Question

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

Answer 1

EXPLANATION.

• GIVEN

From a solid right circular cylinder with

height = 10cm

radius of the base of 6cm

a right circular cone of the same height and the

same base is removed.

FIND THE VOLUME OF THE

REMAINING SOLID.

According to the question,

we have,

height of solid circular cylinder = 10 cm.

Radius of the base = 6 cm.

Volume of remaining solid = volume of

cylinder - volume of cone.

Volume of cylinder = πr²h

volume of cone = ⅓πr²h

=> [ πr²h - ⅓πr²h ]

=> [ π X (6)² X 10 - ⅓ X π X (6)² X 10 ]

=> [ 360π - 120π ]

=> 240πcm³

Therefore,

volume of remaining solid =

=> 240πcm³ or 753.6 cm³

Answer 2

$\Large{\underline{\underline{\red{Given:-}}}}$

•Radius of the cylinder=5cm

•Height of the cylinder =12 cm

$\Large{\underline{\underline{\green{Find:-}}}}$

• The volume and the surface area of the remaining solid

$\Large{\underline{\underline{\red{Solution:-}}}}$

$\textsf{formula used}\begin{cases} \bf volume \: _{cylinder} = \pi r {}^{2} h\\ \bf CSA_{cylinder} = 2 \pi rh\\ \bf CSA\: _{cone}= \pi rl\\ \bf \: volume \: _{cone} = \frac{1}{3} \pi r {}^{2} h</p><p></p><p>\end{cases}$

Now ,

$\bf volume \: _{cylinder} \: = {\bigg (} \frac{22}{7} \times 5 \times 5 \times 12{ \bigg)cm {}^{3} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \bf \frac{6600}{7} cm {}^{3}$

$\bf \: volume \: _{cone} \: ={ \bigg(} \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 { \bigg)}cm {}^{3} \\ \\ \: \: \: \: \: \: \: \: = \bf \frac{2200}{7} cm {}^{3}$

•Volume of the remaining solid =(volume of the cylinder) -(volume of the cone)

$\bf \implies { \bigg(} \frac{6600}{7} - \frac{2200}{7}{ \bigg)} cm {}^{3} = \frac{4400}{7} cm {}^{3} = 628.57 \: cm {}^{3}$

•Slant height of the cone(l)

$\bf \: l = \sqrt{r {}^{2} + h {}^{2} }$

$\bf \implies l = \sqrt{5 {}^{2} + (12) {}^{2} } = \sqrt{169} = 13cm$

$\textsf{CSA} \bf_{cone}={\bigg (} \frac{22}{7} \times 5 \times 13 { \bigg)}cm² = \frac{1430}{7} cm²$

$\textsf{CSA} \bf_{cylinder}= {\bigg (} 2 \times \frac{22}{7} \times 5 \times 12{ \bigg)}= \frac{2640}{7} cm²$

•Now, area of upper circular base of base of cylinder =

$= \bf \pi r {}^{2} sq. \: unit \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \bf{ \bigg(} \frac{22}{7} \times 5 \times 5 { \bigg)}cm {}^{2} = \frac{550}{7} cm {}^{2}$

•Whole surface area of the remaining solid =CSA(cylinder)+CSA(cone)+area of upper base of cylinder

$\implies\bf { \bigg(} \frac{2640}{7} + \frac{1430}{7} + \frac{550}{7} { \bigg)}cm {}^{2} = 660cm {}^{2}$