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nomnom147 Group Title

Think of an equilateral triangle in which the length of a side is s, the perimeter (distance around the triangle) is p, and the area is A. Find the missing parts. s=? p=? area= 9 sqaureroot of 3

  • 2 years ago
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  1. nomnom147 Group Title
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    • 2 years ago
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  2. seashell Group Title
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    ill try:)

    • 2 years ago
  3. nomnom147 Group Title
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    thank you soo much!! i really need someones help! :)

    • 2 years ago
  4. seashell Group Title
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    ok so.. the formula of the area of an equilateral area is

    • 2 years ago
  5. seashell Group Title
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    http://www.mathwords.com/a/a_assets/area%20equilateral%20triangle%20formula.gif

    • 2 years ago
  6. nomnom147 Group Title
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    but how do i get s? if i have to divide it? Im so confused.

    • 2 years ago
  7. seashell Group Title
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    :/ tryna help but i really dont know:/ im soo sorry

    • 2 years ago
  8. myininaya Group Title
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    \[A=9 \sqrt{3}=\frac{1}{2} b h\] So we have that the base, b, of the triangle is s b=s So we have \[9 \sqrt{3}=\frac{1}{2}s h\] So this means we have \[(2) 9 \sqrt{3} =\frac{1}{2}sh (2)\] \[18 \sqrt{3} =sh\] So is there a way to write h in terms of s Well remember we have an equilateral triangle with each side s and h, the height of the equilateral can be found in terms of s by using the Pythagorean Theorem |dw:1337557593984:dw|

    • 2 years ago
  9. myininaya Group Title
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    no look at one of those triangles inside the equilateral |dw:1337557653191:dw|

    • 2 years ago
  10. myininaya Group Title
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    Write h in terms of s

    • 2 years ago
  11. myininaya Group Title
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    Use the Pythagorean thm to do so

    • 2 years ago
  12. myininaya Group Title
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    now not no*

    • 2 years ago
  13. nomnom147 Group Title
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    so 18 square root of 3 using pythagorean thm we get the S?

    • 2 years ago
  14. nomnom147 Group Title
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    @seashell thank you anyways! :) <3

    • 2 years ago
  15. myininaya Group Title
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    I want you to write h in terms of s Use the Pythagorean thm

    • 2 years ago
  16. seashell Group Title
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    lol:) how sweet!

    • 2 years ago
  17. myininaya Group Title
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    Look at the triangle I drew

    • 2 years ago
  18. myininaya Group Title
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    And use the Pythagorean thm to write h in terms of s

    • 2 years ago
  19. nomnom147 Group Title
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    a^2 + b^2 = C^2 so H will eaqul 54?

    • 2 years ago
  20. myininaya Group Title
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    |dw:1337558023888:dw| \[(h)^2+(\frac{s}{2})^2=s^2\]

    • 2 years ago
  21. myininaya Group Title
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    Solve for h

    • 2 years ago
  22. nomnom147 Group Title
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    is H equal 13.5

    • 2 years ago
  23. myininaya Group Title
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    Solve what I have up there for h please \[h^2+\frac{s^2}{4}=s^2\] this right here can you do that?

    • 2 years ago
  24. nomnom147 Group Title
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    im really sorry Im trying to understand it, but i get confused. Im trying my best. Sorry :/

    • 2 years ago
  25. nomnom147 Group Title
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    @myininaya will H equal 4?

    • 2 years ago
  26. myininaya Group Title
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    \[\text{ subtract } \frac{s^2}{4} \text{ on both sides of } h^2+\frac{s^2}{4}=s^2\]

    • 2 years ago
  27. myininaya Group Title
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    \[h^2=s^2-\frac{s^2}{4}\] Combine the fractions on the right hand side of the equation \[h^2=\frac{4s^2}{4}-\frac{s^2}{4} \] \[h^2=\frac{3s^2}{4}\] Take square root of both sides So we have \[h=\sqrt{\frac{3s^2}4}\] \[h=\frac{\sqrt{3s^2}}{\sqrt{4}}\] \[h=\frac{\sqrt{3} \sqrt{s^2}}{2}\] \[h=\frac{\sqrt{3}s}{2}\]

    • 2 years ago
  28. myininaya Group Title
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    I wrote h in terms of s

    • 2 years ago
  29. myininaya Group Title
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    Now we can go back to solving for s Recall we had the equation \[18 \sqrt{3}=sh\] Now we just wrote h in terms of s So we have \[18 \sqrt{3}=s \frac{\sqrt{3} s}{2}\] Can you solve this for s ?

    • 2 years ago
  30. myininaya Group Title
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    \[18 \sqrt{3} = s s \frac{\sqrt{3}}{2}\] \[18 \sqrt{3} = \frac{\sqrt{3}}{2} s^2\] There that might look a bit better to you Try solving this for s

    • 2 years ago
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