Math, asked by MdJaid, 10 months ago

find the volume curved surface area and total surface area of the cylinder whose height and radius of the base are 28 cm and 3 cm respectively​

Answers

Answered by aditya66612
0

Answer:

  • volume of cylinder=pie r2 h
  1. 22/7×8×28
  2. 22×8×4
  3. =684cm
Answered by BrainlyConqueror0901
1

\blue{\bold{\underline{\underline{Answer:}}}}

\green{\therefore{\text{Volume\:of\:cylinder=792\:cm}^{3}}}

\green{\therefore{\text{C.S.A\:of\:cylinder=528\:cm}^{2}}}

\green{\therefore{\text{T.S.A\:of\:cylinder=584.57\:cm}^{2}}}

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

 \green{ \underline \bold{Given : }} \\  :  \implies  \text{Radius(r) = 3 \: cm} \\  \\   : \implies  \text{Height(h) = 28\: cm}  \\  \\  \red{ \underline \bold{To \: Find : }} \\   : \implies   \text{Volume \: of \: cylinder = ? }\\  \\ : \implies   \text{C.S.A \: of \: cylinder = ? }\\ \\  : \implies   \text{T.S.A\: of \: cylinder = ? }

• According to given question :

 \bold{As \: we \: know \: that  } \\  :  \implies  \text{Volume \: of \: cylinder} = \pi {r}^{2} h \\  \\   : \implies  \text{Volume \: of \: cylinder} = \frac{22}{7}  \times  {3}^{2}  \times 28 \\  \\  :  \implies  \text{Volume \: of \: cylinder} =22 \times 9\times 4 \\  \\   \green{ : \implies  \text{Volume \: of \: cylinder} =792  \: {cm}^{3} } \\  \\  \bold{As \: we \: know \: that} \\   : \implies  \text{C.S.A\: of \: cylinder} =2\pi rh \\  \\ : \implies  \text{C.S.A\: of \: cylinder} =2  \times \frac{ 22}{7}  \times 3 \times 28 \\  \\ : \implies  \text{C.S.A\: of \: cylinder} =2 \times 22 \times 12\\  \\ \green{ : \implies  \text{C.S.A\: of \: cylinder} =528 \:  {cm}^{2}} \\  \\  \bold{As \: we \: know \: that} \\   : \implies  \text{T.S.A\: of \: cylinder} =2\pi r(h + r) \\  \\ : \implies  \text{T.S.A\: of \: cylinder} =2 \times  \frac{22}{7}  \times 3(28 + 3) \\  \\ : \implies  \text{T.S.A\: of \: cylinder} =2 \times \frac{22}{7} \times 93 \\  \\ \green{ : \implies  \text{T.S.A\: of \: cylinder} =584.57 \:  {cm}^{2} }

Similar questions