Question

Derive the formula for the curved surface area and total surface area of the frustum of cone.

Answer 1

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let r₁ and r₂ be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

In ΔABG and ΔADF, DF||BG

∴ ΔABG ∼ ΔADF

DF/BG = AF/AG =AD/AB

${\displaystyle \frac{r_2}{r_1}=\frac{h_1-h}{h_1}=\frac{l_1-l}{l_1}}$

${\displaystyle \frac{r_2}{r_1}=1-\frac{h}{h_1}=1-\frac{1}{l_1}}$

${\displaystyle l-\frac{l}{l_1}=\frac{r_2}{r_1}}$

${\displaystyle \frac{l}{l_1}=1-\frac{r_2}{r_1}=\frac{r_1-r_2}{r_1}}$

${\displaystyle \frac{l_1}{l}=\frac{r_1}{r_1-r_2}}$

${\displaystyle l_1=\frac{r_{1}l}{r_1-r_2}}$

CSA of frustum DECB = CSA of cone ABC − CSA cone ADE

${\displaystyle =\pi r_1{l}_1-\pi r_2(l_1-l)}$

${\displaystyle =\pi r_1\left(\frac{l{r}_1}{r_1-r_2}\right)-\pi r_2[\frac{r_{1}l}{r_1-r_2}-l]}$

${\displaystyle =\frac{\pi r_1^{2}l}{r_1-r_2}-\pi r_2\left(\frac{r_{1}l-r_{1}l+r_{2}l}{r_1-r_2}\right)}$

${\displaystyle =\frac{\pi r_1^{2}l}{r_1-r_2}-\frac{\pi r_2^{2}l}{r_1-r_2}}$

${\displaystyle =\pi l\left[\frac{r_1^2-r_2^2}{r_1-r_2}\right]}$

CSA of frustum = Π(r₁ + r₂)l

Total surface area of frustum = CSA of frustum + Area of upper circular end + Area of lower circular end

${\displaystyle =\pi (r_1+r_2)l+\pi r_2^2+\pi r_1^2}$

${\displaystyle =\pi [(r_1+r_2)l+r_1^2+r_2^2]}$