Ex 12.1, 5 - Chapter 12 Class 10 Surface Areas and Volumes

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 12.1, 5
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to edge of the cube. Determine the surface area of the remaining solid.
Given that
Diameter of the hemisphere is equal to the edge of the cube
So, Diameter = Side of cube = l
Here, base of hemisphere would not be included in the total solid area of wooden cube .
Now,
Surface area of solid = Area of cube
+ Curved surface area of hemisphere
– Base area of hemisphere
Area of cube
Here, side = l
Area of cube = 6(Side)2
= 6l2
Curved surface area of hemisphere
Diameter of hemisphere = l
∴ Radius = r = (Diameter )/2 = 𝒍/𝟐
Curved Surface area of hemisphere = 2𝜋𝑟2
= 2𝜋(𝑙/2)^2
=2𝜋 𝑙^2/4
= (𝜋𝑙^2)/2
Curved Surface area of hemisphere = 2𝜋𝑟2
= 2𝜋(𝑙/2)^2
= 2𝜋 𝑙^2/4
= (𝝅𝒍^𝟐)/𝟐
Base area of hemisphere
Base of hemisphere is a circle with
Radius = Radius of hemisphere = r = 𝒍/𝟐
Base area of hemisphere = 𝝅𝒓𝟐
= 𝜋(𝑙/2)^2
= 𝜋 𝑙^2/4
= (𝝅𝒍^𝟐)/𝟒
Now,
Surface area of solid = Area of cube
+ Curved surface area of hemisphere
– Base area of hemisphere
= 6l2 + (𝝅𝒍^𝟐)/𝟐 – (𝝅𝒍^𝟐)/𝟒
= 6l2 + (2𝜋𝑙^2 − 𝜋𝑙^2)/4
= 6l2 + (𝝅𝒍^𝟐)/𝟒
= l2 (6 + 𝜋/4)
= l2 ((6(4) + 𝜋)/4)
= l2 ((24 + 𝜋)/4)
= 𝟏/𝟒 l2 (𝝅 + 24)
Hence, Surface area of solid = 1/4 l2 (𝜋 + 24)

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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