Question

If p. x, y are height, curved surface area and volume of cone respectively, prove
that 3 yp- xp+9y2 = 0​

Answer 1

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!

Answer 2

$\bf\huge{Question:-}$

If p. x, y are height, curved surface area and volume of cone respectively, prove

that 3 yp- xp+9y2 = 0

$\bf\huge{answer:-}$

refer the attachment