Q.1. The difference between the circumference and radius of a circle is 37 cm. Using π = 22/7, find the circumference of the circle.
Given:
Difference between the circumference and the radius of circle = 37 cm
Let the radius of the circle be ‘r’.
Circumference of the circle = 2πr
So, Difference between the circumference and the radius of the circle = 2πr – r = 37
2πr – r = 37
2 × (22/7) × r – r = 37
(44/7) × r – r = 37
r = 37 × (7/37)
r = 7 cm
∴ Circumference of circle = 2 × (22/7) × 7
= 2 × 22
= 44 cm
Hence the circumference of the circle is 44 cm.
Q.2. The circumference of a circle is 22 cm. Find the area of its quadrant.
Given:
Circumference of circle = 22 cm
Let the radius of the circle be ‘r’.
∵ Circumference of circle = 2πr
∴ 22 = 2 × π × r
⇒ 22 = 2 × (22/7) × r
⇒ 22 × (7/22) × (1/2) = r or = r
or r = (7/2)
∵ Area of circle = πr^{2}
∴ Area of its quadrant = (1/4)πr^{2}
=
= 77/8
Hence the area of the quadrant of the circle is 77/8 cm.
Q.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?
Given:
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 of C_{1}, r_{1} = 10/2 = 5cm
and Radius of C_{2}, r_{2} = 24/2 = 12cm
∵ Area of circle = πr^{2} …… (ii)
∴ Area of C_{1 }= πr_{1}^{2}
= (22/7) 5 x 5
= (22/7) x 25
= 550/7 cm^{2}
Similarly, Area of C_{2 }= πr_{2}^{2}
=
= 22/7 × 144
= 3168/7 cm^{2}
∴ Using equation (i), we have
Area of C = +
= cm^{2}
Now, using equation (ii), we have
× r^{2} =
r^{2} =
r^{2} = 169
r =
r = 13 cm
⇒ Diameter = 2 × r
= 2 × 13
= 26 cm
Hence, the diameter of the circle is 26 cm.
Q.4. If the area of a circle is numerically equal to twice its circumference, then what is the diameter of the circle?
Given:
Area of circle = 2 × Circumference of circle …… (i)
Let the radius of the circle be ‘r’.
Then, the area of the circle = πr^{2}
and the circumference of the circle = 2πr
Using (i), we have
πr^{2 }= 2 × 2πr
πr^{2} = 4πr
r = 4 cm
∵ Diameter = 2 × radius
∴ Diameter = 2 × 4
= 8 cm
Hence, the diameter of the circle is 8 cm.
Q.5. What is the perimeter of a square which circumscribes a circle of radius a cm?
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.
Q.6. Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60° at the centre.
Given:
Diameter of circle = 42 cm
⇒ Radius of circle = 42/2 cm = 21 cm
Angle subtended at the centre = 60°
∵ Length of arc = × 2πr
=
= 22 cm
Hence, the length of the arc is 22 cm.
Q.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.
Given:
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 = πr^{2} …… (ii)
∴ Area of C_{1} = πr_{1}^{2}=
= × 16 = cm^{2}
Similarly, Area of C_{2} = πr_{2}^{2}
= × 3 × 3
= × 9 = cm^{2}
So, using (i), we have
Area of C = + = cm^{2}
Now, using (ii), we have
πr^{2} = 550/7
× r^{2} =
r^{2} = × = 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 3cm is 10 cm.
Q.8. Find the area of a circle whose circumference is 8π.
Given:
Circumference of circle = 8π
∵ Circumference of a circle = 2πr
∴ 8π = 2πr
r = 4
∵ Area of circle = πr^{2}
∴ Area of circle = π × 4 × 4
= 16π
Hence, the area of the circle is 16π.
Q.9. Find the perimeter of a semicircular protractor whose diameter is 14 cm.
Given:
Diameter of the semicircular protractor = 14 cm
Radius of the protractor = 14/2 cm = 7cm
∵ Perimeter of semicircle = πr + d
∴ Perimeter of semicircular protractor = (22/7)× 7 + 14 = 22 + 14
= 36 cm
Hence, the perimeter of the semicircular protractor is 36 cm.
Q.10. Find the radius of a circle whose perimeter and area are numerically equal.
Given:
Perimeter of circle = Area of circle …… (i)
∵ Perimeter of circle = 2πr and Area of circle = πr^{2}
∴ Using (i), we have
2πr = πr^{2}
2 = πr^{2}/2πr
2 = r or r = 2
Hence, the radius of the circle is 2 cm.
Q.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.
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 = 2πr
∴ Circumference of C_{1} = 2πr_{1} = 2 × × 19 =
and Circumference of C_{2} = 2πr_{2} = 2 × × 9 =
Using (i), we have
2πr = + =
2 × × r =
r = × × = 28
r = 28 cm
Hence, the radius of the circle is 28 cm.
Q.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.
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 = πr^{2}
∴ Area of C_{1} = πr_{1}^{2} = × 8 × 8 =
and Area of C_{2} = πr_{2}^{2} = × 6 × 6 =
Using (i), we have
πr^{2} = + =
× r^{2} =
r^{2} = × = 100
r^{2} = 100
r = √100 = 10 or r = 10
Hence, the radius of the circle is 10 cm.
Q.13. Find the area of the sector of a circle having radius 6 cm and of angle 30°.
Given:
Radius of circle = 6 cm
Angle of the sector = 30°
∵ Area of sector = × πr^{2}
= × 3.14 × 6 × 6
= 3 × 3.14 = 9.42 cm^{2}
Hence, the area of the sector is 9.42 cm^{2}.
Q.14. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.
Given:
Radius of circle = 21 cm
Angle subtended by the arc = 60°
∵ Length of arc = × 2πr
= × 2 × × 21 = 22 cm
Hence, the length of the arc is 22 cm.
Q.15. The circumferences of two circles are in the ratio 2:3. What is the ratio between their areas?
Given:
Ratio of circumferences of two circles = 2:3
Let the two circles be C1 and C2 with radii ‘r_{1}’ and ‘r_{2}’.
∵ Circumference of circle = 2πr
∴ Circumference of C_{1} = 2πr_{1}
and Circumference of C_{2} = 2πr_{2}
⇒ =
⇒ =
Squaring both sides, we get
⇒ =
Multiplying both sides by ‘π’, we get
⇒ =
∵ Area of circle = πr^{2}
⇒ =
Hence, the ratio between the areas of C_{1} and C_{2} is 4:9.
Q.16. The areas of two circles are in the ratio 4:9. What is the ratio between their circumferences?
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 = πr^{2}
∴ Area of C_{1} = πr_{1}^{2}
and Area of C_{2 }= πr_{2}^{2}
⇒ =
⇒ =
Taking square root on both sides, we get
Multiplying and dividing L.H.S. by ‘π’, we get
Multiplying and dividing L.H.S. by ‘2’, we get
⇒ =
As Circumference of circle = 2πr
⇒ =
Hence, the ratio between the circumferences of C_{1} and C_{2} is 2:3.
Q.17. A square is inscribed in a circle. Find the ratio of the areas of the circle and the square.
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 side of square = = √2r
Area of square = side × side = x × x = √2r × √2r = 2r^{2}
Area of circle = πr^{2}
Ratio of areas of circle and square = = =
Hence, the ratio of areas of circle and square is π:2.
Q.18. The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72°.
Given:
Circumference of circle = 8 cm
Central angle = 72°
∵ Circumference of a circle = 2πr
∴ 2πr = 8
2 × × r = 8
r = 8 × ×
r = cm
∵ Area of sector = × πr^{2}
= 1.02 cm^{2}
Q.19. A pendulum swings through an angle of 30° and describes an arc 8.8 cm in length. Find the length of the pendulum.
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 = × 2πr
∴ We have,
× 2πr = 8.8
(30/360) × 2 × 3.14 × r = 8.8
r = 8.8 ×
r = 16.8 cm
Hence, the length of the pendulum is 16.8 cm.
Q.20. The minute hand of a clock is 15 cm long. Calculate the area swept by it in 20 minutes.
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 = (θ/360) x πr^{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}.
Q.21. A sector of 56°, cut out from a circle, contains 17.6 cm^{2}. Find the radius of the circle.
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 = × πr^{2}
∴ 17.6 = × × r^{2}
r^{2} = × × 17.6
r^{2} = 36
r = √36
r = 6 cm
Hence, the radius of the circle is 6 cm.
Q.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.
Given:
Radius of the circle = 10.5 cm
Area of the sector = 69.3 cm^{2}
∵ Area of the sector = × πr^{2}
∴ 69.3 = × × 10.5 × 10.5
θ = 69.3 × 360 × × ×
θ = 72°
Hence, the central angle is 72°.
Q.23. The perimeter of a certain sector of a circle of radius 6.5 cm is 31 cm. Find the area of sector.
Given:
Radius of circle = 6.5 cm
Perimeter of sector = 31 cm
Now, Perimeter of sector = 2 × radius + Length of arc
∵ Length of arc = × 2r × 2πr
∴ Perimeter of sector = 2 × r + × 2r × π
= 2r × [1 + × π]
31 = 2 × 6.5 × [1 + × ]
31 = 13 × [1 + × ]
= 1 + ×
 1 = ×
= ×
θ = × 360 × …. (i)
∵ Area of sector = × πr^{2}
∴ using (i), we have
Area = × 360 × × × × 6.5 × 6.5
= 18 × 3.25 = 58.5 cm^{2}
Hence, the area of the sector is 58.5 cm^{2}.
Q.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.
Given:
Radius of circle = 17.5 cm
Length of arc = 44 cm
∵ Length of arc = × 2πr
∴ 44 = × 2 × × 17.5
θ = 44 × 360 × × ×
θ = = 144°
Now, Area of sector = × πr^{2}
= × × 17.5 × 17.5 = 385 cm^{2}
Hence, the area of the sector is 385 cm^{2}.
Q.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 the remaining cardboard.
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 = πr^{2}
∴ Area of two circular cut outs = 2 × πr^{2}
= 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}.
Q.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.
Given:
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 × πr2
= 2 × × 1 × 1 = cm^{2}
∵ Area of shaded region = Area of square – Area of 2 circles
= 16  (44/7)
= = cm^{2} = 9.7 cm^{2}
Hence, the area of shaded region is 9.72 cm^{2}.
Q.27. From a rectangular sheet of paper ABCD wit AB = cm and AD = 28 cm, a semicircular portion wit BC as diameter is cut off. Find the area of the remaining paper.
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 = (1/2)πr^{2}
∴ Area of semicircular cut out = × × 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}.
Q.28. In the given figure, OABC is a square of side 7 cm. If COPB is a quadrant of a circle wit centre C find the area of the shaded region.
Given:
Side of square = 7 cm
Radius of the quadrant = 7 cm
Area of square = side × side
= 7 × 7 = 49 cm^{2}
∵ Area of circle = πr^{2}
∴ Area of a quadrant = (1/4) πr^{2}
= × × 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}.
Q.29. In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80° and 40° at the centre are shaded. Find the area of the shaded region.
Given:
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 = (θ/360) × πr^{2}
∴ Area of S_{1} = × × 7 × 7
= (308/9) cm^{2}
Similarly, Area of S_{2} = × × 7 × 7
= 154/6 cm^{2}
and Area of S_{3} = × × 7 × 7
= 154/9 cm^{2}
Thus, using (i), we have
Area of shaded region = + +
=
= 1386/18 = 77cm^{2}
Hence, the area of shaded region is 77 cm^{2}.
Q.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 = (θ/360) x πr^{2}
∴ Area of S_{1} = × × 7 × 7
= 71/6 cm^{2}
Similarly, Area of S_{2} = × × 3.5 × 3.5
= 77/24 cm^{2}
Thus, using (i), we have
Area of shaded region = 
=
= = cm^{2}
Hence, the area of shaded region is 77/8 cm^{2}.
Q.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.
Given:
Side of square = 14 cm
Diameter of each semicircle = 14 cm
Radius of each semicircle = 14/2 = 7 cm
∵ Both the 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 = πr^{2}
= 22/7 × 7 × 7 = 22 × 7 = 154 cm^{2}
Thus, using (i), we have
Area of shaded region = 196 – 154 = 42 cm^{2}
Hence, the area of shaded region is 42 cm^{2}.
Q.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.
Give:
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 = × 2πr
= × 2 × × 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.
Q.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.
Given:
Side of square = 7 cm
Radius of each quadrant = 7 cm
Area of square = side × side = 7 × 7 = 49 cm^{2}
∵ Area of quadrant = 1/4 πr^{2}
∴ Area of 2 quadrants = 2 × 1/4 × πr^{2}
= × × 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}.
Q.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.
Given:
Radius of Circle = 3.5 cm
OD = 2 cm
∵ Area of Quadrant = 1/4 πr^{2}
∴ Area of Quadrant OABC = 1/4 × 22/7 × 3.5 × 3.5
= 9.625 cm^{2}
∵ Area of Triangle = 1/2 × Base × Height
∴ Area of Δ COD = 1/2 × 3.5 × 2
= 3.5 cm^{2}
Area of Shaded Region = Area of Quadrant OABC – Area of ΔCOD
= 38.5 – 3.5 = 35 cm^{2}
Hence, the area of shaded region is 35 cm^{2}.
Q.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.
Given:
Side of square = 14 cm
Diameter of semi circle = 14 cm
⇒ Radius of semi circle = 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 = 2πr
= 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.
Q.36. In a circle of radius 7 cm, a square ABCD is inscribed. Find the area of the circle which is outside the square.
Given:
Radius of the circle = 7 cm
Diameter of the circle = 14 cm
Here, diagonal of square = 14 cm
∵ Side of a square =
⇒ Side = 14/√2 = 7√2 cm
⇒ Area of square = side × side
= 7√2 × 7√2
= 49 × 2 = 98 cm^{2}
Area of circle = πr^{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}.
Q.37. In the given figure, APB and CQD 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.
(i) Given:
Diameter of semicircles APB and CQD = 7 cm
⇒ Radius of semicircles APB and CQD = 7/2cm = 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 = πr …………… (iii)
∴ Perimeter of APB = πr_{1}
= × = 11 cm
Then, using (i), we have
Perimeter of CQD = 11 cm
Now, using (iii), we have
Perimeter of ARC = πr_{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 = (1/2) πr^{2} …. (iv)
∴ Area of APB = (1/2) πr_{1}^{2}
= × × × = cm^{2}
Then, using (i), we have
Area of CQD = 77/4 cm^{2}
Now, using (iv), we have
Area of ARC = 1/2 πr_{2}^{2}
= × × 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 ARCArea of APB) + (Area of BSD Area of CQD)
= (77  ) + (77  )
= () + () = + = = 115.5 cm^{2}
Hence, the area of the shaded region is 115.5 cm^{2}.
Q.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.
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 = πr
∴ Perimeter of semicircle PSR = πr_{1}
= 3.14 × 5 = 15.7 cm
Similarly, Perimeter of semicircle RTQ = πr_{2}
= 3.14 × 1.5 = 4.71 cm
and Perimeter of semicircle PAQ = πr_{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 31.4 cm.
Q.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 π = 3.14]
Given:
OA = Side of square OABC = 20 cm
∵ Area of square = Side × Side
∴ Area of square OABC = 20 × 20 = 400 cm2
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 = 1/4πr^{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}.
Q.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.
Given:
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 AQO = x_{1}
Radius of APB = x_{1}
and Radius of AQO =x_{1}/2 ….. (ii)
Perimeter of shaded region = perimeter of AQO + perimeter APB + diameter of APB …… (iii)
∵ Perimeter of semicircle = πr
∴ Perimeter of semicircle AQO = × = cm
Perimeter of semicircle APB = × x_{1} = cm
Now, using (iii), we have
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 = 1/2 πr^{2}
∴ Area of semicircle APB = 1/2 πr_{1}^{2}
Similarly,
Area of semicircle APB = 1/2 πr_{2}^{2}
Thus, Area of shaded region = Area of APB + Area of AQO
= 77 + 19.25 = 96.25 cm^{2}
Hence, the area of the shaded region is 96.25 cm^{2}.
Q.41. Find the area of a quadrant of a circle whose circumference is 44 cm.
Given:
Circumference of circle = 44 cm
Let the radius of the circle be ‘r’ cm
∵ Circumference of circle = 2πr
∴ 44 = 2πr
= × r
r = 22 × (7/22) = 7 cm
Now, Area of quadrant = 1/4 × πr^{2}
Hence, the area of the quadrant is 38.5 cm^{2}.
Q.42. In the given figure, find the area of the shaded region, where ABCD is a square of side 14 cm and all circles are of the same diameter.
Given:
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
= 7/2
Area of square = side × side
= 14 × 14
= 196 cm^{2}
∵ Area of circle = πr^{2}
∴ Area of 4 circles = 4 × πr^{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}.
Q.43. Find the area of the shaded region in the given figure, if ABCD is a rectangle wit sides 8 cm and 6 cm ad O is the centre of the circle.
Given:
Length of rectangle = 8 cm
Breadth of rectangle = 6 cm
Area of rectangle = length × breadth
= 8 × 6 = 48 cm^{2}
Consider Δ 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 = πr^{2}
= × 5 × 5 = = = 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}.
Q.44. A wire is bent to form a square enclosing an area of 484 m^{2}. Using the same wire, a circle is formed. Find the area of the circle.
Given:
Perimeter of square = Circumference of circle ……. (i)
Area of Square = 484m^{2}
Let the side of square be ‘x’ cm.
∵ Area of Square = side × side
∴ 484 = x × x
x^{2} = 484
x = √484 = 22cm
∵ Perimeter of square = 4 × side
= 4 × 22 = 88 cm
∴ Using (i), we have
Circumference of circle = 88 cm
Also, Circumference of Circle = 2πr
2πr = 88
2 × 22/7 × r = 88
r = 2 × 7 = 14 cm
Area of Circle = πr^{2} = 22/7 × 14 × 14
= 22 × 2 × 14 = 616 cm^{2}
Hence, the area of Circle is 616 cm^{2}.
Q.45. A square ABCD is inscribed in a circle of radius ‘r’. Find the area of the square.
Given: Radius of circle = r
Diagonal of Square = 2r
∵ Side of Square =
∴ Side = = √2r
Area of Square = Side × Side
= √2r × √2r
= 2r^{2}
Hence, the area of square is ‘2r^{2}’ square units.
Q.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 π = 22/7]
Given:
Rate of fencing a circular field = Rs. 25/m
Cost of fencing a circular field = Rs. 5500
Rate of ploughing the field = 50p/m2 = 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 = 5500/25 = 220 m
∵ Circumference of circular field = 2πr
∴ 220 = 2πr
220 = 2 × 22/7 × r
r =
r = 35 m
Area of the circular field = πr^{2}
= 22/7 × 35 × 35
= 22 × 5 × 35
= 3850m^{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.
Q.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 m2. Find the radius of the circular lawn. [Given, π = 3.14]
Given:
Length of the rectangular park = 120 m
Breadth of the rectangular park = 90 m
Area of the park excluding the circular lawn = 2950m^{2}
Area of the rectangular park = length × breadth
= 120 × 90 = 10800m^{2}
Area of circular lawn = Area of rectangular park – Area of park excluding the lawn
= 10800 – 2950
= 7850m^{2}
∵ Area of circle = πr^{2}
∴ 7850 = 3.14 × r^{2}
r^{2 }= 7850/3.14 = 2500
r = √2500 = 50 m
Hence, the radius of the circular lawn is 50m.
Q.48. In the given figure PQSR represents a flower be. If OP = 21 m and OR = 14 m, find the area of the flower bed.
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 = 1/4 πr^{2}
∴ Area of Q1 = 1/4 πr_{1}^{2}
= 1/4 × 22/7 × 21 × 21= 693/2 m^{2}
Similarly, Area of Q_{2} = 1/4 πr_{2}^{2}
= 1/4 × 22/7 × 14 × 14
= 308/2 m^{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}.
Q.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.
Given:
AC = 54 cm
BC = 10 cm
⇒ AB = ACBC = 5410 = 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 = πr^{2}
∴ Area of Bigger Circle = πr_{1}^{2}
= (22/7) × 27 × 27
= 16038/7 cm^{2}
Similarly, Area of Smaller Circle = πr_{2}^{2}
= (22/7) × 22 × 22
= 10648/7 cm^{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}.
Q.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.
Given:
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 BFEC = 1/4 × πr^{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 BFEC
= 15.75 – 9.625 = 6.125 cm^{2}
Hence, the area of the remaining part of metal sheet is 6.125 cm^{2}.
Q.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.
Given:
Radius of Circle = 35 cm
∠AOB = 90°
∵ Area of Sector = × πr^{2}
= × × 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 = πr^{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}.
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