Question

from a solid cylinder of height 12cm and diameter of base 10cm a conical cavity of the same height and same diameter is hollowed out. find the surface area of the remaining solid.​

Answer 1

Answer:

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

r=5 cm   (given)

h=12 cm

Volume of cylinder = πr  

2

h

=3.14×5  

2

×12

=942cm  

2

 

Volume of cone =  

3

1

​  

πr  

2

h=  

3

1

​  

×3.14×5  

2

×12

=314cm  

3

 

Remaining volume = 942-314

=628cm  

3

 

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

Conical sutface area of cone =πr  

r  

2

+h  

2

 

​  

=204.203 cm  

2

 

Conical of cirle = πr  

2

=78.54cm  

2

 

Cylindrical area of cylinder = 2πrh

=376.8cm  

2

 

Total surface area 659.543cm  

2

Step-by-step explanation:

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

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