Question

State and Prove the Volume of Frustum of a Cone:
To Prove:
Volume of Frustum of a Cone:

$\frac{1}{3} \pi h( {r}^{2} + {a}^{2} + ar)$
r - Small Radius
a - Big Radius
a > r
​

Answer 1

Volume of Frustum:

The volume of frustum of a cone is given by:

$\Longrightarrow \sf \dfrac{1}{3} \pi h \ \Big\{r^2 + a^2 + ar\Big\}$

Where:

"r" = Radius of the upper surface of the frustum.

"a" = Radius of the base of the frustum.

"h" = Height of the frustum.  

We also know that:

$\Longrightarrow \sf Volume \ of \ a \ cone = \dfrac{1}{3} \pi r^2 h$

Where:

r = Radius of cone.

h = Height of the cone.

Proof:

Dimensions of the whole cone.

Height - H

Slant Height - L

Radius - a

Dimensions of cone ADE.

Height - h₁

Radius - r

Slant height- l₁

Dimensions of the frustum DECB.

Height - h

Radius - a

Slant height - l

In ΔAFE and ΔAGC:

∠A = ∠A (Common angle)

∠AFE = ∠AGC = 90°

∴ By using AA (Angle-Angle) similarity criterion we can say that:

⇒ ΔAFE ≈ ΔAGC

We know that corresponding parts of similar triangles taken in order are proportional.

$\Longrightarrow \sf \dfrac{GC}{FE} = \dfrac{AG}{AF}$

$\Longrightarrow \sf \dfrac{a}{r} = \dfrac{H}{h_1}$

We know that H = h₁ + h, Substitute this in place of H.

$\Longrightarrow \sf \dfrac{a}{r} = \dfrac{h_1 + h}{h_1}$

$\Longrightarrow \sf \dfrac{a}{r} = \dfrac{h_1}{h_1} + \dfrac{h}{h_1}$

$\Longrightarrow \sf \dfrac{a}{r} = 1 + \dfrac{h}{h_1}$

$\Longrightarrow \sf \dfrac{a}{r} - 1 = \dfrac{h}{h_1}$

$\Longrightarrow \sf \dfrac{a - r}{r} = \dfrac{h}{h_1}$

$\Longrightarrow \sf \dfrac{a - r}{hr} = \dfrac{1}{h_1}$

$\Longrightarrow \sf h_1 = \dfrac{hr}{a - r}$

$\Longrightarrow \sf h_1 = h \Bigg(\dfrac{r}{a - r} \Bigg) \longmapsto Eq\textcircled{\scriptsize \sf 1}$

We know that:

$\Longrightarrow \sf Vol(Frustum) = Vol(Cone \ ABC) - Vol(Cone \ ADE)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi a^2 H - \dfrac{1}{3} \pi r^2 h_1$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 H - r^2h_1 \Bigg)$

Substitute the following values:

⇒ H = h₁ + h

$\Longrightarrow \sf h_1 = h \Bigg(\dfrac{r}{a - r} \Bigg) \Big[ From \ Eq\textcircled{\scriptsize \sf 1} \Big]$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2\big(h_1 + h\big) - r^2 h \bigg(\dfrac{r}{a - r} \bigg) \Bigg)$

Substitute Eq(1) in h₁ again.

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 \Bigg[h \bigg(\dfrac{r}{a - r} \bigg) + h\Bigg] - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 \Bigg[\dfrac{hr}{a - r} + h\Bigg] - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 \Bigg[\dfrac{hr + h(a - r)}{a - r}\Bigg] - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 \Bigg[\dfrac{hr + ha - hr}{a - r}\Bigg] - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(\dfrac{ha^3}{a - r} - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(\dfrac{ha^3 - r^3h}{a - r} \Bigg)$

Take 'h' outside the brackets since it's a common term.

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi h \Bigg(\dfrac{a^3 - r^3}{a - r} \Bigg)$

Using a³ - b³ = (a - b)(a² + ab + b²) we get:

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi h \Bigg(\dfrac{(a - r)(a^2 + ar + r^2)}{a - r} \Bigg)$

(a - r) get's cancelled.

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi h \Bigg(a^2 + ar + r^2\Bigg)$

Hence proved.

Answer 2

Answer:

Volume of Frustum:

The volume of frustum of a cone is given by:

$\Longrightarrow \sf \dfrac{1}{3} \pi h \ \Big\{r^2 + a^2 + ar\Big\}$

Where:

"r" = Radius of the upper surface of the frustum.

"a" = Radius of the base of the frustum.

"h" = Height of the frustum.  

We also know that:

$\Longrightarrow \sf Volume \ of \ a \ cone = \dfrac{1}{3} \pi r^2 h$

Where:

r = Radius of cone.

h = Height of the cone.

Proof:

Dimensions of the whole cone.

Height - H

Slant Height - L

Radius - a

Dimensions of cone ADE.

Height - h₁

Radius - r

Slant height- l₁

Dimensions of the frustum DECB.

Height - h

Radius - a

Slant height - l

In ΔAFE and ΔAGC:

∠A = ∠A (Common angle)

∠AFE = ∠AGC = 90°

∴ By using AA (Angle-Angle) similarity criterion we can say that:

⇒ ΔAFE ≈ ΔAGC

We know that corresponding parts of similar triangles taken in order are proportional.

$\Longrightarrow \sf \dfrac{GC}{FE} = \dfrac{AG}{AF}$

$\Longrightarrow \sf \dfrac{a}{r} = \dfrac{H}{h_1}$

We know that H = h₁ + h, Substitute this in place of H.

$\Longrightarrow \sf \dfrac{a}{r} = \dfrac{h_1 + h}{h_1}$

$\Longrightarrow \sf \dfrac{a}{r} = \dfrac{h_1}{h_1} + \dfrac{h}{h_1}$

$\Longrightarrow \sf \dfrac{a}{r} = 1 + \dfrac{h}{h_1}$

$\Longrightarrow \sf \dfrac{a}{r} - 1 = \dfrac{h}{h_1}$

$\Longrightarrow \sf \dfrac{a - r}{r} = \dfrac{h}{h_1}$

$\Longrightarrow \sf \dfrac{a - r}{hr} = \dfrac{1}{h_1}$

$\Longrightarrow \sf h_1 = \dfrac{hr}{a - r}$

$\Longrightarrow \sf h_1 = h \Bigg(\dfrac{r}{a - r} \Bigg) \longmapsto Eq\textcircled{\scriptsize \sf 1}$

We know that:

$\Longrightarrow \sf Vol(Frustum) = Vol(Cone \ ABC) - Vol(Cone \ ADE)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi a^2 H - \dfrac{1}{3} \pi r^2 h_1$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 H - r^2h_1 \Bigg)$

Substitute the following values:

⇒ H = h₁ + h

$\Longrightarrow \sf h_1 = h \Bigg(\dfrac{r}{a - r} \Bigg) \Big[ From \ Eq\textcircled{\scriptsize \sf 1} \Big]$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2\big(h_1 + h\big) - r^2 h \bigg(\dfrac{r}{a - r} \bigg) \Bigg)$

Substitute Eq(1) in h₁ again.

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 \Bigg[h \bigg(\dfrac{r}{a - r} \bigg) + h\Bigg] - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 \Bigg[\dfrac{hr}{a - r} + h\Bigg] - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 \Bigg[\dfrac{hr + h(a - r)}{a - r}\Bigg] - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(a^2 \Bigg[\dfrac{hr + ha - hr}{a - r}\Bigg] - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(\dfrac{ha^3}{a - r} - \dfrac{r^3h}{a - r} \Bigg)$

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi \Bigg(\dfrac{ha^3 - r^3h}{a - r} \Bigg)$

Take 'h' outside the brackets since it's a common term.

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi h \Bigg(\dfrac{a^3 - r^3}{a - r} \Bigg)$

Using a³ - b³ = (a - b)(a² + ab + b²) we get:

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi h \Bigg(\dfrac{(a - r)(a^2 + ar + r^2)}{a - r} \Bigg)$

(a - r) get's cancelled.

$\Longrightarrow \sf Vol(Frustum) = \dfrac{1}{3} \pi h \Bigg(a^2 + ar + r^2\Bigg)$

Hence proved.