Chapter 18 Areas of Circle, Sector & Segment RS Aggarwal Solutions Exercise 18B 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 Exercise 18B Solutions
1. The circumference of a circle is 39.6 cm. Find its area.
Solution
In order to solve such type of questions we basically need to find the radius of the give circle and simply use it to find the area of the given circle.
Given the circumference of perimeter of the circle = 39.6 cm.
And we know, Perimeter or circumference of circle = 2Ï€r
Where, r = Radius of the circle
Therefore, 2Ï€r = 39.6
⇒ r = 39.6/2Ï€
(Put value of Ï€ = 22/7)
⇒ r = 39.6/(2 × 22/7)
On rearranging we get,
⇒ r = (39.6 × 7)/(2 × 22)
⇒ r = 277.2/44
⇒ r = 6.3 cm
So, the radius of the circle = 6.3 cm
And we also know, Area of the circle = Ï€r^{2}
Where, r = radius of the circle
⇒ Area of the circle = Ï€(6.3)^{2}
(putting value of r)
= 22/7(6.3^{2})
= 22/7 (6.3 × 6.3)
= 22/7 × 39.69
= 22 × 5.67
= 124.74 cm^{2}
The area of the circle = 124.74 cm^{2}.
2. The area of the circle is 98.56 cm^{2}. Find its circumference.
Solution
In order to solve such type of questions we basically need to find the radius of the give circle and simply use it to find the circumference or perimeter of the given circle.
Given the area of the circle = 98.56 cm^{2}
And we know, Area of the circle = Ï€r^{2}
Therefore, Ï€r^{2} = 98.56
⇒ r^{2} = 98.56/Ï€
(put value of Ï€ = 22/7)
⇒ r^{2} = 98.56/(22/7)
On rearranging we get,
⇒ r^{2} = (98.56 × 7)/22
⇒ r^{2} = (689.22)/22
⇒ r^{2} = 31.36
⇒ r = √31.36
⇒ r = 5.6 cm
So, the radius of the circle = 5.6 cm
And we know, Perimeter of circle = 2Ï€r
(put value of r)
⇒ Circumference or Perimeter of circle = 2Ï€(5.6)
= 2 × 22/7 × 5.6 (put Ï€ = 22/7)
= (2 × 22 × 5.6)/7
= (246.4)/7
= 35.2 cm
The circumference or perimeter of the circle is 35.2 cm.
3. The circumference of a circle exceeds its diameter by 45 cm. Find the circumference of the circle.
Solution
Given, the circumference of a circle exceeds its diameter by 45 cm.
⇒ Circumference of circle = Diameter of circle + 45
Let ‘d’ = diameter of the circle
⇒ Circumference = d + 45 ...(i)
And we know, Circumference of circle = 2 Ï€r ...(ii)
Where r = radius of circle
Also, we know that the radius of the circle is half of its diameter.
⇒ r = d/2 ...(iii)
Put value of circumference in equation 1 from equation 2
⇒ 2Ï€r = d + 45 ...(iv)
Put value of r in equation 4 from equation 3
⇒ 2Ï€(d/2) = d + 45
⇒ Ï€d = d + 45
⇒ Ï€d – d = 45
⇒ (Ï€ – 1)d = 45 (taking d common from L.H.S)
⇒ d = 45/(Ï€ – 1) (now put Ï€ = 22/7)
⇒ d = 45/(22/7 – 1)
⇒ d = {45/(22 – 7)/7} (taking 7 as LCM in denominator)
⇒ d = 45/(15/7)
On rearranging, we get
⇒ d = (45 × 7)/15
⇒ d = 315/15
⇒ d = 21 cm
Therefore, the diameter of the circle is 21 cm.
Thus, the radius of the circle r = d/2 (from equation 3)
∴ r = 21/2
⇒ r = 10.5 cm
Now put the value of r in equation 2, we get
⇒ Circumference or Perimeter of circle = 2Ï€(10.5) (put Ï€ = 22/7)
= 2 × 22/7 × 10.5
= (2 × 22 × 10.5)/7
= 462/7
= 66 cm
The circumference of the circle is 66 cm.
4. A copper wire when bent in the form of a square encloses an area of 484 cm^{2}. The same wire is now bent in the form of a circle. Find the area enclosed by the circle.
Solution
In this question the wire is first bent in the shape of square and then same wire is bent to form a circle. The point to be noticed is that the same wire is used both the times which implies that the perimeter of square and that of circle will be equal.
Let the square be of side ‘a’ cm and radius of the circle be ‘r’.
Given the area enclosed by the square = 484 cm^{2}
Also, we know that area of square = Side × Side
Area of the square = a^{2}
⇒ a^{2} = 484
⇒ a = √484
⇒ a = 22 cm
Therefore, side of square, ‘a’ is 22 cm.
Also, circumference of the circle = Perimeter of square ...(i)
Perimeter of square = 4 × side
Perimeter of square = 4 × 22
⇒ Perimeter of square = 88 cm ...(ii)
Also, we know, Circumference of circle = 2Ï€r ...(iii)
Put values in equation 1 from equation 2 & 3, we get
2Ï€r = 88
⇒ r = 88/2Ï€ (put Ï€ = 22/7)
⇒ r = 88/(2 × 22/7)
On rearranging,
⇒ r = (88 × 7)/(2 × 22)
⇒ r = 616/44
⇒ r = 14 cm
So, the radius ‘r’ of the circle is 14 cm.
Area of circle = Ï€r^{2}
where r = radius of the circle
= Ï€(14)^{2 }
= (22/7 × 14 × 14) (put Ï€ = 22/7)
= (22 × 14 × 14)/7
= 4312/7
= 616 cm^{2 }
Area of the circle is 616 cm^{2}.
5. A wire when bent in the form of an equilateral triangle encloses an area of 121√3 cm^{2}. The same wire is bent to form a circle. Find the area enclosed by the circle.
Solution
In this question the wire is first bent in the shape of equilateral triangle and then same wire is bent to form a circle. The point to be noticed is that the same wire is used both the times which implies that the perimeter of equilateral triangle and that of circle will be equal.
Let the equilateral triangle be of side ‘a’ cm and radius of the circle be ‘r’.
Given: Area enclosed by equilateral triangle = 123√3 cm^{2 }
Also, we know that Area of equilateral triangle = √3/4 a^{2}
Where ’a’ = side of equilateral triangle
⇒ √3/4.a^{2} = 121√3
⇒ a^{2 }= 121√3/(√3/4)
⇒ a^{2} = (121√3 × 4)/ √3
⇒ a^{2} = (484√3)/ √3
⇒ a = √484
⇒ a = 22 cm
Therefore, side of equilateral triangle, ‘a’ is 22 cm.
Also, circumference of the circle = Perimeter of equilateral triangle ...(i)
Perimeter of equilateral triangle = 3 × side
= 3 × 22
= 66 cm ...(ii)
Also, we know Circumference of circle = 2Ï€r ...(iii)
Put values in equation 1 from equation 2 & 3, we get
2Ï€r = 66
⇒ r = 66/2Ï€
(put Ï€ = 22/7)
⇒ r = 66/(2 × 22/7)
On rearranging,
⇒ r = (66 × 7)/(2 × 22)
⇒ r = 462/44
⇒ r = 10.5 cm
So, the radius ‘r’ of the circle is 10.5 cm.
Area of circle = Ï€r^{2}
6. The length of a chain used as the boundary of a semicircular park is 108 m. Find the area of the park.
Solution
In this equation the length of chain used as boundary of the semicircular park is the perimeter of the semicircular park. By using this we will first calculate the radius of the semicircular park and then area of semicircle consequently.
Length of chain = 108 m
Length of chain = Perimeter or circumference of semicircle
Therefore, Circumference or Perimeter of semicircle = 108 m
Therefore, Circumference or Perimeter of semicircle = Ï€r
Where r = radius of semicircle
⇒ Ï€r = 108
⇒ r = 108/Ï€
(put Ï€ = 22/7)
⇒ r = 108/(22/7)
On rearranging,
⇒ r = (108 × 7)/22
⇒ r = 756/22
⇒ r = 34.46 m
Therefore, radius of semicircle is 34.36 m.
As, area of semicircle = Ï€r^{2}/2 ...(i)
Put value of ‘r’ in equation 1, we get
Area of semicircle = Ï€(34.36^{2})/2
(put Ï€ = 22/7)
= (22/7 × 34.36 × 34.36)/2
On rearranging,
= (22/ × 34.3636 × 34.3636)/(7 × 2)
= (25973.4112)/14
= 1855.63 m^{2}
The area of the semicircular park is 1855.63 m^{2}.
7. The sum of the radii of two circles is 7 cm, and the difference of their circumferences is 8 cm. Find the circumferences of the circles.
Solution
Given sum of the radius of the circles = 7 cm
The difference of their circumference = 8 cm
Let the radius one circle be ‘r_{1}’ cm and other be ‘r_{2}’ cm and circumference be ‘C_{1}’ and ‘C_{2}’ respectively.
Also, circumference of circle = 2Ï€r
Where r = radius of the circle
C_{1} = 2Ï€r_{1} and C_{2} = 2Ï€r_{2}
r_{1} + r_{2} = 7 ...(i)
C_{1} – C_{2} = 8 ...(ii)
(Note: Her it is considered that r_{1} > r_{2})
We can rewrite equation 2 as,
2Ï€r_{1}  2Ï€r_{2} = 8
⇒ 2Ï€ (r_{1} – r_{2}) = 8
(taking 2Ï€ common from L.H.S)
⇒ r_{1} – r_{2} = 8/2Ï€ – eqn3
⇒ r_{1} – r_{2} = 8/(2 × 22/7)
⇒ r_{1} – r_{2} = (8 × 7)/44
⇒ r_{1} – r_{2} = 56/44
⇒ r_{1} – r_{2} = 14/11
⇒ r_{1} = 14/11 + r_{2} ...(iii)
Put the value of r1 from equation 3 in equation 1
14/11 + r_{2} + r_{2} = 7
⇒ 14/11 + 2r_{2} = 7
⇒ 2r_{2} = 7 – 14/11
⇒ 2r_{2} = (77 – 14)/11
(taking 11 as LCM on R.H.S)
⇒ 2r_{2} = 63/11
⇒ r_{2} = 63/(2 × 11)
⇒ r_{2} = 63/22 cm
Put value of r_{2} in equation 3
∴ r_{1} = 14/11 + 63/22 (from equation 3)
⇒ r_{1} = (28 + 63)/22 (taking 22 as LCM on R.H.S.)
⇒ r_{1} = 91/22 cm
∴ C_{1} = 2Ï€ (91/22) (by putting value of r_{1})
⇒ C_{1} = 2 × 22/7 × 91/22
= (2 × 22 × 91)/(7 × 22)
= (2 × 91)/7
= 182/7
= 26 cm
C_{2} = 2Ï€ (66/22) (by putting value of r_{2})
⇒ C_{1} = 2 × 22/7 × 63/22
= (2 × 22 × 63)/(7 × 22)
= (2 × 63)/7
= 126/7
= 18 cm
The circumference of circles are 26 cm and 18 cm.
8. Find the area of a ring whose outer and inner radii are respectively 23 cm and 12 cm.
Solution
Consider the ring as shown in the figure below:
The inner radius of ring is ‘r’ and the outer radius is ‘R’.
Area of circle = Ï€r^{2} and Area of outer circle = Ï€R^{2}
Where r = 12 cm and R = 23 cm
Area of ring = Area of outer circle – Area of inner circle
Area of ring = Ï€R^{2} – Ï€r^{2} (put values of r & R)
⇒ Area of ring = Ï€(23^{2}) – Ï€(12^{2})
⇒ Area of ring = Ï€(23^{2} – 12^{2}) (taking Ï€ common from R.H.S)
⇒ Area of ring = Ï€(529 – 144)
= (22 × 385)/7
= 8470/7
= 1210 cm^{2}
Area of ring is 1210 cm^{2}
9. A path of 8 m width runs around the outside of a circular park whose radius is 17 m. Find the area of the path.
Solution
Given radius of circular park = R = 17 mWidth of the circular path outside the park = d = 8 m
Therefore, the radius of the outer circle = R’ = R + d
Outer radius = R’ = 17 + 8
R’ = 25 m
Area of inner circle = Ï€R^{2} and,
Area of outer circle = Ï€R’^{2}
Area of path = Area of outer circle = Area of inner circle
= Ï€R’^{2} – Ï€r^{2} (put values of R & R')
= Ï€(25^{2}) – Ï€(17^{2})
= Ï€(25^{2} – 17^{2}) (taking Ï€ common from R.H.S)
= Ï€(625 – 289)
⇒ Area of path = 22/7 × 336
(put Ï€ = 22/7)
= 7392/7
= 1056 m^{2}
The area of the path is 1056 m^{2}.
10. A racetrack 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
Consider the race track as shown below,
The inner and outer radius of track is ‘r’ cm and ‘R’ cm respectively.
Let inner and outer circumference be ‘C_{1}’ and C_{2}’ respectively.
C_{1} = 352 m and C_{2} = 396 m
We know,
Circumference of circle = 2Ï€r
Where r = radius of the circle
C_{1} = 2Ï€r and C_{2} = 2Ï€R
⇒ 2Ï€r = 352 and 2Ï€R = 396
⇒ r = 352/2Ï€ and R = 396/2Ï€ (put Ï€ = 22/7)
⇒ r = 352/(2 × 22/7) and R = 396/(2 × 22/7)
On rearranging,
⇒ r = (352 × 7)/(2 ×22) and R = (396 × 7)/(22 × 22)
⇒ r = 2464/44 and R = 2772/44
⇒ r = 56 m and R = 63 m
So, the width of the race track = R – r,
⇒ Width of the race track = 63  56,
⇒ Width of the race track = 7 m
Area of race track = area of outer circle – area of inner circle
⇒ Area of track = Ï€R^{2} – Ï€r^{2} (put values of r and R)
⇒ Area of track = Ï€(63)^{2} – Ï€(56^{2})
⇒Area of track = Ï€(63)^{2} – (56)^{2} (taking Ï€ common from R.H.S)
⇒ Area of track = Ï€(3969 – 3136)
⇒ Area of track = Ï€ × 833
⇒ Area of track = 22/7 × 833 (put Ï€ = 22/7)
= 22 × 119
= 2618 m^{2 }
The width of track in 7 m and area of track is 2618 m^{2}.
11. A sector is cut from a circle of radius 21 cm. The angle of the sector is 150°. Find the angle of the arc and the area of the sector.
Solution
Consider the circle shown above,
Given radius of the circle = R = 21 cm ...(i)
And angle of the sector = Î¸ = 150° ...(ii)
Length of arc of a sector = Î¸/360 × 2Ï€R ...(iii)
Where ‘R’ =radius of sector (or circle)
Î¸ = Angle subtended by the arc on the centre of the circle
Put the values of R and Î¸ from equation 1 and 2 in equation 3
⇒ Length of arc = 150/360 × 2Ï€(21) (put Ï€ = 22/7)
= 150/360 × 2 × 22/7 × 21
= (150 × 2 × 22 × 21)/(360 × 7)
= 138600/2520
= 55 cm
Area of a sector = Î¸/360 × Ï€R^{2} ...(iv)
Where ‘R’ = radius of sector (or circle)
Î¸ = angle subtended by the arc on the centre of the circle
Put the values of R and Î¸ from equation 1 and 2 in equation 3
⇒ Area of sector = 150/360 × Ï€(212) (put Ï€ = 22/7)
= (150 × 22 × 22 × 21)/(360 × 7)
= 1455300/2520
= 577.5 cm^{2}
The length of arc is 55 cm and area of sector is 577.5 cm^{2}.
12. 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
Consider the circle shown above,
We know, Area of sector = Î¸/360 × Ï€R^{2} ...(i)
Where R = radius of the circle and Î¸ = central angle
Given R = 10.5 cm and Area of sector = 69.3 cm^{2}
Let the angle subtended at centre = Î¸
Put the values of R and area of sector in equation 1
⇒ 69.3 = Î¸/360 × Ï€(10.5^{2}) (put Ï€ = 22/7)
⇒ 69.3 = Î¸/360 × 22/7 × 10.5 × 10.5
⇒ 69.3 = (Î¸ × 22 × 10.5 × 10.5)/(360 × 7)
⇒ 69.3 = (Î¸ × 2425.5)/2520
⇒ (69.3 × 2520)/2425.5 = Î¸
⇒ 174636/2424.5 = Î¸
⇒ Î¸ = 72°
The central angle of the sector is 72°.
13. The length of an arc of a circle, subtending an angle of 54° at the centre is 16.5 cm. Calculate the radius, circumference and area of the circle.
Solution
Consider the circle shown above,We know, Length of arc of sector = Î¸/360 × 2Ï€R ...(i)
Where R = radius of circle and Î¸ = central angle of the sector
Given, Length of arc = l = 16.5 and Î¸ = 54°.
Let the radius be x cm
Put the values of l and Î¸ in equation 1
⇒ 16.5 = 54/360 × 2Ï€x (put Ï€ = 22/7)
⇒ 16.5 = (54 × 2 × 22 × x)/(360 × 7)
⇒ 16.5 = (2376 × x)/2520
On rearranging
⇒ (16.5 × 2520)/2376 = x
⇒ 41580/2376 = x
⇒ x = 17.5 cm
Also, we know circumference of the circle = 2Ï€R
⇒ Circumference of the circle = 2Ï€x (put value of x in this equation)
⇒ Circumference of the circle = 2Ï€(17.5)
⇒ Circumference of the circle = 2 × 22/7 × 17.5 (put Ï€ = 22/7)
= (2 × 22 × 17.5)/7
= 770/7
Circumference of the circle = 110 cm
Also, we know Area of the circle = Ï€R^{2}
⇒ Area of the circle = Ï€x^{2}
⇒ Area of the circle = Ï€(17.5^{2})
⇒ Area of the circle = 22/7 × 17.5 × 17.5 (put Ï€ = 22/7)
⇒ Area of the circle = (22 × 17.5 × 17.5)/7
⇒ Area of the circle = 6737.5/7
⇒ Area of the circle = 962.5 cm^{2}
The radius of circle is 17.5 cm, circumference is 110 cm and area is 962.5 cm^{2}
14. The radius of a circle with centre O is 7 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of minor and major segments.
Solution
Consider the above figure,From here we can conclude that the portion or the segment below the chord AB is the minor segment and the segment above AB is major segment.
Also we know,
Area of minor segment = Area of sector – Area of △AOB ...(i)
Now, Area of sector = Î¸/360 × Ï€R^{2} ...(ii)
Where R = radius of the circle and Î¸ = central angle of the sector
Given, R = 7 cm and Î¸ = 90°
Putting these values in the equation 2, we get
Area of sector = 90/360 × Ï€(7)^{2} (put Ï€ = 22/7)
= 90/360 × 22/7 × 7 × 7
= (90 × 22 × 7 × 7)/(360 × 7)
= 97020/2520
⇒ Area of sector = 38.5 cm^{2} ...(iii)
Area of △AOB = 1/2 × base × height
NOTE : In gene Area of △AOB = 1/2 × OA × OB × sin Î¸
As triangle is isosceles therefore height and base both are 7 cm.
⇒ Area of △AOB = 1/2 × 7 × 7 = 49/2
= 24.5 cm^{2} ...(iv)
Putting values of equation 2 and 4 in equation 1 we get
Area of minor segment = 38.5 – 24.5
⇒ Area of minor segment = 14 cm^{2 }
Area of major segment = Ï€R^{2} – Area of minor segment = eqn5
Put the value of R, and Area of minor segment in equation 5 = Ï€(7^{2}) – 14
= 49Ï€ – 14
⇒ Area of major segment = 22/7 × 49 – 14 (put Ï€ = 22/7)
= (22 × 7) – 14
= 154 – 14
= 140 cm^{2}
Area of minor segment is 14 cm^{2} and of major segment is 130 cm^{2}.
15. Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment.[Take Ï€ = 3.14 and = 1.73.].
Solution
Consider the figure shown above.In this, the triangle AOB is an equilateral triangle as all the sides are equal; therefore it is obvious that the central angle of the sector is 60 degrees. Now by simply applying the formula of length of an arc, we can easily calculate the length of arc of the sector AOB.
Given radius of circle = R = 12 cm,
Length of chord AB = 12 cm
∴ Central angle = Î¸ = 60° (∵ △AOB is an equilateral triangle)
Length of arc = Î¸/360 × 2Ï€(R) ...(i)
Where R = radius of circle and Î¸ = central angle of the sector
Put the values of R and Î¸ in equation 1
⇒ Length of minor arc = 60/360 × 2Ï€(12) (put Ï€ = 3.14)
= 60/360 × 2 × 3.14 × 12
= (60 × 2 × 3.14 × 12)/360
= (2 × 3.14 × 12)/6
= 2 × 3.14 × 2
= 12.56 cm
Now, Length of major arc = 2Ï€R – Length of minor arc
⇒ Length of major arc = 2Ï€(12) – 12.56 (put Ï€ = 3.14)
⇒ Length of major arc = (2 × 3.14 × 12) – 12.56
⇒ Length of major arc = 75.36 – 12.56
⇒ Length of major arc = 62.8 cm
Now, area of minor segment = Area of sector – Area of triangle ...(i)
∴ Area of sector = Î¸/360 × Ï€r^{2} (put the values of R and Î¸)
= 60/360 × Ï€(12^{2})
= 60/360 × 3.14 × 144
= 75.36 cm^{2} eqn2
Area of triangle = √3/4 × a^{2} (put a = 12 cm)
= √3/4 × (12^{2})
⇒ Area of triangle = √3/4 × 144
⇒ Area of triangle = 1.73 × 36
⇒ Area of triangle = 62.28 cm^{2} eqn3
Put the values of equation 2 and 3 in equation 1,
∴ Area of minor segment = 75.36 – 62.28
= 13.08 cm^{2}
Length of major arc is 62.8 cm and minor arc is 12.56 cm and area of minor arc is 12.56 cm and area of minor segment is 13.08 cm^{2}.
16. A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find the areas of both the segments . [Take Ï€ = 3.14].
Solution
Consider the figure shown above.In this, the triangle AOB is an isosceles triangle. So here we will construct a perpendicular bisector from O on AB and as this triangle is isosceles therefore this perpendicular will also act as median and angle bisector.
Therefore,
Draw a perpendicular bisector from O which meets AB at D and bisects AB, as ABO is an isosceles triangle therefore OD acts as a median.
So, consider right angle triangle AOD right angled at D
sin Î¸ = Perpendicular/Hypotenuse
Let ∠AOD = Î¸ ⇒ Perpendicular = AD and Hypotenuse = AO = R
Given Radius of circle = R = 5√2 cm
Length of chord AB = 10 cm, AD = 5 cm
sin Î¸ = AD/AO (put values of AD and AO)
⇒ sin Î¸ = 5/5√2 = 1/√2
⇒ sin Î¸ = sin 45°
(as sin 45° = 1/√2)
⇒ Î¸ = 45°
∴ ∠AOD = 45°
Thus we can say ∠AOB = 90° (As ∠AOD = 1/2∠AOB)
Area of minor segment = Area of sector – Area of right angle triangle ...(i)
Area of sector = Î¸/360 × Ï€R^{2}
Where R = radius of the circle and Î¸ = central angle of the sector
Area of sector = 90/360 × Ï€((5√2)^{2}) (put Ï€= 3.14)
= 90/360 × 3.14 × 50
= (3.14 × 50)/4
∴ Area of sector = 39.25 cm^{2}
Area of right angle triangle = 1/2 × base × height
As this is isosceles rightangle triangle
∴ Height = base = 5√2 cm
Area of right angle triangle = 1/2 × 5√2 × 5√2 = 50/2 = 25 cm^{2}
Put the value of area of sector and area of right angle triangle in equation 1,
⇒ Area of minor segment = 39.25 – 25
= 14.25 cm^{2}
Area of major segment = Ï€R^{2} – area of minor segment
Area of major segment = Ï€((5√2)^{2}) – 14.25
= 3.14 × 5√2 × 5√2 – 14.25
⇒ Area of major segment = 157 – 14.25 = 142.75 cm^{2}
Area of major segment is 142.75 cm^{2} and of minor segment is 14.25 cm^{2}.
17. Find the area of both the segments of a circle of radius 42 cm with central angle 120°. [Given, sin 120°  √3/2 and √3 = 1.73]
Solution
Given R = 42 cm and central angle of sector = 120°
Area of minor segment = Area of sector – Area of triangle ...(i)
Area of sector = Î¸/360 × Ï€R^{2}
Where R = radius of the circle and Î¸ = central angle of the sector
Area of sector = 120/360 × Ï€(42^{2}) (put Ï€ = 22/7)
= 120/360 × 22/7 × 1764
∴ Area of sector = 1848 cm^{2}
Area of right angle triangle = 1/2 × base × height × sin Î¸
Where Î¸ = central angle of the sector
Area of triangle = 1/2 × 42 × 42 × sin 120°
(put the value sin 120° = √3/2)
Area of triangle = 1/2 × 42 × 42 × √3/2
Area of triangle = (42 × 42 × √3)/4
(put √3 = 1.73)
Area of triangle = (42 × 42 × 1.73)/4
= 762.93 cm^{2}
Put the values of area of triangle and area of sector in equation 1
⇒ Area of minor segment = 1848 – 762.93
= 1085.07 cm^{2}
Area of major segment = Ï€R^{2} – Area of minor segment
Put the value of area of minor segment and R in above equation
= Ï€(42^{2}) – 1085.07
⇒ Area of major segment = 22/7 × 42 × 42 – 1085.07 (put Ï€ = 22/7)
⇒ Area of major segment = 5544 – 1085.07
∴ Area of major segment = 4458.93 cm^{2}
Area of major segment is 4458.93 cm^{2} and of minor segment is 1085.07 cm^{2}.
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 Ï€ = 3.14 and √3 = 1.732]
Solution
Area of minor segment = Area of sector – Area of triangle ...(i)
Area of sector = Î¸/360 × Ï€R^{2})
Where R = radius of the circle and Î¸ = central angle of the sector
Area of sector = 60/360 × Ï€(30^{2}) (put Ï€ = 3.14)
Area of sector = 60/360 × 3.14 × 900
Area of sector = (3.14 × 900)/6
∴ Area of sector = 471 cm^{2}
Area of right angle triangle = √3/4 × a^{2}
Where a = side of the triangle
Area of triangle = √3/4 × 30 × 30
Area of triangle = √3/4 × 900
Area of triangle = (900 × (√3))/4
(put √3 = 1.732)
Area of triangle = (1.732 × 900)/4
∴ Area of triangle = 389.7 cm^{2}
Put the values of area of triangle and area of sector in equation 1
Area of minor segment = 471 – 389.7
⇒ Area of minor segment = 81.3 cm^{2}
Area of major segment = Ï€R^{2 }– Area of minor segment
Put the value of area of minor segment and R in above equation
⇒ Area of major segment = Ï€ × (30^{2}) – 81.3 (put Ï€ = 3.14)
⇒ Area of major segment = 3.14 × 30 × 30  81.3
⇒ Area of major segment = 2826 – 81.3
= 2744.7 cm^{2 }
Area of major segment is 2744.7 cm^{2} and of minor segment is 81.3 cm^{2}.
19. In a circle of radius 10.5 cm, the minor arc is onefifth of the major arc. Find the area of the sector corresponding to the major arc.
Solution
Given radius of circle = R = 10.5 cm
Let the area of major sector be ‘A_{1}’ and that of minor sector be ‘A_{2}’.
∴ A_{2} = A_{1}/5 ...(i)
We know, Area of circle = Area of major sector + Area of minor sector
⇒ Area of circle = A_{1} + A_{2}
⇒ Area of circle = A_{1} + A_{1}/5 ...(ii) (from equation 1)
We also know, Area of circle = Ï€R^{2}
Where R = radius of circle, put value of area of circle in equation 2.
⇒ Ï€(10.5^{2}) = (5A_{1} + A_{1})/5
(taking 5 as L.C.M. on R.H.S)
⇒ Ï€ × 10.5 × 10.5 = 6A_{1}/5
⇒ 22/7 × 10.5 × 10.5 = 6A_{1}/5
⇒ (22 × 10.5 × 10.5)/7 = 6A_{1}/5
⇒ 22 × 10.5 × 1.5 = 6A1/5
⇒ 346.5 = 6A_{1}/5
⇒ (5 × 346.5)/6 = A_{1}
= 288.75 cm^{2 }
The area of major sector is 288.75 cm^{2}.
20. The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 2 days. [Take Ï€ = 3.14]
Solution
In an hour the minute hand completes one rotation therefore in 24 hours the minute hand will complete 24 rotations similarly the hour hand completes one rotation in 12 hours therefore in 24 hours it will complete 2 rotations. Now we have to just calculate the perimeter of the circle traced by minute hand and hour hand and multiply it with the number of rotations of minute hand and hour in 2 days respectively.
Length of short/hour hand = r = 4 cm
Length of long/minute hand = R = 6 cm
∴ The perimeter of circle traced by short hand = p = 2Ï€r ...(i)
∴ The perimeter of circle traced by long hand = P = 2Ï€R ...(ii)
Now put the value of ‘r’ and ‘R’ in the equation 1 and 2 respectively.
⇒ p = 2Ï€(4) & P = 2 Ï€(6) (put Ï€ = 3.14)
⇒ p = 2 × 3.14 × 4 & P = 2 × 3.14 × 6
∴ p = 25.12 cm & P = 37.68 cm
Therefore, distance covered by short hand in one rotation = 25.12 cm
Distance covered by long and in one rotation = 37.68 cm
Number of rotation of short hand in one day = 2
Number of rotation of long hand in on day = 24
Therefore number of rotation of small hand in two days = 4
Number of rotation of long hand in two days = 48
Total distance covered by long hand in 2 days = P × no. of rotations in 2 days
⇒ Total distance covered by long hand in 2 days = 37.68 × 48
⇒ Total distance covered by long hand in 2 days = 1808.64 cm ...(iii)
Total distance covered by short hand in 2 days = p × no. of rotations in 2 days
⇒ Total distance covered by short hand in 2 days = 25.12 × 24
⇒ Total distance covered by short hand in 2 days = 100.48 cm ...(iv)
Now, total distance covered by tip of both hands in 2 days = eqn3 + eqn4
⇒ Total distance covered by both hands in 2 days = 1808.64 + 100.48
⇒ Total distance covered by both hands in 2 days = 1909.12 cm
The distance covered by both hands in 2 days is 1909.12 cm
21. Find the area of a quadrant of a circle whose circumference is 88 cm.
Solution
Quadrant is a sector in which the central angle is 90 degrees, and this is the key to solve this question, As we know the central angle of the sector so we can easily calculate the area of quadrant by first calculating the radius of the circle as the circumference of the circle is given and then applying the formula of area of sector.
So, we know Circumference of a circle = 2Ï€R – eqn1
Where R = radius of the circle
Given Circumference of the circle = 88 cm, Î¸ = 90°
Put the given values in equation 1
88 = 2 × 22/7 × R (Ï€ = 22/7)
⇒ 88 = (2 × 22 × R)/7
⇒ 88 = (44 × R)/7
⇒ 88 = 44R/7
⇒ (88 × 7)/44 = R
⇒ 616/44 = R
⇒ R = 14 cm
Now we know, Area of a sector = Î¸/360 × Ï€R^{2}
Put the values of R and Î¸ in the above equation
⇒ Area of quadrant = 90/360 × Ï€(14^{2})
= 90/360 × 22/7 × 14 × 14
= (90 × 22 × 14 × 14)/(360 × 7)
= (22 × 14 × 14)/(4 × 7)
= 4312/28
= 154 cm2
The area of quadrant is 154 cm^{2}.
22. A rope by which a cow is tethered is increased from 16 m to 23 m. How much additional ground does it have now to graze?
Solution
Here the increase in the length of the rope simply means that there is increase in the radius of the circle within which cow can graze. Now find the additional area available for grazing can be easily be found by simply subtracting the initial area available for grazing from the new area available.
Initial radius = r = 16 cm
Increased radius = R = 23 cm
Additional ground available = Area of new ground – Initial area ...(i)
Initial area of ground = Ï€(r^{2})
⇒ Initial area of ground = Ï€(16^{2})
⇒ Initial area of ground = 256Ï€ ...(ii)
Area of new ground = Ï€R^{2}
⇒ Area of new ground = Ï€(23^{2})
⇒ Area of new ground = 529Ï€ ...(iii)
Now put the values of equation 2 and 3 in equation 1
⇒ Additional area of ground available = 529Ï€  256Ï€
⇒ Additional area available 529Ï€  256Ï€
⇒ Additional area available = (529 – 256)Ï€ (Taking Ï€ common)
⇒ Additional ground available = 273Ï€
⇒ Additional ground available = 273 × 22/7 (put Ï€ = 22/7)
= (22 × 273)/7
= 6006/7
= 858 cm^{2}
The additional ground available is 858 cm^{2}.
23. A horse is placed for grazing inside a rectangular field 70 m by 52 m. It is tethered to one corner by a rope 21 m long. On how much area can it graze? How much area is left ungrazed ?
Solution
Here the horse is tethered to one corner implies or means that the area available for grazing is a quadrant of radius 21 m. Now we need to find the area of this quadrant to find out the area available for grazing and then subtract it from the total area of the rectangular field to obtain the area left ungrazed.
Given length of rectangular field = l = 70 m
Breadth of rectangular field = b = 52 m
∴ Area of the field = l × b
⇒ Area of the field = 70 × 52
⇒ Area of the field = 3640 m^{2}
We know in a rectangle all the angles are 90 degrees.
∴ Area available for grazing = area of quadrant
⇒ Area of quadrant/sector = Î¸/360 × Ï€R^{2}
Where R = radius of circle & Î¸ = central angle
Given R = 21 m and Î¸ = 90°
⇒ Area available for grazing = Î¸/360 × Ï€R^{2}
Put the given values in the above equation,
⇒ Area available for grazing = 90/360 × Ï€(21^{2})
(put Ï€ = 22/7)
= 90/360 × 22/7 × 441
= (90 × 22 × 441)/(360 × 7)
= (22 × 63)/4
= 1386/4
⇒ Area available for grazing = 346.5 m^{2}
Area left ungrazed = Area of field – Area available for grazing
⇒ Area left ungrazed = 3640 – 346.5
⇒ Area left ungrazed = 3293.5 m^{2}
The area available for grazing is 346.5 m^{2} and area left ungrazed is 3293.5 m^{2}.
23. A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m. If the length of the rope is 7 m, find the area of the field which the horse cannot graze. Take √3 = 1.732. Write the answer correct to 2 places of decimal.
Solution
Here the horse is tethered to one corner implies or means that the area available for grazing is a sector of radius 21 m with central angle as 60 degrees as the field is in shape of equilateral triangle. Now we need to find the area of this sector to find out the area available for grazing for grazing and then subtract it from the total area of the triangular field to obtain the area left ungrazed.
Given the side of field = a = 12 m
∴ Area of field = Area of equilateral triangle
⇒ Area of field = √3/4 × a^{2}
⇒ Area of field = 1.732/4 × (12^{2})
⇒ Area of field = (1.732 × 144)/4
⇒ Area of field = 62.352 m^{2}
We know in an equilateral triangle all the angles are 60 degrees.
∴ Area available for grazing = Area of the sector
Area of quadrant/sector = Î¸/360 × Ï€R^{2}
Where R = radius of circle and Î¸ = central angle of sector
Given r = 7 m and Î¸ = 60°
Put the given values in the above equation,
⇒ Area available for grazing = 60/360 × Ï€(7^{2}) (put Ï€ = 22/7)
⇒ Area available for grazing = 60/360 × 22/7 × 49
⇒ Area available for grazing = (60 × 22 × 49)/(360 × 7)
⇒ Area available for grazing = (22 × 7)/6
⇒ Area available for grazing = 154/6
⇒ Area available for grazing = 25.666 m^{2}
Area that cannot be grazed = Area of field – Area available for grazing
⇒ Area that cannot be grazed = 62.352 – 25.666
⇒ Area that cannot be grazed = 36.686 m^{2}
The area that cannot be grazed is 36.656 m^{2}.
25. Four cows are tethered at the four corners of a square 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
Here, the 4 cows are tethered to each corner implies or means that the area available for grazing is a quadrant of radius 25 m with central angle as 60 degrees as the field is in shape of square. Now, we need to find the area of this sector to find out the area available for grazing for all the cows and then subtract it from the total area of the square field to obtain the area left ungrazed.
The reason why we have taken the radius as 25 m is, basically we have considered that each cow is tethered to a rope which is equal to half of the side of the square as we had to maximize the area each cow gets to graze without sharing thus the maximum radius within which a cow can graze maximum unshared area is simply the half of the side of square.
Given the side of field which is in shape of square = a = 50 m
∴ Area of the field = Area of square
⇒ Area of field = a^{2}
⇒ Area of field = (50^{2})
⇒ Area of field = 2500 m^{2}
We know in an square all the angles are 90 degrees.
∴ Area available for grazing for one cow = area of sector/quadrant
Area of quadrant/sector = Î¸/360 × Ï€R^{2}
Where R = radius of circle & Î¸ = central angle of sector
Given R = 25 m & Î¸ = 90°
⇒ Area available for grazing for one cow = Î¸/360 × Ï€R^{2}
Put the given values in the above equation,
⇒ Area available for grazing for one cow = 90/360 × Ï€(25^{2}) (put Ï€ = 3.14)
⇒ Area available for grazing for one cow = 90/360 × 3.14 × 625
⇒ Area available for grazing for one cow = (90 × 3.14 × 625)/360
⇒ Area available for grazing for one cow = (3.14 × 625)/4
⇒ Area available for grazing for one cow = (1962.5)/4
⇒ Area available for grazing for one cow = 490.625 m^{2}
⇒ Area available for 4 cows = 4 × Area available for one cow
⇒ Area available for 4 cows = 4 × 490.625
⇒ Area available for 4 cows = 1962.5 m^{2}
Area left ungrazed = Area of field – Area available for grazing for 4 cows
⇒ Area that cannot be grazed = 2500 – 1962.5
⇒ Area that cannot be grazed = 537.5 m^{2}
The area left ungrazed is 537.5 m^{2}.
26. In the given figure, OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is 32√3 cm^{2}, find the radius of the circle.
Here in the given figure ’O’ is the centre on which three vertices of rhombus lie, this implies that OP, OR are both radius of the circle. Also we know that in rhombus all the 4 sides are equal in length. Thus OP, OR, PQ, RQ they all are radii of circle. Also, OQ is equal to radius of circle. Now rhombus being a parallelogram therefore diagonal OQ will divide the rhombus into two equal halves this means that the area of triangle OQR will be equal to half of the area of rhombus. Also we can see that triangle OQR is an equilateral triangle and hence we can easily calculate its area in terms of radius of circle and equate it to half of the area of rhombus and calculate the radius of given circle.
Given area of OPQR = 32√3 cm^{2}
Let the radius of the circle = x cm
Now join OQ
Consider △OQR,
OQ = OR = RQ = x cm
⇒ △OQR is an equilateral triangle
∴ Area of OQR = Area of an equilateral triangle = √3/4 × a^{2} ...(i)
Where a = side of equilateral triangle
Also we know OQ is a diagonal of rhombus OPQR and as in a parallelogram diagonal divides it into two equal area or halves, similarly OQ is also dividing the rhombus into two equal areas therefore,
⇒ Area of △OQR = Area of △OPQ ...(ii)
Area of OPQR = Area of △OQR + Area of △OPQ
Area of OPQR = 2 × Area of △OQR (from eqn2) ...(iii)
Put the values of area of OPQR and equation 1 in equation 3
⇒ 32√3 = 2 × √3/4 × a^{2} (put a = x)
⇒ 32√3 = (2√3)/4 × x^{2}
⇒ 32√3 = √3/2 × x^{2}
⇒ (32√3 × 2)/√3 = x^{2}
⇒ 64 = x^{2}
⇒ x = ±√64
⇒ x = ±8
As every quadratic equation has two roots, similarly x^{2} = 64 also have two roots i.e., x = 8 and x = 8. As we know ‘x’ represents radius of circle therefore it cannot be a negative value, hence we discard the negative root.
Therefore radius of the circle = x = 8 cm
The radius of circle is 8 cm.
27. The side of a square is 10 cm. Find
(i) The area of the inscribed circle, and
(iii) The area of the circumscribed circle. [Take Ï€ = 3.14]
Solution
(i)
Consider the above figure, Join PR,
Now PR = Diameter of the inscribed circle
Also, PR = BC = 10 cm.
So, PR = 10 cm
∴ radius of inscribed circle = r = PR/2
⇒ r = 10/2
⇒ r = 5 cm
∴ Area of inscribed circle = Ï€r^{2} (put value of r in this equation)
⇒ Area of inscribed circle = Ï€(5^{2})
⇒ Area of inscribed circle = 22/7 × 25 (put Ï€ = 22/7)
⇒ Area of inscribed circle = (22 × 25)/7
⇒ Area of inscribed circle = 78.57 cm^{2}
The area of inscribed circle is 78.57 cm^{2}.
(ii)
Consider the above figure, O is the centre and ABCD is a square inscribed. Now OB and OD are radii of circle.
Consider △DBC right angled at c (as C is a vertex of square)
∴ Apply Pythagoras theorem in triangle DBC
Hypotenuse^{2} = Perpendicular^{2} + Base^{2}
In triangle DBC, hypotenuse = DB,
Perpendicular = BC and
Base = DC
⇒ BD^{2} = BC^{2} + DC^{2 }
Put the values of BC and DC i.e., 10 cm
⇒ BD^{2 }= 10^{2} + 10^{2}
⇒ BD^{2} = 200
⇒ BD = √200
⇒ BD = 10√2 cm
Now radius of circle = half of BD
∴ radius of circle = r = BD/2
⇒ r = (10√2)/2
⇒ r = 5√2 cm
Hence, Area of circumscribing circle = Ï€r^{2}
⇒ Area of circumscribing circle = 3.14 × 5√2 × 5√2
(put Ï€ = 3.14 and r = 5√2 cm)
⇒ Area of circumscribing circle = 3.14 × 50
⇒ Area of circumscribing circle = 157 cm^{2}
Area circumscribing circle is 157 cm^{2}.
28. If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.
Solution
Consider the figure shown below where O is centre of circle, join BC which passes through O, let the side of square be ‘a’ and radius of circle be ‘r’.
Now we know OB and OC are radius of circle
So, OB = OC = r
Consider △BDC right angled at D
∴ H^{2} = P^{2} + B^{2} (Pythagoras theorem)
⇒ BC^{2} = BD^{2} + DC^{2} ...(i)
And we know BC = OC + OB
BC = 2r and BD = DC = a (put these values in eqn1)
⇒ (2r)^{2} = a^{2} + a^{2}
⇒ 4r^{2} = 2a^{2}
⇒ r^{2} = 2a^{2}/4
⇒ r^{2} = a^{2}/2 ...(ii)
Area of inscribed square = side × side
Area of inscribed square = a × a
Area of inscribed square = a^{2} ...(iii)
Area of circumscribing circle = Ï€R^{2} where R = radius of circle
⇒ Area of circumscribing circle = Ï€r^{2} ...(iv)
Ration of area of circumscribing circle to that of inscribed circle = (area of circle)/(area of square)
Put the values from equation 3 & 4 in above equation
Ratio = Ï€r^{2}/a^{2}
⇒ Ratio = (Ï€ × a^{2}/2)/a^{2} (from eqn 2)
⇒ Ratio = (Ï€ × a^{2})/(2 × a^{2}) = Ï€/2
So, ratio is Ï€ : 2
29. The area of a circle inscribed in an equilateral triangle is 154 cm^{2}. Find the perimeter of the triangle.[Take √3 = 1.73]
Solution
Consider the figure shown above. AF, BE and CD are perpendicular bisector.Now we know that the point at which all three perpendiculars meet is called incentre, so O is the incentre, thus O divides all three perpendiculars in a ratio 2 : 1.
Let AB = BC = CA = a cm
Therefore let AF = h cm
⇒ ∠AFC = 90° and OF = 1/3 × AF
⇒ OF = h/3 cm (putting value of OF)
⇒ h = 3 × OF ...(i)
And we can see from figure that OF = radius of circle
Now let radius of circle be = r cm
∴ Area of circle = Ï€R^{2}
Where r = radius of circle
Given area of circle = 154 cm^{2}
⇒ Ï€r^{2} = 154
⇒ 22/7 × r^{2} = 154 (put Ï€ = 22/7)
⇒ r^{2} = (154 × 7)/22
⇒ r^{2} = 49
⇒ r = 7 cm
Therefore OF = 7 cm
⇒ h = 3 × 7 (from eqn 1)
⇒ h = 21 cm
We know area of an equilateral triangle = √3/4 × a^{2}
Where a = side of triangle
Also, Area of triangle = 1/2 × base × height
Equating both the areas we get,
√3/4 × a^{2} = 1/2 × BC × AF
Put the values of BC and AF
⇒ √3/4 × a^{2} = 1/2 × a × h
√3/4 × a^{2} = 1/2 × a × 21
(putting value of h = 21 cm)
⇒ √3/4 × a = 21/2
⇒ a = (21 × 4)/(2 × √3)
(rationalize it)
⇒ a = (21 × 4 × √3)/(2 × √3 × √3)
⇒ a = (42 × √3)/3
⇒ a = 14√3 cm
∴ Perimeter of equilateral triangle = 3 × side of triangle
⇒ Perimeter of △ABC = 3 × 14√3 (put √3 = 1.73)
⇒ Perimeter of △ABC = 42 × 1.73
⇒ Perimeter of △ABC = 72.66 cm
The perimeter of triangle is 72.66 cm.
30. The radius of the wheel of a vehicle is 42 cm. How many revolutions will it complete in a 19.8 kmlong journey ?
Solution
In one revolution a wheel will cover a distance equal to its circumference, so in order to find the number of revolutions we have to first calculate the circumference of the wheel and then divide it with the total distance covered to find out the total number of revolutions
Given radius of wheel = r = 42 cm
Circumference of wheel = 2Ï€R where R = radius of the wheel
= 2Ï€(42) (putting value of r)
Circumference of wheel = (2 × 22 × 42)/7 = 264 cm
Therefore distance covered in one revolution = 264 cm
Total distance covered = 19.8 km = 1980000 cm
Total number of revolutions = n
Distance covered on 1 revolution × no. of revolutions = Total distance
264 × n = 1980000
⇒ n = 1980000/264
⇒ n = 7500
Total number of revolutions is 7500.
31. The wheels of the locomotive of a train are 2.1 m in radius. They make 75 revolutions in one minute. Find the speed of the train in km per hour.
Solution
Given radius of wheel = R = 2.1 m
Number of revolutions in one minute = 75
Number of revolutions in 1 hour = 75 × 60
Number of revolutions in 1 hour = 45000
Distance covered in one revolution = Circumference of wheel
Distance covered in one revolution = 2Ï€R (where R = radius of wheel)
Distance covered n 1 revolution = 2Ï€(2.1)
Distance covered in one revolution = 2 × 22/7 × 2.1 (put Ï€ = 22/7)
= 13.2 m
So, distance covered in 4500 revolutions = 4500 × distance covered in 1
Distance covered in 4500 revolution = 4500 × 13.2
Distance covered in 4500 revolutions = 59400 m = 59.4 km
∴ Distance covered in 1 hour = 59.4 km
Hence speed of the locomotive = 59.4 km/hr
The speed of locomotive is 59.4 km/hr
32. The wheels of a car make 2500 revolutions in covering a distance of 4.95 km. Find the diameter of a wheel.
Solution
Let the diameter f the wheel be ‘d’ cm
Total distance covered in 250 revolutions = 49.5 km = 495000 m
Distance covered in one revolution = 495000/2500
⇒ Distance covered in one revolution = 198 cm ...(i)
Also, Distance covered in one revolution = circumference of wheel
∴ Distance covered in one revolution = Ï€D where d = diameter of wheel
Distance covered in one revolution = (22 × d)/7 ...(ii) (put Ï€ = 22/7)
Equate equation 1 and 2 we get,
(22 × d)/7 = 198
⇒ d = (198 × 7)/22
⇒ d = 9 × 7
⇒ d = 63 cm
The diameter of the wheel is 63 cm.
33. A boy is cycling in such a way that the wheels of his bicycle are making 140 revolutions per minute. If the diameter of a wheel is 60 cm, calculate the speed (in km/h) at which the boy is cycling.
Solution
Given number of wheel = d = 60 cm
Number of revolutions in one minute = 140
Number of revolutions in one hour = 140 × 60
Number of revolutions in one hour = 8400
Distance covered in one revolution = circumference of wheel
⇒ Distance covered in one revolution = Ï€d
Distance covered in one revolution = 22/7 × 60 (put Ï€ = 22/7 and value of d)
= 188.57 cm
Distance covered in one hour = Distance in 1 revolution × no. of revolutions
⇒ Total distance covered in one hour = 188.57 × 8400
⇒ Total distance covered in one hour = 1583988 cm = 15.839 km
∴ Speed with which boy is cycling = 15.839 km/hr
The speed with which boy is cycling is 15.839 km/hr.
34. The diameter of the wheels of a bus is 140 cm. How many revolutions per minute do the wheels make when the bus is moving at a speed of 72.6 km per hour ?
Solution
Given diameter of wheel of bus = d = 140 cm
So radius of wheel = R = d/2 = 140/2 = 70 cm
Speed of bus = 72.6 km/hr
∴ Distance covered by bus in one hour = 72.6 km = 7260000 cm
So distance covered by wheels in one minute = 7260000/60
Distance covered in one minute = 121000 cm ...(i)
Let the number of revolutions made by wheel per minute = x
Distance covered by wheel in one revolution = circumference of wheel = 2Ï€R
Distance covered by wheel in one revolution = 2Ï€(70) (putting value of R)
= 2 × 22/7 × 70 (putting value or R and Ï€ = 22/7)
= (2 × 22 × 70)/7
= 2 × 22 × 10
= 440 cm
∴ Total distance = No. of revolution × Distance covered in 1 revolution
On putting the required values we get,
121000 = 440 × (x)
⇒ x = 121000/440
⇒ x = 275
Number of revolutions made per minute is 275.
35. The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that a rear wheel makes to cover the distance which the front wheel covers in 800 revolutions.
Solution
Given diameter of front wheel = d = 80 cm
So, radius of front wheel = r = d/2 = 80/2 = 40 cm
Diameter of rear wheel = D = 2 m = 200 cm
So, Radius of front wheel = R = D/2 = 200/2 = 100 cm.
Distance covered by wheel in 1 revolution = Circumference of wheel
⇒ Distance covered by front wheel = 2Ï€r = 2Ï€(40)
⇒ Distance covered by front wheel = 80Ï€
∴ Distance covered by front wheel in 800 revolutions = 80Ï€ × 800
⇒ Distance covered by front wheel in 800 revolutions = 6400Ï€  eqn1
Similarly
⇒ Distance covered by rear wheel = 2Ï€R = 2Ï€(100)
⇒ Distance covered by rear wheel = 200Ï€ eqn2
Let the number of revolutions made by rear wheel to cover 6400Ï€ cm be “x”
∴ (x) × 200Ï€ = 6400Ï€ (from eqn1 and eqn2)
⇒ x = 64000Ï€/200Ï€
⇒ x = 64000/200
⇒ x = 320
Number of revolution made by rear wheel to cover the distance covered by front wheel in 800 revolutions is 320.
36. Four equal circles are described about the four corners of a square so that each touches two of the others, as shown in the figure. Find the area of the shaded region, if each side of the square measures 14 cm.
SolutionHere the distance between the center of circles touching each other is equal to the side of the square. Therefore, we can say that the radius of each circle is equal to the half of the side of the square. Now by simply calculating the area of the 4 quadrants and then subtracting it from the area of the square we can easily calculate the area of the shaded region.
Given side of square = a = 14 cm
Central angle of each sector formed at corner = Î¸ = 90°
So, radius of 4 equal circles = r = a/2 = 14/2
∴ Radius of 4 circles = r = 7 cm
Area of quadrant formed at each corner = Î¸/360 × Ï€R^{2}
Where R = radius of circle
⇒ Area of one quadrant = 90/360 × Ï€(7^{2})
= 49Ï€/4 cm^{2} ...(i)
Area of all the 4 quadrant = 4 × Area of one quadrant
= 4 × 49Ï€/4 (from eqn 1)
⇒ Area of all 4 quadrants = 49Ï€ ...(ii)
Also, Area of square = side × side = a × a = a^{2} = 14^{2} (putting value of side of square)
⇒ Area of square = 196 cm^{2}  eqn3
∴ Area of shaded region = Area of square – Area of all 4 quadrants
⇒ Area of shade region = 196  49Ï€ (from eqn3 and eqn 2)
⇒ Area of shaded region = 196 – (49 × 22/7) (put Ï€ = 22/7)
= 196 – (7 × 22)
= 196 – 154
= 42 cm^{2}
The area of shaded region is 42 cm^{2}.
37. Four equal circles, each of radius 5 cm, touch each other, as show in the figure. Find the area included between them. [Take Ï€ = 3.14]
Here, first we join the center of all adjacent circles then the distance between the center of circles touching each other is equal to the side of the square formed by joining the center of adjacent circles. Therefore, we can say that the side of the square equal to the twice of the radius of circle. Now by simply calculating the area of the 4 quadrants and then subtracting it from the area of the square we can easily calculate the area of the shaded region.
Given radius of each circle = r = 5 cm
Central angle of each sector formed at corner = Î¸ = 90°
Side of square ABCD = a = 2 × r = 2 × 5 = 10 cm
Area of quadrant formed at each corner = Î¸/360 × Ï€R^{2 }
Where R = radius of circle
⇒ Area of one quadrant = 90/360 × Ï€(5^{2})
(putting value of r and Î¸)
= 25Ï€/4 cm^{2} ...(i)
Area of all 4 quadrants = 4 × Area of one quadrant
= 4 × 25Ï€/4 (from eqn 1)
⇒ Area of all 4 quadrants = 25Ï€ ...(ii)
Area of square = side × side = a × a = a^{2}
⇒ Area of square = 10^{2} (putting value of side of square)
⇒ Area of square = 100 cm^{2} ...(iii)
Area of shaded region = Area of square – Area of all 4 quadrants
Area of shaded region = 100  25Ï€ (from eqn3 and eqn2)
= 100 – (25 × 3.14) (put Ï€ = 3.14)
= 100 – 78/.5
= 21.5 cm^{2 }
The area of shaded region is 21.5 cm^{2}.
38. Four equal circles, each of radius a units, touch each other. Show that the area between them is (6/7.a^{2}) sq. units.
Solution
Given radius of each circle = “a” units
Central angle of each sector formed at corner = Î¸ = 90°
Side of square ABCD = 2 × a units
Area of quadrant formed at each other = Î¸/360 × Ï€R^{2}
Where R = radius of circle
⇒ Area of one quadrant = 90/360 × Ï€(a^{2})
⇒ Area of one quadrant = Ï€a^{2}/4 sq. units ...(i)
∴ Area of all 4 quadrants = 4 × Area of one quadrant
⇒ Area of all the 4 quadrant = 4 × Ï€a^{2}/4 (from eqn 1)
= Ï€a^{2 }sq. units ...(ii)
Area of square = side × side = 2a × 2a = 4a^{2}
⇒ Area of square = 4a^{2} sq. units eqn3
Area of shaded region = Area of square – Area of all 4 quadrants
⇒ Area of shaded region = 4a^{2} – Ï€a^{2} (from eqn3 and eqn2)
⇒ Area of shaded region = 4a^{2} – (a^{2} × 22/7) (put Ï€ = 22/7)
⇒ Area of the shaded region = 4a^{2} – 22a^{2}/7
⇒ Area of shaded region = (28a^{2} – 22a^{2})/7
⇒ Area of shaded region = 6a^{2}/7 sq. units
Area of shaded region is 6a^{2}/7 sq. units
39. Three equal circles, each of radius 6 cm, touch one another as shown in the figure. Find the area enclosed between them. [Take Ï€ = 3.14 and √3 = 1.732]
Solution
Consider the above figure,Here, first we join the centre of all adjacent circles then the distance between the center of circles touching each other is equal to the side of an equilateral triangle formed by joining the center of adjacent circles. Therefore, we can say that the side of the equilateral triangle is equal to the twice of the radius of circle. Now by simply calculating the area of the 3 sectors and then subtracting it from the area of the equilateral triangle we can easily calculate the area of the enclosed region.
Given radius of each circle = r = 6 cm
Central angle of each sector = Î¸ = 60° (∵△ABC is equilateral)
Side of equilateral △ABC = a = 2 × r = 2 × 6
∴ Side of equilateral △ABC = a = 12 cm
Area of sector formed at each other = Î¸/360 × Ï€R^{2} where R = radius of circle
⇒ Area of one sector = 60/360 × Ï€(6^{2})
⇒ Area of one sector = 36 Ï€/6 cm^{2}
⇒ Area of one sector = 6Ï€ cm^{2} ...(i)
Area of all the 3 sector = 3 × Area of one sector
= 3 × 6Ï€ (from eqn1)
= 18Ï€ cm^{2} ...(ii)
Area of equilateral △ABC = √3/4 × a^{2} = √3/4 × (12^{2})
⇒ Area of equilateral △ABC = (√3 × 144)/4
⇒ Area of equilateral △ABC = 36△3 cm^{2} ...(iii)
Area of enclosed region = Area of equilateral △ABC – Area of all 3 sectors
⇒ Area of enclosed region = 36√3  18Ï€ (from eqn 3 and eqn 2)
⇒ Area of enclosed region = (36 × 1.732) – (18 × 3.14)
(put Ï€ = 3.14 & √3 = 1.732)
= 62.352 – 56.52
= 5.832 cm^{2}
The area of enclosed region is 5.832 cm^{2}.
40. If three circles of radius a each, are drawn such that each touches the 4 other two, prove that the area included between them is equal to 4/25.a^{2} [Take = √3 = 1.73 and Ï€ = 3.14]
Solution
Consider the figure shown below
Here, first we join the center of all adjacent circles then the distance between the center of circles touching each other is equal to the side of an equilateral triangle formed by joining the center of adjacent circles. Therefore, we can say that the side of the equilateral triangle is equal to the twice of the radius of circle. Now by simply calculating the area of the 3 sectors and then subtracting it from the area of the equilateral triangle we can easily calculate the area of the enclosed region.
Given radius of each circle = “a” units
Central angle of each sector = Î¸ = 60° (∵△ABC is equilateral)
Side of equilateral △ABC = 2 × a units
Area of sector formed at each corner = Î¸/360 × Ï€R^{2}
⇒ Area of one sector = 60/360 × Ï€(a^{2})
⇒ Area of one sector = Ï€a^{2}/6 sq. units ...(i)
∴ Area of all 3 sectors = 3 × Area of one sector
⇒ Area of all 3 sector = 3 × (Ï€a^{2})/6 (from eqn1)
= (Ï€a^{2})/2 sq. units ...(ii)
Area of equilateral △ABC = √3/4 × (2a)^{2}
= (√3 × 4a^{2})/4
= a^{2}√3 sq. units ...(iii)
Area of enclosed region = Area of equilateral △ABC – Area of all 3 sectors
⇒ Area of enclosed region
= a^{2}√3 – Ï€a^{2}/2 (from eqn 3 and eqn 2)
= a^{2} ×1.73 – (3.14 × a^{2})/2
= (a^{2} ×1.73×2 – 3.14×a^{2})/2
= (3.46 – 3.14)a^{2}/2 (taking a^{2} common)
⇒ Area of the enclosed region = 0.32a^{2}/2
= 32a^{2}/200
= 4a^{2}/25 sq. units
Area of the enclosed region is 4a^{2}/25 sq. units
41. In the given figure, ABCD is a trapezium of area 24.5 cm^{2}. If AD  BC, ∠DAB = 90°, AD = 10 cm, AD = 10 cm, BC = 4 cm, and ABE is quadrant of a circle then find the area of the shaded region.
SolutionHere in order to find the area of the shaded region we have to calculate the area, or the quadrant shown and subtract it from the area of the trapezium. And in order to find the area of the quadrant we have to calculate the radius of the sector EAB by the area of trapezium.
Given Area of trapezium ABCD = 24.5 cm^{2} ...(i)
AD  BC, AD = 10 cm, BC = 4 cm, ∠DAB = 90°
We also now Area of trapezium = 1/2 × (sum of parallel sides) × height
Area of trapezium = 1/2 × (AD + BC) × AB ...(ii)
Putting the values in equation 2, we get,
24.5 = 1/2 × (10 + 4) × AB
⇒ 24.5 = (14 × AB)/2
⇒ 24.5 = 7AB
⇒ AB = (24.5)/7
⇒ AB = 3.5 cm
Therefore radius of the sector EAB = r = 3.5 cm
Area of quadrant EAB = Î¸/360 × Ï€R^{2 }where R = radius of the sector
⇒ Area of quadrant EAB = 90/360 × Ï€(3.5^{2}) (put Ï€ = 22/7)
⇒ Area of the quadrant = 90/360 × 22/7 × 3.5 × 3.5
⇒ Area of the quadrant EAB = (22 × 3.5 × 3.5)/(4 × 7)
⇒ Area of the quadrant EAB = 269.5/28
⇒ Area of the quadrant EAB = 9.625 cm^{2} eqn3
∴ Area of shaded region = Area of trapezium – Area of quadrant EAB
⇒ Area of shaded region = 24.5 – 9.625 (putting values from eqn1 and eqn3)
⇒ Area of shaded region = 14.875 cm^{2}
42. ABCD is a field in the shape of a trapezium, AD BC, ∠ABC = 90° and ∠ADC = 90°. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m.
Find the following:(i) total area of the four sectors,
(ii) area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.
Solution
Here in order to find the area of the shaded region we have to calculate the area, or the quadrant shown and subtract it from the area of the trapezium. And in order to find the area of the quadrant we have to calculate the radius of the sector EAB by the area of trapezium.
Given AB = 30 m, AD = 55m, BC = 45 m
Î¸_{A} = 90°, Î¸_{B} = 90° Î¸_{C} = 120°, Î¸_{D} = 60°
Radius of each sector = r = 14 m
(i) total area of 4 sectors
Area of sector = Î¸_{i}/360 × Ï€R^{2} ...(i)
= Î¸A/360 × Ï€R^{2}
Area of sector at corner A = 90/360 × Ï€ × 14^{2} (putting values in eqn 1)
Area of sector at corner A = 196Ï€/4
Area of sector at corner A = 49Ï€ m^{2} ...(ii)
As we know that central angle at A and B are both 90 degrees and radius is also same i.e., 14 m therefore the area of the sector at B will be exactly same as that of sector at A.
∴ Area of sector at corner B = Area of sector at corner A
⇒ Area of sector at corner B = 49Ï€ ...(iii)
Similarly,
Area of sector = C/360 × Ï€R^{2}
Area of sector at corner C = 120/360 × Ï€ × 14^{2} (putting values in eqn 1)
Area of sector at corner C = (196Ï€)/3
Area of sector at corner C = 65.33Ï€ m^{2} ...(iv)
Similarly,
Area of sector = Î¸D/360 × Ï€R^{2}
Area of sector at corner D = 60/360 × Ï€ × 14^{2} (putting values in eqn 1)
Area of sector at corner D = 196Ï€/6
Area of sector at corner D = 32.67Ï€ ...(v)
Total area of 4 sectors = eqn2 + eqn3 + eqn4 + eqn5
⇒ Total area = 49Ï€ + 49Ï€ + 65.33Ï€ + 32.67Ï€
⇒ Total area of 4 sectors = 196Ï€
Total area of 4 sectors = 196Ï€
Total area of 4 sectors = 196 × 22/7 (put Ï€ = 22/7)
∴ Total area of 4 sectors = 616 m^{2}
Total area of 4 sectors is 616 m^{2}.
(ii) Area of the remaining portion.
Here in order to find the area of the remaining portion of the trapezium we have to subtract the area of the 4 sectors from the area of the trapezium.
Area of trapezium = 1/2 × (sum of parallel sides) × height
Area of trapezium = 1/2 × (AD + BC) × AB
On putting the values,
Area of trapezium = 1/2 × (55 + 45) × 30
= (100 × 30)/2
= 50 × 30
Area of trapezium = 1500 m^{2} ...(i)
Area of remaining portion = Area of trapezium – Area of the 4 sectors
⇒ Area of remaining potion = 1500 – 616 (from eqn 1 and part (i))
∴ Area of remaining portion = 884 m^{2}
The area of the remaining portion is 884 m^{2}.
43. Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex of an equilateral triangle of side 12 cm as centre and a sector of circle of radius 6 cm with centre B is made. [Use √3 = 1.73 and Ï€ = 3.14]
Solution
Area of shaded region can be calculated by subtracting the area of minor sector at O and area of equilateral triangle.
Given Radius of circle at O = r = 6 cm
Side of equilateral triangle = a = 12 cm
Central angle at O = 360 – 60 = 300°
Central angle at B = 60°
Area of the equilateral triangle = √3/4 × a^{2}
Where a = side of equilateral triangle
Area of the equilateral triangle = √3/4 × (12)^{2} (putting the value of a)
Area of the equilateral triangle = (144 × √3)/4
Area of the equilateral triangle = 36√3 cm^{2} ...(i)
Area of sector = Î¸/360 × Ï€R^{2} where r = radius of the sector
Area of minor sector at B = 60/360 × Ï€ × (6^{2}) (given)
∴ Area of minor sector at B = 6Ï€ cm^{2} ...(ii)
Similarly,
Area of major at O = 300/360 × Ï€(6^{2})
∴ Area of major sector at O = 30Ï€ cm^{2} ...(iii)
Area of shaded region = eqn1 + eqn3 – eqn 2
On putting values
⇒ Area of shaded region = 36√3 + 30Ï€  6Ï€
Area of shaded region = 36√3 + 24Ï€
(put Ï€ = 3.14 and √3 = 1.73)
∴ Area of shaded region = (36 × 1.73) + (24 × 3.14)
⇒ Area of shaded region = 62.28 + 75.36
∴ Area of shaded region = 137.64 cm^{2}
Area of the shaded region is 137.64 cm^{2}.
44. In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm, ∠AED = 90° and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the area of the shaded region.
SolutionHere in order to find the area of the shaded region we have to subtract the area of the semicircle and the triangle from the area of the rectangle.
Given AB = 80 cm, BC = 70 cm, DE = 42 cm, ∠AED = 90°
Here we see that the triangle AED is right angle triangle, therefore, we can apply Pythagoras theorem i.e.,
H^{2} = P^{2} + B^{2} (Pythagoras theorem)
AD^{2} = DE^{2} + AE^{2}
⇒ 70^{2} = 42^{2} + AE^{2} (putting the given values)
⇒ 4900 = 1764 + AE^{2}
⇒ 4900 – 1764 = AE^{2}
⇒ 3136 = AE^{2}
AE = √3136
∴ AE = 56 cm
Area of △ = 1/2 × AE × DE
(Area of triangle = 1/2 × base × height)
On putting values we get,
Area of △AED = 1/2 × 56 × 42
⇒ Area of △AED = 28 × 42
∴ Area of △AED = 1176 cm^{2} ...(i)
Area of semicircle = Ï€R^{2}/2
Here radius of semicircle = BC/2 = 70/2
R = 35 cm
∴ Area of semicircle = (Ï€ × 35^{2})/2
Area of semicircle = (22 × 1225)/(2× 7) (putting Ï€ = 22/7)
⇒ Area of semicircle = 11× 175
∴ Area of semicircle = 1925 cm^{2} ...(ii)
Area of rectangle = l × b (l = length of rectangle, b = breadth of rectangle)
⇒ Area of rectangle = 80 × 70 = 5600 cm^{2 }...(iii)
Area of shaded region = Area of rectangle – Area of semicircle – Area of △
⇒ Area of shaded region = 5600  1925 – 1176 (from eqn1, eqn2 and eqn3)
∴ Area of shaded region = 2499 cm^{2 }
Area of the shaded region is 2499 cm^{2}.
45. In the given figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. [Use Ï€ = 3.14]
SolutionHere in order to find the area of the shaded region (region excluding the triangle) we have to subtract the area of the triangle from the area of the rectangle and then add the area of the semicircle.
Given AB = 20 cm, DE = 12 cm, AE = 9 cm and ∠AED = 90°
Here we see that the triangle AED is right angle triangle, therefore, we can apply Pythagoras theorem i.e.,
H^{2} = P^{3} + B^{2} (Pythgoras theorem)
AD^{2} = DE^{2} + AE^{2}
AD^{2} = 12^{2} + 9^{2} (putting given values)
⇒ AD^{2} = 144 + 81
⇒ AD^{2} = 225
⇒ AD = √225
∴ AD = 15 cm
Area of △AED = 1/2 × AE × DE
(Area of triangle = 1/2 × base × height)
On putting values we get,
Area of △AED = 1/2 × 9 × 12
⇒ Area of △AED = 9 × 6
∴ Area of △AED = 54 cm^{2} ...(i)
Area of semicircle = Ï€R^{2}/2
Here radius of semicircle = BC/2 = 15/2
⇒ R = 7.5 cm
∴ Area of semicircle = (Ï€ × 7.5^{2})/2
Area of semicircle = (3.14 × 56.25) (putting Ï€ = 3.14)
⇒ Area of semicircle = 1.07 × 56.25
∴ Area of semicircle = 88.3125 cm^{2} ...(ii)
Area of rectangle = l × b (l = length of rectangle, b = breadth of rectangle)
⇒ Area of rectangle = 20 × 15 (putting the values of l & b)
∴ Area of rectangle = 300 cm^{2} ...(iii)
Area of shaded region = Area of rectangle + Area of semicircle – Area of △
⇒ Area of shaded region = 300 + 88.3125 – 53 (from eqn1, eqn2, eqn3)
∴ Area of shaded region = 334.3125 cm^{2}
Area of shaded region is 334.3125 cm^{2}.
46. In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of shaded region. [Use Ï€ = 3.14].
SolutionHere in order to find the area of the shaded region (region excluding the area of segment AC and Quadrant OCD) can be calculated by subtracting the area of triangle and quadrant OBD from the area of the circle.
Given AC = 24 cm, AB = 7 cm and ∠BOD = 90°.
Here we see that the triangle ACB is right angle triangle, therefore, we can apply Pythagoras theorem i.e.,
H^{2} = P^{2} + B^{2} (Pythagoras theorem)
BC^{2} = AC^{2} + AB^{2}
⇒ BC^{2} = 24^{2} + 7^{2} (putting the given values)
⇒ BC^{2} = 576 + 49
⇒ BC^{2} = 625
BC = √625
∴ BC = 25 cm
Area of △ACB = 1/2 × AB × AC (Area of triangle = 1/2 × base × height)
On putting values we get,
Area of △ACB = 1/2 × 7 × 24
⇒ Area of △AED = 7 × 12
∴ Area of △AED = 84 cm^{2} ...(i)
Area of circle = Ï€R^{2} (R = radius of circle)
Here radius of circle = BC/2 = 25/2 (because ABCD is a rectangle)
⇒ R = 12.5 cm
∴ Area of circle = Ï€ × 12.5^{2}
⇒ Area of circle = 156.25 × 3.14 (put Ï€ = 3.14)
∴ Are of circle = 490.625 cm^{2} ...(ii)
Area of quadrant OBD = Î¸/360 × Ï€R^{2}
Area of quadrant OBD = 90/360 × Ï€ × 12.5^{2} (put Ï€ = 3.14)
Area of quadrant = (3.14 × 156.25)/4
⇒ Area of quadrant OBD = 122.65625 cm^{2} ...(iii)
Area of shaded region = Area of circle – Area of quadrant – Area of △
⇒ Area of shaded region = 490.625 – 84 – 122.65625 (from eqn1, 2 and 3)
⇒ Area of shaded region = 283.96875 cm^{2}
Area of shaded region is 283.96875 cm^{2}.
47. In the given figure, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region. [Use √3 = 1.73 and Ï€ = 3.14.]
SolutionHere we will draw median from all the vertices of the equilateral triangle and the point at which they intersect will be the centre of the circle. Then with the help of which find out the height of the triangle and subsequently the radius of the circle and ultimately the area of the shaded region (region of equilateral triangle excluding the area of circle inscribed.)
As AD = BF = CE = h
Consider △ADB, ∠ADB = 90°, BD = 6 cm
AB^{2} = AD^{2} + BD^{2} (Pythagoras theorem)
12^{2} = AD^{2} + 6^{2} (putting the given values)
144 = AD^{2} + 36
144 – 36 = AD^{2}
So, OD = h/3
OD = (6√3)/3
OD = 2√3 cm
∴ Radius of the circle = r = 2√3 cm
Area of the circle = Ï€r^{2}
Area of the circle = Ï€ × (2√3)^{2} (putting the value of r)
∴ Area of the circle = 12Ï€ cm^{2} ...(i)
Area of △ABC = √3/4 × a^{2} where a = side of equilateral triangle
Area of △ABC = √3/4 × 12^{2}
Area of △ABC = (144 × √3)/4
Area of △ABC = 36√3 cm^{2} ...(ii)
Area of shaded region = area of triangle – area of circle
Area of the shaded region = 36√3  12Ï€ (put Ï€ = 3.14 & √3 = 1.73)
⇒ Area of the shaded region = (36 × 1.73) – (12 × 3.14)
⇒ Area of the shaded region = 62.28 – 37.68
∴ Area of the shaded region = 24.6 cm^{2}
The radius of the circle is 2√3 cm and area of shaded region is 24.6 cm^{2}
48. On a circular table cover of radius 42 cm, a design is formed by a girl leaving an equilateral triangle ABC in the middle, as shown in the figure. Find the covered area of the design. [Use √ = 1.73].
SolutionHere we will first find the sides of equilateral triangle ant then subtract the area of the triangle from the area of the circle.
Given radius of circle = r = 42 cm
∴ Area of the circle = Ï€R^{2}, where R = radius of the circle
⇒ Area of the circle = Ï€(42^{2})
∴ Area of circle = 22/7 × 1764 (putting Ï€ = 22/7)
⇒ Area of the circle = 22 × 252
∴ Area of the circle = 5544 cm^{2} ...(i)
In △ABD, ∠ADB = 90°
∴ AB^{2} = AD^{2} + BD^{2} ...(ii) (Pythagoras theorem)
Let the sides of the equilateral triangle = a cm
And as we know AD is a median therefore it will bisect the side BC into two equal parts i.e.,
BD = DC ...(iii)
Also, BC = BD + DC
⇒ BC = BD + BD (from eqn3)
⇒ a = 2BD (BC = a)
BD = a/2 cm
So, a^{2} = AD^{2} + (a/2)^{2} (putting values of AC and BD in eqn2)
⇒ a^{2 }= AD^{2} + a^{2}/4
⇒ a^{2 }– a^{2}/4 = AD^{2}
⇒ (4a^{2 }– a^{2})/4 = AD^{2}
= 3a^{2}/4 = AD^{2}
Now, we also know that the point ‘O’ which is the intersection of all the three medians i.e., centroid of the triangle. Also we know that the centroid divides the median in the ratio 2: 1,
So, we can say that AO = 2AD/3
Also, we know AO = radius = r = 42 cm
∴ 42 = 2AD/3
⇒ (42 × 3)/2 = AD
⇒ AD = 63 cm
Putting the value in equation 4,
63 = (a√3)/2
⇒ (63 × 2)/ √3 = a
⇒ 126/√3 = a
⇒ 126 × √3)/( √3 × √3) = a (rationalizing L.H.S.)
⇒ 126√3)/3 = a
⇒ a = 42√3 cm
O, area of equilateral triangle ABC = √3/4 × a^{2} (where a = side of triangle)
⇒ Area of triangle ABC = √3/4 × (42√3)^{2} (putting the value of a)
⇒ Area of triangle ABC = √3/4 × (1764 × 3)
⇒ Area of triangle ABC = √3/4 × 5292
⇒ Area of triangle ABC = 1323√2 cm^{2} ...(v)
Area of covered by design = Area of circle – Area of triangle ABC
Area covered by design = 5544  1323√3 (from eqn1 and eqn5)
⇒ Area covered by design = 5544 – (1323 × 1.73) (putting √3 = 1.73)
⇒ Area covered by design = 5544 – 2288.79
∴ Area covered by design = 3255.21 cm^{2}
Area covered by design is 3255.21 cm^{2}.
49. The perimeter of the quadrant of a circle is 25 cm. Find its area.
Solution
We know perimeter of a sector = Length of its arc + 2R ...(i)
Where R = Radius of the sector.
Perimeter = 25 cm
Also, length of arc of sector = Î¸/360 × 2Ï€R
Î¸ = 90°
∴ 25 = 90/360 × 2Ï€R + 2R (putting the values in eqn1)
⇒ 25 = 2Ï€R/4 + 2R
⇒ 25 = Ï€R/2 + 2R
⇒ 25 (Ï€R + 4R)/2 (taking 2 as L.C.M on R.H.S)
⇒ 25 × 2 = (Ï€ + 4)R (taking R common)
⇒ 50 = (22/7 + 4)R (putting Ï€ = 22/7)
⇒ 50 = {(22 + 28)/7}R (taking 7 as L.C.M. on R.H.S)
⇒ 50 × 7 = 50R
⇒ (50 × 7)/50 = R
⇒ R = 7 cm ...(ii)
Area of a sector = Î¸/360 × Ï€R^{2}
⇒ Area of quadrant = 90/360 × Ï€(7^{2}) (putting the value Î¸ and R)
⇒ Area of quadrant = 49Ï€/4 (put Ï€ = 22/7)
⇒ Area of quadrant = (49 × 22)/(4 × 7)
⇒ Area of the quadrant = (7 × 11)/2
∴ Area of the quadrant = 38.5 cm^{2}
Area of the quadrant is 38.5 cm^{2}.
50. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use Ï€ = 3.14.]
Solution
Given the radius of the circle = 42 cm
Central angle of the sector = Î¸ = 90°
Area of the minor segment = Area of sector – area of the right angle triangle
Area of the sector = 90/360 × Ï€(10^{2}) (putting the values of Î¸ and R)
⇒ Area of the sector = 100Ï€/4 (put Ï€ = 3.14)
⇒ Area of the sector = (100 × 3.14)/4
⇒ Area of sector = 25 × 3.14
∴ Area of the sector = 78.5 cm^{2} ...(i)
Area of triangle = 1/2 × base × height
⇒ Area of triangle = 1/2 × 10 × 10
∴ Area of triangle = 50 cm^{2} ...(ii)
Area of the minor segment = 78.5 – 50 (from eqn1, eqn2)
∴ Area of the minor segment = 28.5 cm^{2}
51. The radius of a circular region is 100 m. There is a road 10 m wide, running all around it. Find the area of the road and the cost of levelling it at Rs 20 per m^{2}. [Use Ï€ = 3.14]
Solution
Here we will first out the area of the road running around the circular garden and then multiplying it with rate per square meter to calculate the cost of leveling.Here we see in the figure there are two concentric circles so,
Area of road = Area of outer circle – Area of circular garden
Area of circle = Ï€R^{2} (where R = radius of circle) ...(i)
Let the radius of inner circle = r = 100 m
Also, radius of outer circle = R = 110 m (R = r + 10)
Area of outer circle = Ï€(110)^{2} ...(ii) (putting R in eqn1)
Area of inner circle = Ï€(100)^{2} ...(iii) (putting r in eqn1)
∴ Area of road = Ï€(110)^{2} – Ï€(100)^{2} (from eqn2 and 3)
⇒ Area of road = Ï€(12100 – 10000)
⇒ Area of road = 2100Ï€ (put Ï€ = 3.14)
⇒ Area of road = 2100 × 3.14
∴ Area of road = 6594 m^{2}
Cost of levelling = Rate of levelling × Area of road
⇒ Cost of travelling = 20 × 6594
∴ Cost of levelling = Rs 131880
Area of road is 6594 m^{2} and cost of levelling is Rs. 131880.
52. The area of an equilateral triangle is 49√3 cm^{2}. Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. [Take √3 = 1.73]
Solution
Each angle of triangle = Î¸ = 60°
Area of triangle not included in circles = Area of triangle – Area of all sectors
Area of all 3 sectors area equal as all the three circles are having same radius which is equal to the half of the side of the equilateral triangle.
Let the side of equilateral triangle be = a cm
Also, area of equilateral triangle = √3/4 × a^{2} (a = side of triangle)
49√3 = √3/4 × a^{2} (equating the value of area to the above equation)
⇒ (49√3 × 4)/√3 = a^{2}
⇒ a^{2} = 49 × 4
⇒ a = 7 × 2
⇒ a = 14 cm
So, radius of the circles = 7 cm
Area of sector = Î¸/360 × Ï€r^{2}
Area of sector = 60/360 × Ï€(7)^{2}
Area of sector = (22 × 49)/(6 × 7) (put Ï€ = 22/7)
Area of sector = (11 × 7)/3
Area of sector = 77/3 cm^{2}
Area of 3 sector = 3 × area of one sector
Area of 3 sector = 3 × 77/3
∴ Area of all 3 sectors = 77 cm^{2} ...(i)
Area of triangle not included in circles = 49√3 – 77 (putting value of area of triangle and sector)
Area of triangle not included = (49 × 1.73) – 77
⇒ Area of triangle not included = 84.77 – 77
∴ Area of triangle not included = 7.77 cm^{2}
Area of triangle not included in circles is 7.77 cm^{2}.
53. A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP ⟂ CD, HQ ⟂ FI and EL ⟂ DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure. [Take Ï€ = 3.14]
SolutionArea of whole figure = ar ABCD + arFGHI + ar DCIF + ar△DEF + Area of semicircle
CD = 8 cm, BP = HQ = 4 cm, DF = EF = 5 cm, CI = 8 cm
ar ABCD = arFGHI = base × height
⇒ ar ABCD = arFGHI = BP × DC
⇒ arABCD = arFGHI = 4 × 8
⇒ arABCD = arFGHI = 32 cm^{2} ...(i) & (ii)
ar DCIF = area of square = side × side
⇒ ar DCIF = DC × CI
⇒ ar DCIF = 8 × 8
⇒ ar DCIF = 64 cm^{2} ...(iii)
Consider △DEF, EF ⟂ DF and △DEF is isosceles
So, FL = LD
FL = LD = 8/2
⇒ FL = LD = 4 cm
In △DEL, ∠DLE = 90°
ED^{2} = EL^{2} + LD^{2} (Pythagoras theorem)
⇒ 5^{2} = EL^{2} + 4^{2} (putting the values)
⇒ 25 = EL^{2} + 16
⇒ 25 – 16 = EL^{2}
⇒ EL^{2} = 9
⇒ EL = √9
∴ EL = 3 cm
Area of △DEF = 1/2 × base × height
⇒ Area of △DEF = 1/2 × DF × EL
⇒ Area of △DEF = 1/2 × 8 × 3
⇒ Area of △DEF = 4 × 3
∴ Area of △DEF = 12 cm^{2} ...(iv)
Area of semicircle = Ï€R^{2}/2 where R = radius of the semicircle
R = 4 cm
Area of semicircle = Ï€(4)^{2}/2 (put Ï€ = 3.14)
Area of semicircle = (3.14 × 16)/2
⇒ Area of semicircle = 3.14 × 8
∴ Area of semicircle = 25.12 cm^{2} ...(v)
Area of whole figure = (i) + (ii) + (iii) + (iv) + (v)
⇒ Area of whole figure = 32 + 32 + 64 + 12 + 25.12
∴ Area of whole figure = 165.12 cm^{2 }
Area of the whole figure is 165.12 cm^{2}.
54. A circular disc of radius 6 cm is divided into three sectors with central angle 90°, 120° and 150°. What part of the whole circle is the sector with central angle 150°? Also, calculate the ratio of the areas of the three sectors.
Solution
Î¸_{1} = 90°, Î¸_{2} = 120°, Î¸_{3} = 150°
Radius of circle = r = 6 cm
Area of sector = Î¸/360 × Ï€R^{2} ...(i)
Area of circle = Ï€R^{2}
⇒ Area of circle = Ï€ × 6^{2}
⇒ Area of circle = 36Ï€ ...(ii)
Area of sector (Î¸_{3}) = 150/360.Ï€ × 6^{2} (from i)
Area of sector (Î¸_{3}) = 15Ï€/36 × 36
Area of sector (Î¸_{3}) = 15/36 × 36Ï€
Area of sector (Î¸_{3}) = 15/36 × Area of circle (from ii)
Area of sector (Î¸_{3}) = 5/12 × Area of circle
Area of sector (Î¸_{3}) = 15Ï€ cm^{2}
Area of sector (Î¸_{2}) = 120/360 × Ï€6^{2} (putting values in i)
Area of sector (Î¸_{2}) = 12/36 × 36 Ï€
Area of sector (Î¸_{2}) = 12Ï€ cm^{2 } ...(iii)
Area of sector (Î¸_{1}) = 90/360 × Ï€6^{2} (putting values in i)
Area of sector (Î¸_{1}) = 9/36 × 36Ï€
Area of sector (Î¸_{1}) = 9Ï€ cm^{2} ...(iv)
Ratio of three sectors = 9Ï€ : 12Ï€ : 15Ï€
Ratio of three sectors : : 3 : 4 : 5
Area of sector (Î¸_{3}) is 5/12 × Area of circle and Ratio of three sectors : : 3 : 4: 5.
55. A round table covers has six equal designs as shown in the given figure. If the radius of the cover is 35 cm then find the total area of the design. [Use = √3 = 1.732 and Ï€ = 3.14.]
SolutionTotal area of design = Area of all the minor segments
Here we will find out the area of one segment and then multiply it with 6 to get the total area of design. And as the figure inscribed in the circle is a regular hexagon this implies that it will be having all edges of same length. Therefore we can say that the angle subtended by each chord which are actually the edges of regular hexagon are equal (theorem).
Let angle subtended by chord AB on centre O be Î¸
So angle subtended = Î¸ = 360/6
∴ Angle subtended = Î¸ = 60°
Radius of circle = 35 cm
Area of sector = Î¸/360 × Ï€R^{2} ...(i)
Area of one sector = 60/360 × Ï€(35)^{2} (putting values in i)
Area of one sector = 1225Ï€/6 cm^{2 }
Area of △OAB = 1/2 × 35 × 35 × sin Î¸ (where Î¸ = central angle of sector)
Area of △OAB = 1/2 × OA × OB × sin 60
Area of △OAB = 1/2 × 1225 × √3/2 (sin 60 = √3/2)
Area of △OAB = (1225√3)/4 cm^{2}
Area of minor segment OAB = Area of sector – Area of △OAB
Area of minor segment OAB = (1225Ï€)/6 – (1225√3)/4 (put Ï€ = 3.14 & √3 = 1.732)
Area of minor segment OAB = (1225 × 3.14)/6 – (1225 × 1.732)/4
Area of minor segment OAB = (3846.5)/6 – (2121.7)/4
Area of minor segment OAB = 641.0833333 – 530.425
∴ Area of minor segment OAB = 110.65833333 cm^{2} ...(ii)
Total area of design = 6 × Area of minor segment OAB
⇒ Total area of design = 6 × 110.65833333 (from ii)
∴ Total area of design = 663.95 cm^{2}
Total area of design is 663.95 cm^{2}.
56. In the given figure, PQ = 24 cm, PR = 7 cm and 0 is the centre of the circle. Find the area of the shaded region. [Take = Ï€ = 3.14]
SolutionHere we will subtract the area of right angle triangle PQR and semicircle from the area of entire circle.
Given PQ = 24 cm, PR = 7 cm
Consider △PQR, ∠QPR = 90°
RQ^{2} = PQ^{2} + PR^{2} (Pythagoras theorem)
RQ^{2} = 24^{2 }+ 7^{2 }
⇒ RQ^{2} = 576 + 49
⇒ RQ2 = 625
∴ RQ = 25 cm
Therefore Radius of the circle = half of RQ
Let radius be ‘r’
r = 25/2
∴ r = 12.5 cm
Area △PQR = 1/2 ×base × height
Area of △PQR = 1/2 × PR × PQ
Area of △PQR = 1/2 × 7 × 24
Area of △PQR = 7 × 12
∴ Area of △PQR = 84 cm^{2} ...(i)
Area of semicircle = Ï€r^{2}/2
Area of semicircle = (3.14 × (12.5^{2})/2 (putting Ï€ = 3.14)
Area of semicircle = (3.14 × 156.25)/2
∴ Area of semicircle = 245.3125 cm^{2} ...(ii)
Area of circle = Ï€r^{2}
Area of circle = Ï€(12.5^{2})
⇒ Area of circle = 3.14 × 156.25 (putting Ï€ = 3.14)
∴ Area of circle = 490.625 cm^{2} ...(iii)
Area of shaded region = (iii)  (ii)  (i)
⇒ Area of shaded region = 490.625 – 245.3125 – 84
∴ Area of shaded region = 161.3125 cm^{2}
Area of shaded region is 161.3125 cm^{2}.
57. In the given figure, △ABC is rightangled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and 0 is the centre of the incircle of △ABC . [Take Ï€ = 3.14]
SolutionGiven AB = 6 cm, BC = 10 cm
Consider △ABC, ∠BAC = 90°
BC^{2} = AB^{2} + AC^{2} (Pythagoras theorem)
⇒ 10^{2} = 6^{2} + AC^{2} (putting given values)
⇒ 100 = 36 + AC^{2}
⇒ 100 – 36 = AC^{2}
⇒ AC^{2} = 64
∴ AC = 8 cm
Join OB, OA, OC, OE, OF, OD
Here OE = OF = OD = radius of circle = r cm
∠OEC = ∠ODB = ∠OFB = 90° (angle at the point of contact of radius & tangent)
Area △ABC = Area of △OAC + Area of △OCB + Area of △OAB ...(i)
Area of △OAC = 1/2 × OE × AC
Area of △OAC = 1/2 × r × 8
∴ Area of △OAC = 4r ...(ii)
Area of △OCB = 1/2 × OD × BC
Area of △OCB = 1/2 × r × 10
∴ Area of △OCB = 5r ...(iii)
Area of △OAB = 1/2 × OF × AB
Area of △OAB = 1/2 × r × 6
∴ Area of △OAB = 3r ...(iv)
Area of △ABC = 1/2 × AB × AC
Area of △ABC = 1/2 × 6 × 8
⇒ Area of △ABC = 3 × 8
∴ Area of △ABC = 24 cm^{2} ...(v)
Putting all the values in equation we get:
⇒ 24 = 4r + 5r + 3r
⇒ 24 = 12r
r = 24/12
∴ r = 2 cm
Area of circle = Ï€r^{2}
Put the value of r, we get,
⇒ Area of circle = Ï€ × 2^{2}
⇒ Area of circle = 3.14 × 4 (putting Ï€ = 3.14)
∴ Area of circle = 12.56 cm^{2} ...(vi)
Area of shaded region = Area of triangle – Area of circle
⇒ Area of shaded region = 24 – 12.56 (from v and vi)
∴ Area of shaded region = 11.44 cm^{2}
Area of shaded region is 11.44 cm^{2}.
58. In the given figure, △ABC is rightangled at A. A semicircle are drawn on AB, AC and BC as diameters. It is given that AB = 3 cm and AC = 4 cm. Find the area of the shaded region.
SolutionHere we will first find out the area of semicircle whose diameter is BC and then subtract the area of right angle triangle ABC from it and then we will subtract this result from the area of semicircles whose diameters area AB and AC.
Consider △ABC, ∠BAC = 90°
BC^{2} = AC^{2} + AB^{2} (Pythagoras theorem)
⇒ BC^{2} = 4^{2} + 3^{2}
⇒ BC^{2} = 16 + 9
⇒ BC^{2} = 25
∴ BC = 5 cm
Area of semicircle = Ï€r^{2}/2 ...(i)
where r = radius of the semicircle
Area of semicircle whose diameter is AC.
Radius = AC/2
Radius = 4/2
∴ Radius = 2 cm
Area of semicircle = Ï€(2)^{2}/2 (from i)
∴ Area of semicircle = 2Ï€ cm^{2}...(ii)
Area of semicircle whose diameter is AB
Radius = AB/2
Radius = 3/2
∴ Radius = 1.5 cm
Area of semicircle = Ï€(1.5)^{2}/2 (from i)
∴ Area of semicircle = 1.125Ï€ cm^{2} ...(iii)
Area of semicircle whose diameter is BC
Radius = BC/2
Radius = 5/2
∴ Radius = 2.5 cm
Area of semicircle = Ï€(2.5)^{2}/2 (from i)
∴ Area of semicircle = 3.125Ï€ cm^{2} ...(iv)
Area of triangle PQR = 1/2 × AB×AC
Area of triangle PQR = 1/2 × 3 × 4
⇒ Area of triangle PQR = 3 × 2
∴ Area of triangle PQR = 6 cm^{2} ...(v)
Now subtract equation 5 from equation 4.
⇒ Area of semicircle excluding △ABC = (iv)  (v)
⇒ Area of semicircle excluding △ABC = 3.125Ï€ – 6
⇒ Area of semicircle excluding △ABC = 3.125 × 22/7 – 6 cm^{2}
⇒ Area of semicircle excluding △ABC = (3.125 × 22)/7 – 6
⇒ Area of semicircle excluding △ABC = (68.75 – 42)/7
⇒ Area of semicircle excluding △ABC = (26.75)/7
∴ Area of semicircle excluding △ABC = 3.8214 cm^{2} ...(vi)
Area of shaded region = (iii) + (ii)  (vi)
⇒ Area of shaded region = 2Ï€ + 1.125Ï€ – 3.8214
Area of shade region = 3.125Ï€ – 3.8214
Area of shaded region = 3.125 × 22/7 – 3.8214
Area of shaded region = 68.75/7 – 3.8214
⇒ Area of shaded region = 9.8214 – 3.8214
∴ Area of shaded region = 6 cm^{2}
Area of shaded region 6 cm^{2}.
59. PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semicircles are drawn with PQ and QS as diameters, as shown in the given figure. If PS = 12 cm, find the perimeter and area of the shaded region. [Take Ï€ = 3.14]
SolutionHere we will subtract the area of semicircle whose diameter is QS from the area of the semicircle whose diameter PS and add the area of semicircle whose diameter is PQ so as to find out the area of the shaded region.
Given PS = 12 cm.
Radius of the circle = 6 cm
PQ = QR = RS
So, let PQ = QR = RS = k cm
Also, PQ + QR + RS = PS
⇒ k + k + k = 12
⇒ 3k = 12
k = 12/3
∴ k = 4 cm
So, PQ = QR = RS = 4 cm
Area of semicircle = Ï€r^{2}/2 eqn1 where r = radius of the semicircle
Area and perimeter of semicircle whose diameter is PS
Radius = PS/2
Radius = 12/2
∴ Radius = 6 cm
Area of semicircle = Ï€(6)^{2}/2 (from i)
∴ Area of semicircle = 18Ï€ cm^{2} ...(ii)
Perimeter of semicircle = Ï€r
⇒ Perimeter of semicircle = Ï€ × 6
∴ Perimeter of semicircle = 6Ï€ cm ...(iii)
Area of semicircle whose diameter is QS
Radius = QS/2
Radius = 8/2
⇒ Radius = 4 cm
Area of semicircle = Ï€(4)^{2}/2 (from i)
∴ Area of semicircle = 8Ï€ cm^{2} ...(iv)
Perimeter of semicircle = Ï€r
⇒ Perimeter of semicircle = Ï€ × 4
∴ Perimeter of semicircle = 4Ï€ cm ...(v)
Area of semicircle whose diameter is PQ
Radius = PQ/2
Radius = 4/2
∴ Radius = 2 cm
Area of semicircle = Ï€(2)^{2}/2 (from i)
∴ Area of semicircle = 2Ï€ cm^{2} ...(vi)
Perimeter of semicircle = Ï€r
⇒ Perimeter of semicircle = Ï€ × 2
∴ Perimeter of semicircle = 2Ï€ cm ...(vii)
Area of the shaded region = eqn2 – eqn4 + eqn6
Area of shaded region = 18Ï€  8Ï€ + 2Ï€
⇒ Area of shaded region = 12Ï€
⇒ Area of shaded region = 12 × 3.14 (putting Ï€ = 3.14)
∴ Area of shaded region = 37.68 cm^{2}
Perimeter of shaded region = (iii)  (v) + (vii)
⇒ Perimeter of shaded region = 6Ï€  4Ï€ + 2Ï€
⇒ Perimeter of shaded region = 4Ï€
⇒ Perimeter of shaded region = 4 × 3.14 (put Ï€ = 3.14)
∴ Perimeter of shaded region = 12.56 cm
Perimeter of the shaded region is 12.56 cm and Area of shaded region is 37.68 cm^{2}.
60. The inside perimeter of a running track shown in figure is 400 m. The length of each of the straight portion is 90 m, and the ends are semicircles. If the track is 14 m wide everywhere, find the area of the track. Also, find the length of the outer boundary of the track. Also, find the length of the outer boundary of the track.
Solution
Consider the figure as a combination of two semicircles on the ends of the rectangle.
Let the length of rectangle be ‘L’ m and breadth be ‘B’ cm
Given L = 90 m, W = 14 m
Perimeter of running track = 400 m
Perimeter of inside of running track = 2L + Arc of two semicircles ...(i)
Arc length of a semicircle = Ï€r where r = radius
⇒ 400 = (2 × 90) + (2 × Ï€r) (putting values in eqn1)
⇒ 400 = 18 + 2Ï€r
⇒ 400  180 = 2Ï€r
⇒ 220 = 2Ï€r
⇒ 220 = 2 × 22/7 × r (Ï€ = 22/7)
⇒ 220 = 44/7 × r
⇒ (220 × 7)/44 = r
∴ r = 35 m
Area of inner of running track = Area of rectangle + 2 × area of semicircles ...(ii)
Area of rectangle = L × B
Here B = 2r
B = 70 m
⇒ Area of inner rectangle = 90 × 70
∴ Area of inner rectangle = 6300 m^{2} ...(iii)
Area of inner semicircles = 2 × Ï€r^{2}/2
Area of inner semicircles = 2 × Ï€(35)^{2}/2
⇒ Area of inner semicircles = 1225Ï€
Area of inner semicircles = 1125 × 22/7
⇒ Area of inner semicircle = 175 × 22
∴ Area of inner semicircle = 3850 m^{2} ...(iv)
Area of inner of running track = 6300 + 3850 (from 3& 4)
∴ Area of inner of running track = 10150 m^{2}
Now radius of semicircles of outer of the running track = R = r + W
⇒ R = 35 + 14
∴ R = 49 m
Area of outer semicircles = 2 × Ï€R^{2}/2
Area of outer semicircles = 2 × Ï€(49)^{2}/2
⇒ Area of outer semicircles = 2401Ï€
⇒ Area of outer semicircles = 2401 × 22/7
⇒ Area of outer semicircles = 343 × 22
∴ Area of outer semicircle = 7546 m^{2} ...(v)
Breadth of outer running track = B’ = 2R
⇒ B’ = 2 × 49
∴ B’ = 98 m
Area of outer rectangle = L × B’
⇒ Area of outer rectangle = 90 × 98
∴ Area of outer rectangle = 8820 m^{2} ...(vi)
Area of entire ground = 8820 + 7546 (from 5 & 6)
∴ Area of entire ground = 16366 m^{2}
Area of running track = Area of entire ground – Area of inner ground
⇒ Area of running track = 16366  10150
∴ Area of running track = 6216 m^{2 }
Perimeter of outer boundary = 2L + Arc of outer semicircles
Arc length of an outer semicircle = Ï€R, where R = outer radius
Perimeter of outer boundary = (2 × 90) + (2 × 22/7 × 49) (putting values)
⇒ Perimeter of outer boundary = 180 + (2 × 22 × 7)
⇒ Perimeter of outer boundary = 180 + 308
∴ Perimeter of outer boundary = 488 m
Area of running track is 6216 m^{2} and perimeter of outer boundary is 488 m.