Question

measures of frustum of cone diameter are 4 and 2 cm and height 14 cm. find the volume of frustum of the cone.​

Answer 1

Answer:

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Answer 2

$\large\tt\underline\bold\red{Given :-}$

• D1 $\dashrightarrow$4 cm

• D2 $\dashrightarrow$ 2 cm

• H $\dashrightarrow$14 cm

$\setlength{\unitlength}{1cm}\begin{picture}\linethickness{}\qbezier( - 1, 0)( - 1,0)(1,3)\qbezier(5.2, 0)(5.2,0)(3,3)\qbezier(1, 3)(2,2.5)(3,3)\qbezier(1, 3)(2,3.5)(3,3)\qbezier( - 1, 0)(1.8, 0.8)(5.2,0)\qbezier( - 1, 0)(1.8, - 1)(5.2,0)\qbezier(4.8, 0)( - 1, 0)(5.2,0)\qbezier(3, 3)(1, 3)(3,3)\put(2,0){\dashbox{0.1}(1,3)}\put(2,0){\circle*{0.19}}\put(2,2.99){\circle*{0.19}}\put(1.2,1.3){\bf 14cm}\put(3.2,-1){\bf2cm}\put(2.3,3.4){\bf\large1cm}\end{picture}$

ㅤ $\small\underline\bold\blue{Frustum}$

____________________________

$\large\tt\underline\bold\red{To\:find}$

• The Volume of frustum of the cone

$\large\red{\underline{{\boxed{\textbf{Formula\: Used}}}}}$

• Volume of Frustum of cone

$\bullet\:\: \underline{\boxed{\bold{\bf{Volume = \dfrac{1}{3} \times \pi \times h \times [( R_1 )^2+(R_2)^2+ (R_1 \times R_2) }}}}$

$\huge\tt\underline\orange{Solution}$

$\implies\:\:\sf{\dfrac{1}{3} \times \dfrac{22}{7} \times 14 [ (2)^2 + (1)^2+(2\times 1)]}$

$\implies\:\:\sf{\dfrac{22\times 14}{7\times3}\times [ (2)^2 + (1)^2+(2\times 1)]}$

$\implies\:\:\sf{\dfrac{308}{7\times3}\times [ 4 + 1 +2]}$

$\implies\:\:\sf{\dfrac{308}{21}\times 7}$

$\implies\:\:\sf{\dfrac{44}{3}\times 7}$

$\implies\:\:\sf {102.66\:cm^3}$