Prove that th^2=ch^2+4v^2+8v^2rh where t,c,v,h and r is total surface area curved surface area volume hieght and radius of a cylinder
Answers
Answer:
For A cylinder
The curved surface area = c
The total surface area = t
The volume = v
To Proved :
t h² = c h² + 4 v² + 8 v² r h
Solution :
We know that
Curved surface area of cylinder = 2 π r h
where r = radius
h = height
So, c = 2 π r h .........1
And
Total surface area of cylinder = 2 π r h + 2 π r²
where r = radius
h = height
So, t = 2 π r h + 2 π r² ,............2
And
The volume of cylinder = π r² h
where r = radius
h = height
So, v = π r² h ........3
Now,
The dimensions of both t h² and c h² are L^{4} M^{0} T^{0}L
4
M
0
T
0
And The dimensions of 4 v² is L^{6} M^{0} T^{0}L
6
M
0
T
0
And The dimensions of 8 v²r h is L^{8} M^{0} T^{0}L
8
M
0
T
0
So, we have to prove t h² = c h² + 2 v h
Therefor
From Right Hand Side of equation
i.e c h² + 2 v h = ( 2 π r h ) × h² + 2 × ( π r² h ) ×h
= 2 π r h × h² + 2 π r² h²
= 2 π r h³ + 2 π r² h²
From Left Hand Side of equation
i.e t h² = ( 2 π r h + 2 π r² ) × h²
= 2 π r h × h² + 2 π r² × h²
= 2 π r h³ + 2 π r² h²
Hence, c h² + 2 v h = t h²