Question

A frustum of cone has top and bottom diameters as 14cm and 10cm respectively and a depth of 6cm. Find the volume of the frustum

Answer 1

Given :-

• Diameter 1 = 14cm
• Diameter 2 = 10cm.
• Depth = 6cm.

To Find :-

• Volume of The frustum ?

Formula used :-

• Volume of Frustum of cone = (1/3) * π * h * (r1² + r2² + (r1 * r2)).
• radius = (Diameter/2) .

Solution :-

→ r1 = (D1/2) = 14/2 = 7cm.

→ r2 = (D2/2) = 10/2 = 5cm.

→ Depth = Height = h = 6cm.

Putting All values we get :-

→ V = (1/3) * (22/7) * 6 * [ 7² + 5² + (7*5)]

→ V = (44/7) * [ 49 + 25 + 35 ]

→ V = (44/7) * 109

→ V ≈ 685.14. cm³ (Ans.)

______________________

Extra :-

→ CSA of the frustum of a cone = πl( r1 + r2).

→ Total surface area of a conical frustum: A = π * (r1² + r2² + (r1 + r2) * l)

→ slant Height = l = √[(r1 - r2)² + h²]

Answer 2

Answer:

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

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

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

$\rule{150}{1}$

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

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

⠀

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