Question

The total surface area of a frustum of cone is calculated as ________________.

• A. SA = $\pi (r+R)\sqrt{(R-r{)}^2+h^2}+\pi r^2+\pi R^2$
• B. SA = $\pi (r-R)\sqrt{(R-r{)}^2+h^2}+\pi r^2+\pi R^2$
• C. SA = $\pi (r+R)\sqrt{(R-r{)}^2+h^2}+\pi r^2-\pi R^2$
• D. SA = $\pi (r+R)\sqrt{(R-r{)}^2+h^2}-\pi r^2+\pi R^2$

Answer 1

The correct option is A SA = $\pi (r+R)\sqrt{(R-r{)}^2+h^2}+\pi r^2+\pi R^2$

Image 1

Let us assume the frustum is cut an open

Image 2

By ratio and proportion

$\frac{l_1}{R}=\frac{l}{R-r}\Rightarrow l_1=\frac{Rl}{R-r}$

From figure,

$l_2=l_1-l$

$l_2=\frac{Rl}{R-r}-l\Rightarrow \frac{Rl(R-r)l}{R-r}$

$l_2=\frac{rl}{R-r}$

The length of the arc is circumference of base

$S_1=2\pi r,S_2=2\pi r$

From figure,

$A=\frac{1}{2}{S}_1{l}_1-\frac{1}{2}{S}_2{l}_2$$=\frac{1}{2}\left(2\pi R\right)\left(\frac{Rl}{R-r}\right)-\frac{1}{2}\left(2\pi r\right)\left(\frac{rl}{R-r}\right)$

$\Rightarrow \frac{\pi R^{2}l-\pi r^{2}l}{R-r}$

$A=\frac{\pi (R^2-r^2)l}{R-r}$

$A=\frac{\pi (R-r)(R+r)l}{(R-r)}$

$A=\pi (R+r)l$

Now, $l=\sqrt{{(R-r)}^2+h^2}$

$A=\pi (R+r)\sqrt{{(R-r)}^2+h^2}$ (Lateral surface area)

Total area

$A=\pi (R+r)\sqrt{{(R-r)}^2+h^2}+\pi r^2+\pi R^2$