Question

$\tt\large\purple{\underbrace{Question:-}}$
If S denotes the area of curved surface of right circular cone of height h and semivertical angle (alpha) then S equal to = ?​

Answer 1

→ Hey Mate:-

→ Given Question:-

→ If S denotes the area of the curved surface of a right circular cone of height h and semivertical angle (alpha) then S equal to = ?​

→ To Find:-

→ S equal to?

→ Solution:-

⇒  s = πrl ,

⇒  r = radius

⇒  $l$  = slant height

⇒  h = Height

⇒ Now , $\frac{r}{h}$ = tan α ⇒ r = h tan α

⇒  $l =\sqrt{h^2 + r^2}$

⇒    $= \sqrt{h^2 + r^2 tan ^2 a}$

⇒   $= \sqrt{h^2(1 + tan^2a)}$

⇒   $= \sqrt{h^2 sec^2 a}$  $= h sec a$

⇒  ∴ s = π .h tan α. h sec α = πh² sec α tan α.

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Answer 2

Consider a cylindrical element of radius r and slant width dl at a vertical distance of x units and slant distance of l units along the curved surface from the vertex of the right circular cone of height h and semivertical angle α, as shown in the figure.

The curved surface area of the element is,

$\small\longrightarrow\sf{dA=2\pi r\ dl\quad\dots(1)}$

But in the right triangle ABC,

$\small\text{ \longrightarrow\sf{\sin\alpha=\dfrac{r}{l}} }$

$\small\longrightarrow\sf{r=l\sin\alpha}$

Then (1) becomes,

$\small\longrightarrow\sf{dA=2\pi l\sin\alpha\ dl}$

Now the total curved surface area of the cone is,

$\small\displaystyle\longrightarrow\sf{A=\int\limits_0^L2\pi l\sin\alpha\ dl}$

$\small\displaystyle\longrightarrow\sf{A=2\pi\sin\alpha\int\limits_0^Ll\ dl}$

$\small\displaystyle\longrightarrow\sf{A=\pi\sin\alpha\left[l^2\right]_0^L}$

$\small\displaystyle\longrightarrow\sf{A=\pi L^2\sin\alpha\quad\dots(2)}$

But in the right triangle ADE,

$\small\text{ \longrightarrow\sf{\cos\alpha=\dfrac{h}{L}} }$

$\small\text{ \longrightarrow\sf{L=\dfrac{h}{\cos\alpha}} }$

Then (2) becomes,

$\small\text{ \displaystyle\longrightarrow\sf{A=\pi\cdot\dfrac{h^2}{\cos^2\alpha}\cdot\sin\alpha} }$

$\small\text{ \displaystyle\longrightarrow\sf{A=\pi h^2\cdot\dfrac{1}{\cos\alpha}\cdot\dfrac{\sin\alpha}{\cos\alpha}} }$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{\sf{A=\pi h^2\sec\alpha\tan\alpha}}} }$