Area of square?

- anonymous

Area of square?

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- anonymous

|dw:1350208678915:dw|

- anonymous

What is that?

- anonymous

is that 5,7,10

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## More answers

- anonymous

A square, a point inside of it distances from 3 vertices as shown

- anonymous

9,10,7

- anonymous

This is for experimentX, who loves algebra..:-)

- experimentX

haha ... you are kidding!!

- anonymous

It's not too hard..(it's still too early)

- experimentX

is that all info??

- anonymous

if we call ne side a
then the area is
\[A=a^2\]

- anonymous

and the side is less than 17

- anonymous

according to triangle equality

- experimentX

|dw:1350209046775:dw|

- anonymous

\[9^2=a^2+10^2-2(10)(a)\cos \theta\]
\[762=10^2+a^2-2(a)(10)\cos (90-\theta)\]

- anonymous

2 var 1 equation suggests simultaneous equations as the way to go....

- hartnn

|dw:1350209083317:dw|

- anonymous

i meant 7^2|dw:1350176811646:dw|

- anonymous

JonasK is on the right track

- anonymous

what about using the area rule

- experimentX

probably @hartnn method would work out too ... since they are all independent equations.

- anonymous

Probably...

- anonymous

"what about using the area rule"
U just want to get rid of the 90- bit and then you can get rid of theta

- anonymous

yes

- experimentX

still that's quite a bit of algebra since they are all non linear equations ... let's try something better.

- anonymous

okay

- anonymous

Cos of angle opposite 7 is sin theta....

- experimentX

|dw:1350209708034:dw|

- anonymous

What I mean is that if angle opposite 9 is theta, then cos of angle opposite 7 is sin theta

- anonymous

As JonasK had it at first...

- anonymous

expX, now eliminate theta....

- experimentX

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

- anonymous

\[32=-20(a)\cos \theta+20(a)\sin \theta\]

- anonymous

Don't be lazy....:-)

- experimentX

haha ..

- anonymous

Sin and Cos should auto response.....

- anonymous

\[\frac{ 32 }{ -20 a}=\cos \theta -\sin \theta\]

- anonymous

\[\cos \theta -\sin \theta =\sqrt{2} \sin(45-\theta)\]

- anonymous

with - before

- anonymous

@Jonask An identity (well known) in Sin and Cos?

- experimentX

haa ... i was being fool trying to eliminate theta directly.

- experimentX

|dw:1350210145598:dw|

- anonymous

A bit messy but sufficient...:-)

- anonymous

seems great

- experimentX

still quartic equation is pretty awful ... i must say!!

- anonymous

Yes, squaring and adding gets shot of theta...

- anonymous

U get after simplification...400x^2 = (x^2 +19)^2 + (x^2 +51)^2 ->
x^4 -130 x^2 +1481

- anonymous

Which is quadratic in x^2

- experimentX

Oh .. that's pretty nice!!

- experimentX

direct answer.

- anonymous

let \[-10^2-a^2=b\]
\[\frac{ (7^2 -b)^2}{ (-20a )^2}+\frac{ (9^2-b)^2 }{ (20a)^2 }=1\]

- anonymous

@estudier seems better

- shubhamsrg

|dw:1350210434675:dw|
x+2y = 7 ..(1)
y^2 + (x+y)^2 = 81 ..(2)
hope that does it..

- shubhamsrg

correction :
x+2y = 10

- shubhamsrg

solving shouldnt be much problem since from (1) , x+y = 10-y
and we plug that directly in (2)

- anonymous

the line with lenght 10 is not necessarily a diagonal when extrapolated ??

- anonymous

not sure

- anonymous

Eh? Some extra ssumptions there....

- shubhamsrg

ohh..lol..so sorry! :P

- anonymous

no prob

- anonymous

\[a=10,834, a=3,552\]

- anonymous

can we eliminate according to triangle inequality 3,552
such that \[A=(10,834)^2\]

- anonymous

65 + 14 sqrt 14 I think.

- anonymous

same thing i just approaximated it

- anonymous

\[A=(\sqrt{64+14\sqrt{14}} )^2\]

- anonymous

\[A=65+14\sqrt{14}\]

- anonymous

@shubhamsrg if we consider the triangles to be equilateral then how wuld we solve it
|dw:1350178935156:dw|
with common side 10,

- shubhamsrg

equilateral ?

- anonymous

|dw:1350179036162:dw| such that the legth is now diagonal

- shubhamsrg

you mean like this:
|dw:1350211506937:dw|
??

- anonymous

yes

- shubhamsrg

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..

- anonymous

nice work @sauravshakya

- anonymous

Thanx @Jonask

- anonymous

what disqualifies 65-14root14

- anonymous

@Jonask That's for x^2, remember....

- anonymous

I think I have written reason for that too.

- anonymous

@sauravshakya Yes, good job (I think he just forgot it was for x^2)

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