Ex 12.1,8
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Since cylinder is solid, it has a base
also whose area is also to be calculated.
Total surface area of remaining solid
= Curved Surface Area of cylinder + Area of cylinder base
+ Curved surface area of cone
Curved Surface Area of cylinder
Diameter = 1.4 cm
So, radius = /2 = 1.4/2 = 0.7 cm
& Height = 2.4 cm
Curved surface area of cylinder = 2
= 2 22/7 (0.7) 2.4
= 2 22 0.1 2.4
= 10.56 cm2
Area of cylinder base
Base of cylinder is a circle with
radius = radius of cylinder = 0.7 cm
So, Area of base = 2
= 22/7 (0.7)2
= 22/7 0.7 0.7
= 22 0.1 0.7
= 1.54 cm2
Curved Surface area of cone
Curved Surface area of cone =
Radius of cone = Radius of cylinder = 0.7 cm
& Height of cone = h = 2.4 cm
Now, we find slant height (l)
We know that
l2 = h2 + r2
l2 = (2.4)2 + (0.7)2
l2 = (2.4 2.4 )2 + (0.7 0.7)2
l2 = 5.76 + 0.49
l2 = 6.25
l = 6.25
l = (625/100)
l = ((5 5 5 5)/(10 10))
l = (5 5)/10
l = 25/10
l = 2.5 cm
Curved surface area of cone =
= 22/7 0.7 2.5
= 22 0.1 2.5
= 5.5 cm2
Hence,
Total surface area of remaining solid
= Curved Surface Area of cylinder + Area of cylinder base
+ curved surface area of cone
= 10.56 + 1.54 + 5.5
= 17.6 cm2
Hence, area of remaining solid surface = 17.6 cm2
= 18 cm2 (approximately)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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