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

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

• D1 $\dashrightarrow$ 4 cm

• D2 $\dashrightarrow$2 cm

• H $\dashrightarrow$14 cm

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

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

____________________________

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

• 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$${102.66\:cm^3}$

Answer 2

Step-by-step explanation:

So the radius of the upper base is r1 and the radius of the lower base is r2. Since diameters of upper and lower bases are 14 and 10 respectively the r1=7 and r2=5. The depth is H=6. So the volume of the frustum is:

V=H\3 x (r1^2 + r1xr2 + r1^2) π

V=2 x (49 + 35 + 25)π

V= 218 cm3