In the xy-plane, there is a rectangle ABCO shown as the above figure.

OB = 5, this can be any point with a length of 5 from point O. - Insufficent.

B(3,4) = Clearly sufficient to get the points of A and C, thus perimeter can be found. - Sufficient

Option B

==> If you modify the original condition and the question, the quadrilateral ABCD is a rectangle, so in order to find the perimeter, you only need to know B, hence con 2) is sufficient.

Therefore, the answer is B.
Answer: B
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Why is not A sufficient?. Given B is a point and O the origin will the diagonal length not be 5? And can't we calculate the sides of the rectangle from the diagonal? Can someone pls explain.


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You cannot find the area of a rectangle just knowing the length of its diagonal. Area of a rectangle is length*width and knowing the length of diagonal is not enough to calculate these values. In other words knowing the length of hypotenuse (diagonal) in a right triangle (created by length and width), is not enough to find the legs of it (length and width).
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Thanks Bunuel. Got it.

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if the diagonal is 5 then the two sides can be 3 and 4. so we can find the perimeter.
I think even statement 1 is sufficient to find the perimeter.
So I think the answer is D

i also have the similar doubt doesnt matter if 4=OA / AB WE CAN FIND THE PERIMETER HELP?

Bunuel MathRevolution Please help !!

Lets start with STATEMENT 1:-
Give OB = 5 .

Let the co-ordinates of B be (x,y).
Hence as per the distance formula \((x-0)^2 + (y-0)^2 = 5^2\)

\(x^2 + y^2 = 5 ^2\)

So , x can be 3 or 4 , and hence y can be 4 or 3.

Once we get the co-ordinates of B , its easy to find the perimeter.

STATEMENT 2 is sufficient itself.

Hence OPTION D.

Thanks

This is a common trap. You assume that x and y are integers but we don't know that.

Knowing that the hypotenuse equals to 5 does NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 3:4:5. Or in other words: \(x^2+y^2=5^2\) does NOT mean that \(x=3\) and \(y=4\) (or vise-versa). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=5^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=3\) and \(y=5\).

For example: \(x=1\) and \(y=\sqrt{24}\) or \(x=2\) and \(y=\sqrt{21}\)...

I collected the questions which use this trap:
https://gmatclub.com/forum/the-circular ... 67645.html
https://gmatclub.com/forum/figure-abcd- ... 48899.html
https://gmatclub.com/forum/in-right-tri ... 63591.html
https://gmatclub.com/forum/m22-73309-20.html
https://gmatclub.com/forum/points-a-b-a ... 84423.html
https://gmatclub.com/forum/if-vertices- ... 59-20.html
https://gmatclub.com/forum/if-p-is-the- ... 35832.html
https://gmatclub.com/forum/if-the-diago ... 04205.html
https://gmatclub.com/forum/what-is-the- ... 05414.html
https://gmatclub.com/forum/what-is-the- ... 96381.html
https://gmatclub.com/forum/pythagorean- ... 31161.html
https://gmatclub.com/forum/given-that-a ... 27051.html
https://gmatclub.com/forum/m13-q5-69732 ... l#p1176059
https://gmatclub.com/forum/m20-07-trian ... 71559.html
https://gmatclub.com/forum/what-is-the- ... 96381.html

Hope this helps.
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