Find the volume of the frustum cone whose base and top radius is 4 mm and 2 mm respectively. The height of the cone is 27 mm. (Use $π$ = 3.14).
781.28 mm $^{3}$
791.28 mm $^{3}$
761.28 mm $^{3}$
711.28 mm $^{3}$

Question

Find the volume of the frustum cone whose base and radius is 4 mm and 2 mm respectively- The height of the cone is 27 mm- -Use pi - 3-14-781-28 mm-3791-28 mm-3761-28 mm-3711-28 mm-3

Toppr Admin · Accepted Answer

Volume of the frustum cone is V-x3C0-h3-R2-Rr-r2-R-4mm-r-2mm-h-27mmV-3-14-xD7-27-3-42-4-xD7-2-22-xA0- -xA0- -3-14-xD7-9-16-8-4-xA0- -xA0- -28-26-28-xA0- -xA0- -791-28mm3

Find the volume of the frustum cone whose base and top radius is 4 mm and 2 mm respectively. The height of the cone is 27 mm. (Use π = 3.14).781.28 mm3791.28 mm3761.28 mm3711.28 mm3

Volume of the frustum cone is V=πh3(R2+Rr+r2)R=4mm,r=2mm,h=27mmV=(3.14×27)/(3)[42+4×2+22] =3.14×9[16+8+4] =28.26[28] =791.28mm3

Find the volume of the frustum cone whose base and top radius is 4 mm and 2 mm respectively. The height of the cone is 27 mm. (Use $π$ = 3.14).
781.28 mm $^{3}$
791.28 mm $^{3}$
761.28 mm $^{3}$
711.28 mm $^{3}$

Volume of the frustum cone is $V = \frac{π h}{3} (R^{2} + R r + r^{2})$
$R = 4 m m, r = 2 m m, h = 27 m m$
$V = (3.14 \times 27) / (3) [4^{2} + 4 \times 2 + 2^{2}]$
$= 3.14 \times 9 [16 + 8 + 4]$
$= 28.26 [28]$
$= 791.28 m m^{3}$