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vertices must be on the origin?

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by sanju09 » Fri May 23, 2014 3:36 am
Good question, worth a try!
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by theCodeToGMAT » Fri May 23, 2014 4:41 am
Nice one...

=> a^2 + b^2 = 100

Considering only one vertice.

Possible = 10,0 or 6,8 or 8,6 or 0,10 or -6,8 or -8,6 ==> 6 in 1 & 2 quadrant

Similarly for 3 & 4th quadrant

So, 12

[spoiler]ANswer {E}???[/spoiler]
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by sanju09 » Fri May 23, 2014 5:02 am
theCodeToGMAT wrote:Nice one...

=> a^2 + b^2 = 100

Considering only one vertice.

Possible = 10,0 or 6,8 or 8,6 or 0,10 or -6,8 or -8,6 ==> 6 in 1 & 2 quadrant

Similarly for 3 & 4th quadrant

So, 12

[spoiler]ANswer {E}???[/spoiler]
Will the answer be same if we first draw the possibilities and then count?
The mind is everything. What you think you become. -Lord Buddha



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