chancemorris123
  • chancemorris123
Find the surface area of the composite solid. Round your answer to the nearest hundredth.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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chancemorris123
  • chancemorris123
UnkleRhaukus
  • UnkleRhaukus
The total surface area of the composite solid, can be decomposed into three areas: 1 the circular base 2 the lateral surface area of the cylinder (a rectangle) & 3 the lateral area of the cone (a sector)
UnkleRhaukus
  • UnkleRhaukus
The area of a circle (or radius \(r\)) is : \[A_\text{circle} = \pi r^2\]

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More answers

UnkleRhaukus
  • UnkleRhaukus
The area of a rectangle (of width \(w\), height \(h\)) is: \[A_\text{rectangle} = w\times h\]
chancemorris123
  • chancemorris123
so i tryed it and got c is that right
UnkleRhaukus
  • UnkleRhaukus
i don't know,
chancemorris123
  • chancemorris123
that makes to of us
UnkleRhaukus
  • UnkleRhaukus
The area of the sector (of radius \(r\), and slope \(s\)) is: \[A_\text{sector} = \pi r s\]
UnkleRhaukus
  • UnkleRhaukus
add them all together, what do you get ? (before plugging in the numbers)
chancemorris123
  • chancemorris123
i put c and i only got 30 min :( and got 6 more questions
UnkleRhaukus
  • UnkleRhaukus
i dont know what the answer is, we have to work it out
UnkleRhaukus
  • UnkleRhaukus
\[SA_\text{total} = A_\text{circle}+A_\text{rectangle}+A_\text{sector}\\ \qquad\quad= \quad. . .\]
chancemorris123
  • chancemorris123
https://www.youtube.com/watch?v=Qmb9cnx__hA
chancemorris123
  • chancemorris123
never mind that one i guessed cuz ii need to do the rest befopre time runs out
UnkleRhaukus
  • UnkleRhaukus
NB: the width of the rectangle the circumference of a circle \[w =C = 2\pi r\]
UnkleRhaukus
  • UnkleRhaukus
\[SA_\text{total} = A_\text{circle}+A_\text{rectangle}+A_\text{sector}\\ \qquad\quad= \pi r^2+2\pi r h+\pi rs\\ \qquad\quad= \pi r(r+2h+s)\\ \qquad\quad= \quad. . .\]

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