estudier 3 years ago Area of square?

1. estudier

|dw:1350208678915:dw|

2. sauravshakya

What is that?

is that 5,7,10

4. estudier

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

5. estudier

9,10,7

6. estudier

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

7. experimentX

haha ... you are kidding!!

8. estudier

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

9. experimentX

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

13. experimentX

|dw:1350209046775:dw|

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

15. estudier

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

16. hartnn

|dw:1350209083317:dw|

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

18. estudier

JonasK is on the right track

what about using the area rule

20. experimentX

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

21. estudier

Probably...

22. estudier

"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

24. experimentX

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

okay

26. estudier

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

27. experimentX

|dw:1350209708034:dw|

28. estudier

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

29. estudier

30. estudier

expX, now eliminate theta....

31. experimentX

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

33. estudier

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

34. experimentX

haha ..

35. estudier

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

39. estudier

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

40. experimentX

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

41. experimentX

|dw:1350210145598:dw|

42. estudier

A bit messy but sufficient...:-)

seems great

44. experimentX

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

45. estudier

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

46. estudier

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

47. estudier

48. experimentX

Oh .. that's pretty nice!!

49. experimentX

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

@estudier seems better

52. shubhamsrg

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

53. shubhamsrg

correction : x+2y = 10

54. shubhamsrg

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

57. estudier

Eh? Some extra ssumptions there....

58. shubhamsrg

ohh..lol..so sorry! :P

no prob

$a=10,834, a=3,552$

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

62. estudier

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,

67. shubhamsrg

equilateral ?

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

69. shubhamsrg

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

yes

71. 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..

72. sauravshakya

nice work @sauravshakya

74. sauravshakya

what disqualifies 65-14root14

76. estudier

77. sauravshakya

I think I have written reason for that too.

78. estudier

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