Ex 12.1, 8 - Chapter 12 Class 10 Surface Areas and Volumes

Last updated at April 16, 2024 by Teachoo

Transcript

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.
Now,
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
∴ Radius = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 = 1.4/2 = 0.7 cm
& Height = h = 2.4 cm
Now,
Curved surface area of cylinder = 2𝜋𝑟ℎ
= 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 = 𝝅𝒓𝟐
= 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 = 𝝅𝒓𝒍
Now,
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
Now,
Curved surface area of cone = 𝜋𝑟𝑙
= 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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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