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Vinvika [58]
2 years ago
6

The radius of a cylinder is 7 c m and its height is 10 c m. Find curved surface area and volume.​

Mathematics
2 answers:
erma4kov [3.2K]2 years ago
8 0

Answer:

  • CSA of the cylinder = 440 sq. cm

  • Volume of the cylinder = 1540 cu. cm

\\

Step-by-step explanation:

Given:

  • Radius of the cylinder = 7 cm
  • Height of the cylinder = 10 cm

\\

To Find:

  • Curved surface area
  • Volume

\\

Solution:

\\

Using formula:

\dashrightarrow \:  \:  { \underline{ \boxed{ \pmb{ \sf{ \purple{CSA  \: of  \: cylinder = 2\pi rh}}}}}}  \:  \star \\  \\

<em>Substituting the required values: </em>

\\

\dashrightarrow \:  \:  \sf CSA {(cylinder)} = 2 \times \dfrac{22}{7} \times 7 \times  10 \\  \\  \\  \dashrightarrow \:  \:  \sf CSA {(cylinder)}  = 2 \times 22 \times 10 \\  \\ \\   \dashrightarrow \:  \:  \sf CSA {(cylinder)} = 44 \times 10 \\  \\  \\  \dashrightarrow \:  \:  \sf CSA {(cylinder)}  = 440 \:  {cm}^{2}  \\ \\ \\

Now,

\dashrightarrow \:  \:  { \underline{ \boxed{ \pmb{ \sf{ \purple{Volume {(cylinder)}=  \pi {r}^{2} h}}}}}}  \:  \star \\  \\ \\

<em>Substituting the required values, </em>

\\

\dashrightarrow \:  \:   \sf Volume {(cylinder)}=  \dfrac{22}{7}  \times  {(7)}^{2}  \times 10 \\ \\  \\  \dashrightarrow \:  \:   \sf Volume {(cylinder)}=  \frac{22}{7}  \times 49  \times 10 \\ \\ \\  \dashrightarrow \:  \:   \sf Volume {(cylinder)}= 22 \times 7 \times 10 \\ \\  \\  \dashrightarrow \:  \:   \sf Volume {(cylinder)}= 1540 {cm}^{3}  \\ \\ \\

Hence,

  • CSA of the cylinder = 440 sq. cm

  • Volume of the cylinder = 1540 cu. cm
tiny-mole [99]2 years ago
6 0

Step-by-step explanation:

CSA of cylinder 440 cm sq.

Volume 1540 cm cube.

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Find the equation of a line passing through (-4,-3) and perpendicular to<br> 3 x + 2 y = 14.
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3x+2y=14\implies 2y=-3x+14\implies y=\cfrac{-3x+14}{2}\implies y = \cfrac{-3x}{2}+\cfrac{14}{2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{3}{2}}x+7\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

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\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{-3}\implies \cfrac{2}{3}}}

so we're really looking for the equation of a line whose slope is 2/3 and passes through (-4 , -3)

(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{2}{3}}(x-\stackrel{x_1}{(-4)}) \implies y+3=\cfrac{2}{3}(x+4) \\\\\\ y+3=\cfrac{2}{3}x+\cfrac{8}{3}\implies y=\cfrac{2}{3}x+\cfrac{8}{3}-3\implies y=\cfrac{2}{3}x-\cfrac{1}{3}

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Answer:

Step-by-step explanation:

Ans:A.

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