If p. x, y are height, curved surface area and volume of cone respectively, prove
that 3 yp- xp+9y2 = 0
Answers
Answer:
If p,\ xp, x and yy are height, curved surface area and volume of a cone respectively, prove that 3\pi yp^3-x^2p^2+9y^2=0.3πyp
3
−x
2
p
2
+9y
2
=0.
Solution:-
Let rr and ll be base radius and slant height of the cone respectively.
Then we get,
\longrightarrow l^2=p^2+r^2\quad\quad\dots(1)⟶l
2
=p
2
+r
2
…(1)
Volume of the cone,
\longrightarrow y=\dfrac{1}{3}\,\pi r^2p⟶y=
3
1
πr
2
p
From this, radius is given by,
\longrightarrow r^2=\dfrac{3y}{\pi p}\quad\quad\dots(2)⟶r
2
=
πp
3y
…(2)
Putting (2) in (1),
\longrightarrow l^2=p^2+\dfrac{3y}{\pi p}⟶l
2
=p
2
+
πp
3y
\longrightarrow l^2=\dfrac{\pi p^3+3y}{\pi p}⟶l
2
=
πp
πp
3
+3y
\longrightarrow \pi pl^2=\pi p^3+3y\quad\quad\dots(3)⟶πpl
2
=πp
3
+3y…(3)
Curved surface area of the cone,
\longrightarrow x=\pi rl⟶x=πrl
Squaring both sides,
\longrightarrow x^2=\pi^2r^2l^2⟶x
2
=π
2
r
2
l
2
From (2),
\longrightarrow x^2=\pi^2\cdot\dfrac{3y}{\pi p}\cdot l^2⟶x
2
=π
2
⋅
πp
3y
⋅l
2
\longrightarrow x^2=\dfrac{3y\pi l^2}{p}⟶x
2
=
p
3yπl
2
\longrightarrow x^2p=3y\pi l^2⟶x
2
p=3yπl
2
Multiplying both sides by p,p,
\longrightarrow x^2p^2=3y\pi pl^2⟶x
2
p
2
=3yπpl
2
From (3),
\longrightarrow x^2p^2=3y(\pi p^3+3y)⟶x
2
p
2
=3y(πp
3
+3y)
\longrightarrow x^2p^2=3\pi yp^3+9y^2⟶x
2
p
2
=3πyp
3
+9y
2
\longrightarrow\underline{\underline{3\pi yp^3-x^2p^2+9y^2=0}}⟶
3πyp
3
−x
2
p
2
+9y
2
=0
Hence Proved!
If p. x, y are height, curved surface area and volume of cone respectively, prove
that 3 yp- xp+9y2 = 0
refer the attachment
