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Dallasb22

  • 3 years ago

AD=4√3 in.; AB = 8 in. Area =? times the square root of ? sq. in.

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  1. Dallasb22
    • 3 years ago
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  2. Dallasb22
    • 3 years ago
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    @timo86m Help with mine?

  3. timo86m
    • 3 years ago
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    do you know trig? or does this only require pythogoras theorem?

  4. Dallasb22
    • 3 years ago
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    It should only require Pythagoras theorem. It is merely Geometry.

  5. Dallasb22
    • 3 years ago
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    But the Sq Rt is what confuses me.

  6. Dallasb22
    • 3 years ago
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    I tried using the 30-60-90 ratio and i got the answer: 2 times sqrt of 1.5 sq in which is wrong.

  7. timo86m
    • 3 years ago
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    |dw:1346873004579:dw|

  8. Dallasb22
    • 3 years ago
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    Don't i take half of 4√3 since its the hypotunese?

  9. timo86m
    • 3 years ago
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    |dw:1346873167383:dw| once you have a then a*2 would be height

  10. Dallasb22
    • 3 years ago
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    And how do i find A? A is what?

  11. timo86m
    • 3 years ago
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    a is 4 i think the height is 8

  12. Dallasb22
    • 3 years ago
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    SO it would be area= 8 times the sqrt of 3?

  13. timo86m
    • 3 years ago
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    then i think it is just 8*8 the area Something seems odd tho. THe figure is not to scale or i am wrong.

  14. timo86m
    • 3 years ago
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    i did it wrong

  15. timo86m
    • 3 years ago
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    all you have to do is find height then it is simply 8*h

  16. Dallasb22
    • 3 years ago
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    But how will that get me a sq rt?

  17. timo86m
    • 3 years ago
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    i found where i gone wrong lol

  18. Dallasb22
    • 3 years ago
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    Lol

  19. timo86m
    • 3 years ago
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    |dw:1346874388605:dw|

  20. timo86m
    • 3 years ago
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    a=2*sqrt(3) So h = 2*sqrt(3)*sqrt(3) and A=h*b can you do that?

  21. Dallasb22
    • 3 years ago
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    The height is √3 correct?

  22. timo86m
    • 3 years ago
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    a=2*sqrt(3) So h = 2*sqrt(3)*sqrt(3) and A=h*b can you do that?

  23. Dallasb22
    • 3 years ago
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    19-3= 16 times √3

  24. Dallasb22
    • 3 years ago
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    Thanks man!

  25. Dallasb22
    • 3 years ago
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    I got it.

  26. timo86m
    • 3 years ago
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    :)

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