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# DS- Geometry

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Senior Manager
Joined: 30 Oct 2004
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19 Oct 2005, 20:09
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See attached

[EDITED THE DIAGRAM]
Attachments

triangle_403.gif [ 5.79 KiB | Viewed 1136 times ]

_________________

-Vikram

Last edited by vikramm on 19 Oct 2005, 21:07, edited 1 time in total.
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19 Oct 2005, 21:03
1) insuff coz you can form many triangles ABC with C and B are intersection points between any tangents to the circle with two sides of the angle A.
2) sorry, I can not attach my illustration ( in paint file) here.
As vikramm just edited the picture, A in my solution is indeed C in his picture and vice versa

You can easily prove that AE is indeed AD.
We have angle FOD= 360- (90+90+45) = 135 --->AOF= 180-135=45--->OAF= 445 or DAB= 45
We also have EAB=45
DAB=EAB= 45 -----> AD is indeed AE.
AD=1/2 BC ( Coz AB=AC ---> AD is also the median of right triangle ABC) ----> BC= 2*AD= 2+2*sqrt2
Area of ABC= AD*BC/2= (1+sqrt 2)^2

1 and 2 must be together to be sufficient to provide answer. C is correct.

NOTE: You'd better open the doc file in seperate page so that you can understand my explanation. I'm badly sorry for my clumsy illustration coz I know no other means to draw some illustration.
Attachments

ABC.doc [24.5 KiB]

Last edited by laxieqv on 19 Oct 2005, 21:19, edited 3 times in total.
Senior Manager
Joined: 30 Oct 2004
Posts: 284
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19 Oct 2005, 21:08
My bad... I edited the diagram.
_________________

-Vikram

19 Oct 2005, 21:08
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