Curved Surface Area and Total Surface Area of Hemisphere
Curved surface area = 2Πr^{2} Hollow hemisphere = 2Π(R^{2} + r^{2}) sq. units Total surface area of = 3Πr^{2} Hollow hemisphere = Π(3R^{2} + r^{2}) sq. units |
Problem 1 :
The radius of a sphere increases by 25%. Find the percentage increase in its surface area.
Solution :
Radius of hemisphere = r
Increased percentage of radius = 125% of r
= 125r/100
= 5r/4
Surface area of hemisphere = 2Πr^{2}
Surface area of increased hemisphere = 2Π(5r/4)^{2}
= 25Πr^{2}/8
= [(25Πr^{2}/8) - (2Πr^{2})]/(2Πr^{2})
= 9Πr^{2}/8(2Πr^{2})
= 9Πr^{2}/16 Πr^{2}
= 0.5625
= 0.5625 (100)
= 56.25%
Problem 2 :
The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost to paint the vessel all over at ₹0.14 per cm^{2}
Solution :
internal radius (r) = 20/2 = 10 cm
External radius (R) = 28/2 = 14 cm
Curved surface area of hemisphere = 2Π(R^{2} + r^{2})
= 2 (22/7)(14^{2} + 10^{2})
= 2 (22/7)(196 + 100)
= 1860.57
Cost of painting = ₹0.14 per cm^{2}
Required cost = 0.14 (1860.57)
= ₹260.48
Problem 3 :
What will be the diameter of a sphere whose surface area is 55.44 m^{2}?
(a) 4.6 m (b) 4.2 m (c) 3.4 m (d) 2.4 m
Solution :
Surface area of sphere = 4Πr^{2}
4Πr^{2 }= 55.44
4 x (22/7) x r^{2 }= 55.44
r^{2 }= 55.44 x (1/4) x (7/22)
r^{2 }= 4.41
r = √4.41
r = 2.1
Diameter = 2r
= 2 (2.1)
= 4.2 m
So, the required diameter is 4.2 m.
Problem 4 :
The curved surface area of a sphere of radius 7 cm is
Solution :
Curved surface area of the sphere = 4Πr^{2}
Here radius (r) = 7 cm
= 4 x (22/7) x 7 x 7
= 616 cm^{2}
Problem 5 :
Find the total surface area of a hemisphere of radius r/2 unit
Solution :
Total surface area of hemisphere = 3Πr^{2}
radius = r/2
= 3 x Π x (r/2)^{2}
= 3 x Π x (r^{2}/4)
= (3/4) Π r^{2 }
Problem 6 :
The radius of a spherical balloon increases from 7 cm to 14 cm when air is being pumped into it. Then find the ratio of surface area of the balloon in the two cases.
Solution :
Old radius = 7 cm and new radius = 14 cm
Surface area of balloon = 4Πr^{2}
Let old radius be r and new radius be R.
= 4ΠR^{2 }: 4Πr^{2}
= 4Π(14)^{2 }: 4Π(7)^{2}
= (14)^{2 }: (7)^{2}
= 4 : 1
So, the required ratio is 4 : 1.
Problem 7 :
A cylinder, a cone and a sphere are of the same radius and same height. Find the ratio of their curved surface.
Solution :
Radius of cylinder = radius of cone = radius of sphere = r = h
Surface area of cylinder = 2Πrh
Surface area of sphere = 4Πr^{2}
Surface area of cone = Πrl
l = √r^{2} + h^{2}
l = √r^{2} + r^{2}
l = √2r^{2}
l = r√2
Surface area of cylinder : Surface area of cone : Surface area of sphere
= 2Πrh : Πrl : 4Πr^{2}
= 2r(r) : r (r√2) : 4r^{2}
= 2r^{2} : r^{2}√2 : 4r^{2}
= 2 : √2 : 4
= 1 : √2 : 2
= 1 : 2 : 4
So, the required ratio is 1 : 2 : 4.
Problem 8 :
Find the radius of a sphere with a surface area of 1024π square inches.
Solution :
Surface area of sphere = 4Πr^{2}
1024π = 4Πr^{2}
1024 = 4r^{2}
r^{2 }= 1024/4
r^{2 }= 256
r = 16
So, the radius of the sphere is 16 inches.
Problem 9 :
You friend claims that if the radius of a sphere is doubled, then the surface area of the sphere will also be doubled. Is your friend correct? Explain your reasoning.
Solution :
Let radius of the sphere be r, then when it is double the new radius will be 2r.
Surface area of sphere = 4Πr^{2}
When r = 2r
Surface area of new sphere = 4Π(2r)^{2}
= 16Πr^{2}
= 4 (4Πr^{2})
= 4(surface area of old sphere)
So, it is incorrect.
C.S.A. of a frustum = Π(R +r)l sq. units T.S.A. of a frustum = Π(R + r)l + Π(R^{2}+ r^{2}) where, l = √[h^{2} +(R −r)^{2}] |
Problem 10 :
Th e frustum shaped outer portion of the table lamp has to be painted including the top part. Find the total cost of painting the lamp if the cost of painting 1 sq.cm is ₹2.
Solution :
Curved surface area of lamp = Π(R +r)l sq. units
R = 12 m, r = 6 m and h = 8 m
l = √[h^{2} +(R −r)^{2}]
l = √[8^{2} +(12 −6)^{2}]
l = √[64 + 6^{2}]
l = √(64 + 36) = √100
l = 10
C.S.A of frustum shaped lamp = Π(R +r)l
Area to be painted = Π(R +r)l + Πr^{2}
= (22/7)[(12 + 6)(10) + 6^{2}]
= (22/7)[180+36]
= 678.85
Cost per sq.cm is ₹2
Required cost = 678.85 (2)
= ₹1357.71
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