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Area of square?

Mathematics
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|dw:1350208678915:dw|
What is that?
is that 5,7,10

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Other answers:

A square, a point inside of it distances from 3 vertices as shown
9,10,7
This is for experimentX, who loves algebra..:-)
haha ... you are kidding!!
It's not too hard..(it's still too early)
is that all info??
if we call ne side a then the area is \[A=a^2\]
and the side is less than 17
according to triangle equality
|dw:1350209046775:dw|
\[9^2=a^2+10^2-2(10)(a)\cos \theta\] \[762=10^2+a^2-2(a)(10)\cos (90-\theta)\]
2 var 1 equation suggests simultaneous equations as the way to go....
|dw:1350209083317:dw|
i meant 7^2|dw:1350176811646:dw|
JonasK is on the right track
what about using the area rule
probably @hartnn method would work out too ... since they are all independent equations.
Probably...
"what about using the area rule" U just want to get rid of the 90- bit and then you can get rid of theta
yes
still that's quite a bit of algebra since they are all non linear equations ... let's try something better.
okay
Cos of angle opposite 7 is sin theta....
|dw:1350209708034:dw|
What I mean is that if angle opposite 9 is theta, then cos of angle opposite 7 is sin theta
As JonasK had it at first...
expX, now eliminate theta....
you know i rely on wolf for this http://www.wolframalpha.com/input/?i=2*10*a+sin%28theta%29+%3D+10^2+%2B+a^2++-+9^2%2C+2*10*a+cos%28theta%29+%3D+10^2+%2B+a^2++-+7^2
\[32=-20(a)\cos \theta+20(a)\sin \theta\]
Don't be lazy....:-)
haha ..
Sin and Cos should auto response.....
\[\frac{ 32 }{ -20 a}=\cos \theta -\sin \theta\]
\[\cos \theta -\sin \theta =\sqrt{2} \sin(45-\theta)\]
with - before
@Jonask An identity (well known) in Sin and Cos?
haa ... i was being fool trying to eliminate theta directly.
|dw:1350210145598:dw|
A bit messy but sufficient...:-)
seems great
still quartic equation is pretty awful ... i must say!!
Yes, squaring and adding gets shot of theta...
U get after simplification...400x^2 = (x^2 +19)^2 + (x^2 +51)^2 -> x^4 -130 x^2 +1481
Which is quadratic in x^2
Oh .. that's pretty nice!!
direct answer.
let \[-10^2-a^2=b\] \[\frac{ (7^2 -b)^2}{ (-20a )^2}+\frac{ (9^2-b)^2 }{ (20a)^2 }=1\]
@estudier seems better
|dw:1350210434675:dw| x+2y = 7 ..(1) y^2 + (x+y)^2 = 81 ..(2) hope that does it..
correction : x+2y = 10
solving shouldnt be much problem since from (1) , x+y = 10-y and we plug that directly in (2)
the line with lenght 10 is not necessarily a diagonal when extrapolated ??
not sure
Eh? Some extra ssumptions there....
ohh..lol..so sorry! :P
no prob
\[a=10,834, a=3,552\]
can we eliminate according to triangle inequality 3,552 such that \[A=(10,834)^2\]
65 + 14 sqrt 14 I think.
same thing i just approaximated it
\[A=(\sqrt{64+14\sqrt{14}} )^2\]
\[A=65+14\sqrt{14}\]
@shubhamsrg if we consider the triangles to be equilateral then how wuld we solve it |dw:1350178935156:dw| with common side 10,
equilateral ?
|dw:1350179036162:dw| such that the legth is now diagonal
you mean like this: |dw:1350211506937:dw| ??
yes
for that i had given the solution above.. x+2y = 10 ..(1) y^2 + (x+y)^2 = 81 ..(2) from (1), x+y = 10-y and we plug this in (2) to solve for y, and x and ultimately area..
2 Attachments
nice work @sauravshakya
Thanx @Jonask
what disqualifies 65-14root14
@Jonask That's for x^2, remember....
I think I have written reason for that too.
@sauravshakya Yes, good job (I think he just forgot it was for x^2)

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