Question

From a solid cylinder of height 12 centimetre and diameter 10 cm a conical cavity of same height and same diameter is hollowed out find the total surface area of remaining solid​

Answer 1

Given

Height = 12 cm

Diameter = 10 cm

—————————

Volume of remaining solid = volume of cylinder - volume of cone.

Remaining volume = 942-314 ( solve in rough)

=628cm³

Total surface of remaining solid = conical sutface area of cone + Area Of circle + Cylindrical area of cylinder.

Conical sutface area of cone = πr

r² +h² =204.203 cm²

Conical of cirle = πr²

=78.54cm²

Cylindrical area of cylinder = 2πrh

=376.8cm ²

= we would be plus the all value shown in underline Statement .

so, Total surface area 659.543cm ²

Answer 2

GIVEN:-

•Radius of the cylinder=5cm

•Height of the cylinder =12 cm

TO FIND OUT:--

• The volume and the surface area of the remaining solid

SOLUTION :-

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

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

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