ChampakDas Posted November 12, 2012 Report Posted November 12, 2012 sure man... [quote name='cavaliers' timestamp='1352744898' post='1302784079'] ] Bhayya I think this should work. I will check and get back to you. Thanks [/quote]
ChampakDas Posted November 12, 2012 Report Posted November 12, 2012 adhi solve avuthey post the relation here [quote name='cavaliers' timestamp='1352744898' post='1302784079'] ] Bhayya I think this should work. I will check and get back to you. Thanks [/quote]
cavaliers Posted November 12, 2012 Author Report Posted November 12, 2012 Leedhu bhayya... Any way I try I am getting... X = t ... here is how I did that [img]http://i46.tinypic.com/bje9f7.jpg[/img]
powerstar02 Posted November 12, 2012 Report Posted November 12, 2012 [quote name='GatisKandis' timestamp='1352743183' post='1302783917'] since you got many replies and looks like its no straight..so I would try with the area concept...just check if this gives some reasonable relation of t in terms of x and y area of triangle = area of inside two triangles. 1/2 * x * y = 1/2 * t* y + (find the area of the other triangle 1 / 2 *[size=5] sqrt(x^2 + y ^2)[/size] * height) check if this works [/quote] bhayya..it has to be (x-t) ....which cancels"t" on both sides...it doesnt work..
tupaki007 Posted November 13, 2012 Report Posted November 13, 2012 you cannot answer t based on the given information. Because, other than the fact that the smaller right angle triangle is inside the bigger right angle triangle there is no other relationship between the two triangles. So they are just two independent right angled triangles, and one of the sides is equal length for both triangles.
Chakram12 Posted November 13, 2012 Report Posted November 13, 2012 [quote name='cavaliers' timestamp='1352736066' post='1302783125'] Guys, I have the coordinates X and Y. Is there a way we can find t? Thanks. [img]http://i48.tinypic.com/6nuohi.jpg[/img] [/quote]nothing is impossible
ChampakDas Posted November 13, 2012 Report Posted November 13, 2012 got it? [quote name='cavaliers' timestamp='1352752548' post='1302785835'] Leedhu bhayya... Any way I try I am getting... X = t ... here is how I did that [img]http://i46.tinypic.com/bje9f7.jpg[/img] [/quote]
ChampakDas Posted November 13, 2012 Report Posted November 13, 2012 (x-t) ardam kale? in which area are you talking about? [quote name='powerstar02' timestamp='1352752795' post='1302785872'] bhayya..it has to be (x-t) ....which cancels"t" on both sides...it doesnt work.. [/quote]
powerstar02 Posted November 13, 2012 Report Posted November 13, 2012 [quote name='GatisKandis' timestamp='1352768030' post='1302787509'] (x-t) ardam kale? in which area are you talking about? [/quote] aa x-t length traingle area...
vikuba Posted November 13, 2012 Report Posted November 13, 2012 Similar triangles principle use chesi cheyochu
ChampakDas Posted November 13, 2012 Report Posted November 13, 2012 mama I haven't considered (x-t) as base rather I considered sqrt(x^2+y^2) as base. You can consider (x-t) as the base also but the problem is on simplification you will get the universal truths that is xy = xy....andhuku I chose hypotenuse as the base. [quote name='powerstar02' timestamp='1352768244' post='1302787511'] aa x-t length traingle area... [/quote]
I am back Posted November 13, 2012 Report Posted November 13, 2012 It can be done.. Use angles... Not ares.... Oka linear equation ostadi... And t will have infinite solutions... But in terms of x and y, it will be 1 equation.... Maa daddy k2s ni adugu... 2 mins la Answer cheptadu...
powerstar02 Posted November 13, 2012 Report Posted November 13, 2012 [quote name='GatisKandis' timestamp='1352768694' post='1302787524'] mama I haven't considered (x-t) as base rather I considered sqrt(x^2+y^2) as base. You can consider (x-t) as the base also but the problem is on simplification you will get the universal truths that is xy = xy....andhuku I chose hypotenuse as the base. [/quote] well,for a right angled traingle a side which has the 90degree angle on it is considered to be base..and then only you can use 1/2bh to calculate the area......and yes xy = xy vasthadhi thats what i mean by saying "t" is being eliminated... and bhayya are you sure sqrt(x^2+y^2) as base ga consider chesi area 1/bh vaadachu ani..?may be i am missing something.....?
ChampakDas Posted November 13, 2012 Report Posted November 13, 2012 angle tho sambandham ledhu mitrama afaik, lets say you have a parallelogram oka diagonal gisavu anuko then the area of parallelogram is the sum of areas of two triangles kada needless to say none of them is right angled triangle. [quote name='powerstar02' timestamp='1352776580' post='1302787963'] well,for a right angled traingle a side which has the 90degree angle on it is considered to be base..and then only you can use 1/2bh to calculate the area......and yes xy = xy vasthadhi thats what i mean by saying "t" is being eliminated... and bhayya are you sure sqrt(x^2+y^2) as base ga consider chesi area 1/bh vaadachu ani..?may be i am missing something.....? [/quote]
powerstar02 Posted November 13, 2012 Report Posted November 13, 2012 [quote name='GatisKandis' timestamp='1352776955' post='1302787983'] angle tho sambandham ledhu mitrama afaik, lets say you have a parallelogram oka diagonal gisavu anuko then the area of parallelogram is the sum of areas of two triangles kada needless to say none of them is right angled triangle. [/quote] yes..sum of the two triangles ee bhayya...but as far as i remember 1/2*base*height valids only for Rht triangles ani...the simples is square and rectangle and break them along the diagonal which sums up to 2(1/2*b*h)....
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