#help

38 messages · Page 1 of 1 (latest)

limpid mauve
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i need some assistance solving this exercice

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serene cloudBOT
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limpid mauve
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#helper-nominations

olive junco
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Hm...
Well, triangles EFG and BCG are similar. So:
CG/FG = BC/EF
As BC = a, EF = x sin(α):
CG/FG = a/(x sin(α))
Then we have:
CG + FG + FD = CD
As FD = x cos(α), CD = a:
CG + FG + x cos(α) = a
So, we get a system:
CG/FG = a/(x sin(α))
CG + FG = a - x cos(α)
So, solve this system, then you'll be able to find EB using the pythagorean theorem. Remember that a = √(S(ABCD)).

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dim plaza
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or just note that angle EDB is 60° and DB is 4

olive junco
dim plaza
olive junco
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Oh, the angle... Well, I like to solve generally.

dim plaza
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fair enough

olive junco
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In that case the cosine theorem will be:
EB^2 = 2a^2 + x^2 - 2√(2)ax cos(α + π/4)
Or, in terms of area of the square S:
EB^2 = 2S + x^2 - 2√(2S)x cos(α + π/4)

clear raft
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mmmmmmmmmmmmmmmmmmmmmmmmmmmmm

olive junco
clear raft
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nothing

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I was bored \

limpid mauve
limpid mauve
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@clear raft

dim plaza
limpid mauve
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i dont understand it

dim plaza
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why

limpid mauve
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how is DB=4

dim plaza
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because DC = 2sqrt(2)

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because A(ABCD) = 8

limpid mauve
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how is DB=4

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is there a rule

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applied

dim plaza
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DB^2 = DC^2 + CB^2

limpid mauve
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aah yes

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thannks

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then what should i do next

dim plaza
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DE = 2, DB = 4, EDB = 60°

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use cosine law or find easier way

limpid mauve
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i dont know that law

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hol on ill google it

shell fractal
dim plaza
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15 + 45

shell fractal
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oh nice