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

Diameters of frustum of cone = 14 cm & 10 cm

Depth or height = 6 cm

To find

Volume of frustum of cone

Solution

Here,

⟼ d₁ = 14 cm

⟼ d₂ = 10 cm.

➝ r₁ = 14/2 = 7 cm.

➝ r₂ = 10/2 = 5 cm.

As we know,

➨ Volume of frustum of cone = ⅓ × π × h (r₁² + r₂² + r₁ × r₂)

Putting values :

➻ Volume = ⅓ (22/7 × 6)(7² + 5² + 7 × 5)

➻ Volume = ⅓ (132/7)(49 + 25 + 35)

➻ Volume = ⅓ (132/7)(109)

➻ Volume = (132 × 109)/(3 × 7)

➻ Volume = 14388/21

➻ Volume = 685.14 cm³

Therefore,

Volume of frustum of cone = 685.14 cm³

Answer 2

Answer:

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

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

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

$\rule{150}{1}$

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

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

⠀

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