Dallasb22
AD=4√3 in.; AB = 8 in. Area =? times the square root of ? sq. in.



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Dallasb22
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Dallasb22
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@timo86m Help with mine?

timo86m
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do you know trig? or does this only require pythogoras theorem?

Dallasb22
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It should only require Pythagoras theorem. It is merely Geometry.

Dallasb22
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But the Sq Rt is what confuses me.

Dallasb22
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I tried using the 306090 ratio and i got the answer: 2 times sqrt of 1.5 sq in which is wrong.

timo86m
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dw:1346873004579:dw

Dallasb22
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Don't i take half of 4√3 since its the hypotunese?

timo86m
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dw:1346873167383:dw
once you have a then
a*2 would be height

Dallasb22
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And how do i find A?
A is what?

timo86m
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a is 4 i think the height is 8

Dallasb22
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SO it would be area= 8 times the sqrt of 3?

timo86m
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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.

timo86m
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i did it wrong

timo86m
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all you have to do is find height then it is simply 8*h

Dallasb22
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But how will that get me a sq rt?

timo86m
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i found where i gone wrong lol

Dallasb22
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Lol

timo86m
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dw:1346874388605:dw

timo86m
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a=2*sqrt(3)
So h = 2*sqrt(3)*sqrt(3)
and A=h*b can you do that?

Dallasb22
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The height is √3 correct?

timo86m
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a=2*sqrt(3)
So h = 2*sqrt(3)*sqrt(3)
and A=h*b can you do that?

Dallasb22
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193= 16 times √3

Dallasb22
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Thanks man!

Dallasb22
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I got it.

timo86m
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:)