anonymous
  • anonymous
AD=4√3 in.; AB = 8 in. Area =? times the square root of ? sq. in.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
@timo86m Help with mine?
anonymous
  • anonymous
do you know trig? or does this only require pythogoras theorem?

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anonymous
  • anonymous
It should only require Pythagoras theorem. It is merely Geometry.
anonymous
  • anonymous
But the Sq Rt is what confuses me.
anonymous
  • anonymous
I tried using the 30-60-90 ratio and i got the answer: 2 times sqrt of 1.5 sq in which is wrong.
anonymous
  • anonymous
|dw:1346873004579:dw|
anonymous
  • anonymous
Don't i take half of 4√3 since its the hypotunese?
anonymous
  • anonymous
|dw:1346873167383:dw| once you have a then a*2 would be height
anonymous
  • anonymous
And how do i find A? A is what?
anonymous
  • anonymous
a is 4 i think the height is 8
anonymous
  • anonymous
SO it would be area= 8 times the sqrt of 3?
anonymous
  • anonymous
then i think it is just 8*8 the area Something seems odd tho. THe figure is not to scale or i am wrong.
anonymous
  • anonymous
i did it wrong
anonymous
  • anonymous
all you have to do is find height then it is simply 8*h
anonymous
  • anonymous
But how will that get me a sq rt?
anonymous
  • anonymous
i found where i gone wrong lol
anonymous
  • anonymous
Lol
anonymous
  • anonymous
|dw:1346874388605:dw|
anonymous
  • anonymous
a=2*sqrt(3) So h = 2*sqrt(3)*sqrt(3) and A=h*b can you do that?
anonymous
  • anonymous
The height is √3 correct?
anonymous
  • anonymous
a=2*sqrt(3) So h = 2*sqrt(3)*sqrt(3) and A=h*b can you do that?
anonymous
  • anonymous
19-3= 16 times √3
anonymous
  • anonymous
Thanks man!
anonymous
  • anonymous
I got it.
anonymous
  • anonymous
:)

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