Question

From a right circular cylinder with height 10cm and radius of base 6cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid.

Answer 1

$\underline{\bold{Solution:-}}$

Height of cylinder (H) = 10 cm

Radius of cylinder (R) =6 cm

Volume of cylinder

$= \pi {R}^{2} H$

$= \pi \times {6}^{2} \times 10$

Height of cone (h) = 10 cm

Radius of cone (r) = 6 cm

Volume of cone

$= \frac{1}{3} \pi {r}^{2} h \\ \\ = \frac{1}{3} \pi \times {6}^{2} \times 10$
a right circular cone of the same height and base is removed.

Remaining volume

= Volume of Cylinder - volume of cone

$= \pi {6}^{2} \times 10- \frac{1}{3} \pi {6}^{2} \times 10 \\ \\ = \pi \times {6}^{2} \times 10 \times (1 - \frac{1}{3} ) \\ \\ = 3.14 \times 36 \times 10 \times ( \frac{2}{3}) \\ \\ = 753.6 {cm}^{2}$

So, 753.6 cm2 is the remaining volume of the cylinder.

Answer 2

Answer:

Let $\sf{V_1}$ and $\sf{V_2}$ be the volumes of the right circular cylinder and cone respectively.

Then,

$\sf\qquad{V_1=}$$\sf\frac{22}{7}×6×6×10\:cm^3$

and, $\sf\qquad{V_2}$$\sf\frac{1}{3}×$$\sf\frac{22}{7}×6×6×10\:cm^3$

$\therefore$ Volume of the remaining solid = $\bf{V_1-V_2}$

$\leadsto$Volume of the remaining solid= ${22}{7}×6×6×10-$$\sf\frac{1}{3}×{22}{7}×6×6×10\:cm^3$

$\leadsto$$[tex]\sf\frac{22}{7}×6×6</p><p>×10×1-$$\sf\frac{1}{3}cm^3$

$\leadsto$ $\sf\frac{22}{7}×6×6×10</p><p>×$$\sf\frac{2}{3}cm^3$

$\implies$ $\sf\pink{Volume\:\:of\:\:the\:\:remaining\:\:solid=754.28\:\:cm^3}$