anonymous
  • anonymous
Help! I give medals Find the surface are and volume of a regular tetrahedron with a slant height of 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1370298509999:dw|
anonymous
  • anonymous
anonymous
  • anonymous
hope this helps @ganeshie8

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ganeshie8
  • ganeshie8
yes, but really as u said il have to look it in mirror lol :)
anonymous
  • anonymous
LOL you have my drawing as a reffrence to look at
ganeshie8
  • ganeshie8
okay
ganeshie8
  • ganeshie8
lets find surface area first
ganeshie8
  • ganeshie8
surface area = 1/2 * (perimeter of base) * (slant height) + (area of base)
ganeshie8
  • ganeshie8
|dw:1370296184124:dw|
ganeshie8
  • ganeshie8
perimeter of base = ?
anonymous
  • anonymous
would it be the 1/3?
ganeshie8
  • ganeshie8
perimeter is just the length around a shape. since u have 3 sides for base, its 3*side
ganeshie8
  • ganeshie8
|dw:1370296354829:dw|
ganeshie8
  • ganeshie8
surface area = 1/2 * (perimeter of base) * (slant height) + (area of base) = \(1/2 * (4 * \sqrt{3} )* 2 \) + (area of base)
ganeshie8
  • ganeshie8
so to find surface area, we need to figure out the area of base still.
ganeshie8
  • ganeshie8
you have any ideas how to find area of base ?
anonymous
  • anonymous
would it be the same as the area of a triangle?
ganeshie8
  • ganeshie8
yes !
ganeshie8
  • ganeshie8
since base is an equilateral triangle, u need to use the formula for area of equilateral triangle
anonymous
  • anonymous
would it beA=1/2(4rt2/3)(4rt3/3)
ganeshie8
  • ganeshie8
close ! use this :- area of equilateral triangle wid side s = \(s^2\sqrt{3}/4\)
ganeshie8
  • ganeshie8
plugin s = \(4\sqrt{3}/3\)
anonymous
  • anonymous
so it would be\[\frac{ 4\sqrt{3} }{ 3 } \frac{ \sqrt{3} }{ 4 }\]
ganeshie8
  • ganeshie8
you forgot to square s... \(\large (\frac{ 4\sqrt{3} }{ 3 })^2\times \frac{ \sqrt{3} }{ 4 } \)
ganeshie8
  • ganeshie8
simplify - that gives area of base
anonymous
  • anonymous
im a litte confused bcuz of the square root sign
anonymous
  • anonymous
would it be\[\frac{ 16\sqrt{9} }{ 6 }\times \frac{ \sqrt{3} }{4}\] then cross multiply?? @ganeshie8
ganeshie8
  • ganeshie8
\(\large (\frac{ 4\sqrt{3} }{ 3 })^2\times \frac{ \sqrt{3} }{ 4 } \) \(\large (\frac{ 16*3}{ 9 })\times \frac{ \sqrt{3} }{ 4 } \) \(\large (\frac{ \cancel{16}^4*3}{ 9 })\times \frac{ \sqrt{3} }{ \cancel{4} } \) \(\large (\frac{ 4*3}{ 9 })\times \sqrt{3} \) \(\large (\frac{ 4*3}{ 9 })\times \sqrt{3} \) \(\large \frac{ 4\sqrt{3} }{ 3 } \)
ganeshie8
  • ganeshie8
thats the area of base, add it
ganeshie8
  • ganeshie8
surface area = 1/2 * (perimeter of base) * (slant height) + (area of base) = \(1/2∗(4\sqrt{3})∗2\) + (area of base) = \(1/2∗(4\sqrt{3})∗2\) + \(\frac{4\sqrt{3}}{3} \)
ganeshie8
  • ganeshie8
simplify - that gives surface area
ganeshie8
  • ganeshie8
you wud get surface area = \(4\sqrt{3} + \frac{4\sqrt{3}}{3}\) =\( \frac{16\sqrt{3}}{3}\)
anonymous
  • anonymous
wouldn't it be surface area=\[\frac{ 1 }{ 2 }(4\sqrt{3})\times2+\frac{ 4\sqrt{3} }{ 3 }\]
ganeshie8
  • ganeshie8
yes, you can canel 2
ganeshie8
  • ganeshie8
and simplify, you would get, \(\frac{16\sqrt{3}}{3}\)
anonymous
  • anonymous
ok so woud the volume be the same as a pyramids volume?
ganeshie8
  • ganeshie8
yes, volume = \(\frac{1}{3}\) * (base area) * (height)
ganeshie8
  • ganeshie8
plugin : base area = \(\frac{4\sqrt{3}}{3}\) height = \(\frac{4\sqrt{2}}{3}\)
anonymous
  • anonymous
so would it be\[144\sqrt{54}\]?? the square roots consfuse me
ganeshie8
  • ganeshie8
its okay, see my steps next
ganeshie8
  • ganeshie8
volume = \(\frac{1}{3}\) * (base area) * (height) = \(\frac{1}{3} * \frac{4\sqrt{3}}{3} * \frac{4\sqrt{2}}{3}\) = \(\frac{16\sqrt{6}}{27}\)
anonymous
  • anonymous
Ok :) thanks so much for your help much apreacheated
ganeshie8
  • ganeshie8
np :)

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