeter of semicircle PSR = Ï€r_{1}
= 3.14 × 5
= 15.7 cm
Similarly, Perimeter of semicircle RTQ = Ï€r_{2 }
= 3.14 × 1.5
= 4.71 cm
and Perimeter of semicircle PAQ = Ï€r_{3}
= 3.14 × 3.5
= 10.99 cm
Perimeter of shaded region = Perimeter of semicircle PSR + Perimeter of semicircle RTQ + Perimeter of semicircle PAQ
= 15.7 + 4.71 + 10.99
= 31.4 cm
Hence, the perimeter of the shaded region is 3.14 cm.
39. In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle. IF OA = 20 cm, find the area of the shaded region. [Use Ï€ = 3.14]
SolutionGiven,
OA = Side of square OABC = 20 cm
∵ Area of square = Side × Side
∴ Area of square OABC = 20 × 20 = 400 cm^{2}
Now,
∵ Length of diagonal of square = √2 × Side of square
∴ Length of diagonal of square OABC = √2 × 20
= 20√2 cm
⇒ Radius of the quadrant = 20√2 cm
∵ Area of quadrant = 1/4 Ï€r^{2 }
∴ Area of quadrant OPBQ = 1/4 × 3.14 × 20√2 × 20√2
= 3.14/4 × 400 × 2
= 3.14 × 200
= 628 cm^{2 }
Area of shaded region = Area of quadrant OPBQ – Area of square OABC
= 628 – 400
= 228 cm^{2 }
Hence, the area of the shaded region is 228 cm^{2}
40. In the given figure, APB and AQO are semicircles and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region.
SolutionGiven,
AO = OB
Perimeter of the figure = 40 cm …(i)
Let the diameters of semicircles AQO and APB be ‘x_{1}’ and ‘x_{2}’ respectively.
Then, using (1), we have
AO = OB
Also, AB = AO + OB
= AO + AO
= 2AO
⇒ x_{2} = 2x_{1}
So, diameter of APB = 2x_{1}
and diameter of AOQ = x_{1}
Radius of APB = x_{1}
and Radius of AOQ = x_{1}/2 …(ii)
Perimeter of shaded region = Perimeter of AQO + perimeter APB + diameter of APB …(iii)
∵ Perimeter of semicircle = Ï€r
∴ Perimeter of semicircle AOQ = 22/7 × x_{1}/2
= 11x_{1}/7 cm
Perimeter of semicircle APB = 22/7 × x_{1} = 22x_{1}/7 cm
Now, using (iii), we have
40 = 11x_{1}/7 + 22x_{1}/7 + x_{1}
⇒ 40 = (11x_{1} + 22x_{1} + 7x_{1})/7
⇒ 40 × 7 = 40x_{1}
⇒ 280 = 40x_{1}
⇒ x_{1 }= 280/40 = 7 cm
∴ Using (ii), we have
Radius of APB = 7 cm = r_{1}
And Radius of AQO = 7/2 cm
= 3.5 cm
= r_{2}
Now,
∵ Area of semicircle = 1/2Ï€r^{2}
Thus, Area of shaded region = Area of APB + Area AQO
= 77 + 19.25
= 96.25 cm^{2}
Hence, the area of the shaded region is 96.25 cm^{2}.
41. Find the area of a quadrant of a circle whose circumference is 44 cm.
Solution
Given,
Circumference of circle = 44 cm
Let the radius of the circle be ’r’ cm
∵ Circumference of circle = 2Ï€r
∴ 44 = 2Ï€r
⇒ 44/2 = 22/7 × r
⇒ r = 22 × 7/22 = 7 cm
Now, Area of quadrant = 1/4 × Ï€r^{2}
= 1/4 × 22/7 × 7 × 7
= (11 × 7)/2
= 22/7
= 38.5 cm^{2}
Hence, the area of the quadrant is 38.5 cm^{2}.
42. In the given figure, find the area of the shaded region, where ABCD is a square of side 14 cm and all circle are of the same diameter.
SolutionGiven:
Side of square = 14 cm
Let the radius of each circle be ‘r’ cm
Then, 2r + 2r = 14 cm
4r = 14 cm
⇒ r = 14/4
⇒ r = 7/2
Area of square = side × side
= 14 × 14
= 196 cm^{2}
∵ Area of circle = Ï€r^{2}
∴ Area of 4 circles = 4 × Ï€r^{2}
= 4 × 22/7 × 7/2 × 7/2
= 22 × 7
= 154 cm^{2}
Area of shaded region = Area of the square – Area of 4 circles
= 196 – 154
= 42 cm^{2}
Hence, the area of the shaded region is 42 cm^{2}.
43. Find the area of the shaded region in the given figure, if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of the circle.
SolutionGiven,
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 = 78.7 – 48
= 30.57 cm^{2}
Hence, the area of shaded region is 30.57 cm^{2}
44. A wire is bent to from a square enclosing an area of 484 m^{2}. Using the same wire, a circle is formed. Find the area of the circle.
Solution
Given,
Perimeter of square = Circumference of circle …(i)
Area of square = 484 m^{2}
Let the side of square be ‘x’ cm.
∵ Area of Square = side × side
∴ 484 = x × x
⇒ x^{2} = 484
⇒ x = √484 = 22 cm
∵ Perimeter of square = 4 × side
= 4 × 22
= 88 cm
∴ Using (i), we have
Circumference of circle = 88 cm
Also, Circumference of Circle = 2Ï€r
2Ï€r = 88
⇒ 2 × 22/7 × r = 88
⇒ r = 88 × 1/2 × 7/22
⇒ r = 2 × 7 = 14 cm
Area of Circle = Ï€r^{2} = 22/7 × 14 × 14
= 22 × 2 × 14
= 616 cm^{2}
Hence, the area of Circle is 616 cm^{2}.
45. A square ABCD is inscribed in a circle of radius ‘r’. Find the area of the square.
Solution
Given,
Radius of circle = r
Diagonal of Square = 2r
∵ Side of square = (length of diagonal)/√2
∴ Side = 2r/√2 = √2r
Area of square = Side × Side
= √2r × √2r
= 2r^{2}
Hence, the area of square is ‘2r^{2}’ square units.
46. The cost of fencing a circular field at the rate of Rs 25 per meter is Rs 5500. The field is to be ploughed at the rate of 50 paise per m^{2}. Find the cost of ploughing the field. [Take Ï€ = 22/7]
Solution
Given,
Rate of fencing a circular field = Rs 25/m
Cost of fencing a circular field = RS 5500
Rate of ploughing the field = 50p/m^{2 }= RS 0.5 m^{2}
Let the radius of circular field be ‘r’ and the length of the field fenced be ‘x’ m.
Then, 25 × x = 5500
x = 5500/225
= 220 m
∵ Circumference of circular field = 2 Ï€r
∴ 220 = 2Ï€r
220 = 2 × 22/7 × r
⇒ r = (220 × 7)/(2 × 22)
⇒ r = 35 m
Area of the circular field = Ï€r^{2}
= 22/7 × 35 × 35
= 22 × 5 × 35
= 3850 m^{2}
Now, cost of ploughing the field = Rate of ploughing the field × Area of the field
= 0.5 × 3850
= Rs 1925
Hence, the cost of Ploughing the field is Rs 1925.
47. A park is in the form of a rectangle 120 m by 90 m. At the centre of the park, there is a circular lawn as shown in the figure. The area of the park excluding the lawn is 2950 m^{2}. Find the radius of the circular lawn. [Given, Ï€ = 3.14]
SolutionGiven,
Length of the rectangular park = 120 m
Breadth of the rectangular park = 90 m
Area of the park excluding the circular lawn = 2950 m^{2}
Area of rectangular park = length × breadth
= 120 × 90
= 10800 m^{2}
Area of circular lawn = Area of rectangular park – Area of park excluding the lawn
= 10800 – 2950
= 7850 m^{2 }
∵ Area of circle = Ï€r^{2}
∴ 7850 = 3.14 × r^{2}
r^{2} = 7850/3.14 = 2500
⇒ r = √2500 = 50 m
Hence, the radius of the circular lawn is 50 m.
48. In the given figure PQRS represents a flower be. If OP = 21 m and OR = 14 m, find the area of the flower bed.
Solution
Given,
OP = 21 m = r_{1}
OR = 14 m = r_{2 }
Let the quadrants made by outer and inner circles be Q_{1} and Q_{2}, with radius r_{1} and r_{2} respectively.
Then, Area of flower bed = Area of Q_{1} – Area of Q_{2}
∵ Area of Quadrant = 1/4Ï€r^{2}
Thus, Area of flower bed = 693/2 – 308/2
= 385/2
= 192.5 m^{2}
Hence, the area of the flower bed is 192.5 m^{2}.
49. In the given figure, O is the centre of the bigger circle, and AC is its diameter. Another circle with AB as diameter is drawn. If AC = 54 cm and BC = 10 cm, find the area of the shaded region.
SolutionGiven,
AC = 54 cm
BC = 10 cm
⇒ AB = AC – BC = 54 – 10
= 44 cm
Radius of bigger circle = AC/2 = 54/2 = 27 cm = r_{1}
Radius of Smaller circle = AB/2 = 44/2 = 22 cm = r_{2}
∵ Area of circle = Ï€r^{2}
Area of shaded region = Area of Bigger Circle – Area of smaller circle = 16038/7 – 10648/7
= 5390/7
= 770 cm^{2}
Hence, area of shaded region is 770 cm^{2}.
50. From a thin metallic piece in the shape of a trapezium ABCD in which AB CD and ∠BCD = 90°, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet.
SolutionGiven,
AB  CD
∠BCD = 90°,
AB = BC = 3.5 cm = EC
DE = 2 cm
DC = DE + EC = 2 + 3.5 = 5.5 cm
Area of Trapezium = 1/2 × Sum of Parallel Sides × h
= 1/2 × (AB + DC) × BC
= 1/2 × (3.5 + 5.5) × 3.5
= 1/2 × 9 × 3.5
= 15.75 cm^{2}
Area of quadrant BEFC = 1/4 × Ï€r^{2}
= 1/4 × 22/7 × 3.5 × 3.5
= 9.625 cm^{2}
Thus, Area of remaining part of metal sheet = Area of Trapezium – Area of Quadrant BEFC
= 15.75 – 9.625
= 6.125 cm^{2 }
Hence, the area of the remaining part of metal sheet is 6.125.
51. Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°, as shown in the given figure.
SolutionGiven,
Radius of Circle = 35 cm
∠AOB = 90°
∵ Area of Sector = Î¸/360 × Ï€r^{2 }
= 90/360 × 22/7 × 35 × 35
= 1925/2 cm^{2 }
∵ △AOB is rightangled triangle.
∴ Area of △AOB = 1/2 × OA × OB
= 1/2 × 35 × 35
= 1225/2 cm^{2 }
Now, Area of Minor Segment ACB = Area of sector – Area of △AOB
= 1925/2 – 1225/2
= 700/2
= 350 cm^{2 }
Area of circle = Ï€r^{2 }
= 22/7 × 35 × 35
= 22 × 5 × 35
= 3850 cm^{2}
Thus, Area of Major Segment = Area of Circle – Area of Minor segment
= 3850 – 350
= 3500 cm^{2 }
Hence, the area of the major segment is 3500 cm^{2}.