Question

(1) A frusttum of a cone has top and diameters of 14cm and 10cm respectively and a depth of 6cm find the volume of the frusttum. (take =22\7 )

Answer 1

Answer:

Appropriate Question :-

• A frustum of a cone has top and bottom diameters of 14 cm and 10 cm respectively and a depth of 6 cm. Find the volume of the frustum. (Take π = 22/7).

Given :-

• A frustum of a cone has top and bottom diameters of 14 cm and 10 cm respectively and a depth of 6 cm.

To Find :-

• What is the volume of the frustum.

Formula Used :-

$\clubsuit$ Volume Of A Conical Frustum Formula :

$\mapsto \sf\boxed{\bold{\pink{Volume_{(Frustum)} =\: \dfrac{1}{3} \times {\pi} \times h\bigg\lgroup (r_1)^2 + (r_2)^2 + (r_1 \times r_2)\bigg\rgroup}}}$

where,

• π = pie or 22/7
• h = Height
• r₁ = Upper or top radius
• r₂ = Lower or bottom radius

Solution :-

First, we have to find the radius :

As we know that,

$\clubsuit$ Radius Formula :

$\mapsto \sf\boxed{\bold{\pink{Radius =\: \dfrac{Diameter}{2}}}}$

$\implies \sf Radius (r_1) =\: \dfrac{\cancel{14}}{\cancel{2}}$

$\implies \sf \bold{\purple{Radius (r_1) =\: 7\: cm}}$

Again,

$\implies \sf Radius (r_2) =\: \dfrac{\cancel{10}}{\cancel{2}}$

$\implies \sf \bold{\purple{Radius (r_2) =\: 5\: cm}}$

Given :

• Radius (r₁) = 7 cm
• Radius (r₂) = 5 cm
• Depth (h) = 6 cm

According to the question by using the formula we get,

$\longrightarrow \sf Volume_{(Frustum)} =\: \dfrac{1}{3} \times \dfrac{22}{7} \times 6\bigg\lgroup (7)^2 + (5)^2 + (7 \times 5)\bigg\rgroup$

$\longrightarrow \sf Volume_{(Frustum)} =\: \dfrac{22}{21} \times 6\bigg\lgroup (7 \times 7) + (5 \times 5) + (35)\bigg\rgroup$

$\longrightarrow \sf Volume_{(Frustum)} =\: \dfrac{132}{21}\bigg\lgroup (49 + 25 + 35)\bigg\rgroup$

$\longrightarrow \sf Volume_{(Frustum)} =\: \dfrac{132}{21}\bigg\lgroup (74 + 35)\bigg\rgroup$

$\longrightarrow \sf Volume_{(Frustum)} =\: \dfrac{132}{21}\bigg\lgroup 109\bigg\rgroup$

$\longrightarrow \sf Volume_{(Frustum)} =\: \dfrac{132}{21} \times 109$

$\longrightarrow \sf Volume_{(Frustum)} =\: \dfrac{14388}{21}$

$\longrightarrow \sf \bold{\red{Volume_{(Frustum)} =\: 685.14\: cm^3}}$

$\therefore$ The volume of frustum is 685.14 cm³.

Answer 2

$\begin{gathered}\frak{ Given}\begin{cases}\sf{Diameter_1=10 \:cm}\\\sf{Diameter_2=14 \:cm}\\\sf{Height=6 \:cm}\end{cases}\end{gathered}$

$\bullet\:\sf Radius_1=\frac{Diameter_1}{2}=\frac{10\:cm}{2}=5\:cm$

$\bullet\:\sf Radius_2=\frac{Diameter_2}{2}=\frac{14\:cm}{2}=7\:cm$

$\boxed{\bf{\mid{\overline{\underline{\bigstar\:According\:to\:the\:Question :}}}}\mid}$

$\begin{gathered}:\implies\sf Volume=\dfrac{\pi \times Height}{3} \times \Bigg\lgroup (R_1)^2+(R_2)^2+(R_1 \times R_2)\Bigg\rgroup\\\\\\:\implies\sf Volume = \dfrac{22 \times 6}{7 \times 3} \times \Bigg\lgroup (5)^2+(7)^2+(5\times 7)\Bigg\rgroup\\\\\\:\implies\sf Volume = \dfrac{22 \times 2}{7} \times \Bigg\lgroup 25+49+35\Bigg\rgroup\\\\\\:\implies\sf Volume = \dfrac{44}{7} \times 109\\\\\\:\implies\underline{\boxed{\sf Volume = 685.14\:cm^3}}\end{gathered}$

⠀

$\therefore\:\underline{\textsf{Volume of the Frustum will be \textbf{685.14 cm ^\text3 }}}$