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superloz
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how to find the second one?
OpenSessame
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The diagonals are in the middle of the triangle so its basically it is breaking it up in 4 equal peices.
superloz
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pok
superloz
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ok
superloz
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then
OpenSessame
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So if you are trying to find ac it would what half of it twice.
OpenSessame
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so (x+4)(x+4)
superloz
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simplify?
OpenSessame
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Since you have the sum of the length of the diaganols you have to make an equation equal to 20.
OpenSessame
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(2x-3)(2x-3)+(x+4)(x+4)=20
OpenSessame
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You have to solve for x before moving on
superloz
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is the answer 7?
OpenSessame
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Uh one sec still solving!
OpenSessame
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Plug it in to check!
OpenSessame
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I dont think thats it?
superloz
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idk
superloz
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i solve this for x (2x-3)(2x-3)+(x+4)(x+4)=20
OpenSessame
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You have to do the quadratic eqaution...
superloz
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idk how to do quad
OpenSessame
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did you get 5x^2-4x+25=20 when you simplified?
superloz
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yep
OpenSessame
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Okay you need quad to find x give me a minute and ill do it...
superloz
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ok
OpenSessame
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Its an imaginary number...
OpenSessame
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Assuming the rhombus is ABCD and E is the intersection of the diagonals, then we know the diagonals perpendicularly bisect each other, so AE is half of one diagonal and BE is half the other diagonal. That allows us to solve for x:
(x+4) + (2x-3) = 20/2
3x + 1 = 10
3x = 9
x = 3
and since
BE = 2x - 3 = 3 is half of BD,
BD = 2*3 = 6 (answer C).
OpenSessame
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Figured out the better way to solve it lol
superloz
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thank you
OpenSessame
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DO you understand how i got there?
superloz
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so BE is half and BD is the other half?
superloz
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each is 3?
OpenSessame
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Wait!!! i messed up lol you plug the three in x+4 and multiply it by 2 because you are finding ac
OpenSessame
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For some reason i thought it was the other one. But it will be 14 in that case.
superloz
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thanks