What is the difference between the lengths of diagonals of

I dont follow - how do you move one point about but not another? If it's a parallelogram, then surely moving one point will move one of the other's if it is to maintain two parallel sides?

Agreed - but how do you move a single point around? Im confused.

I am confused to. How can diagonals of a parallelogram be different in length.
They have to be equal in length to satisfy the definition of a parallelogram which is
" A quadrilateral with both pairs of opposite sides parallel and equal in length."

What is the difference between the lengths of diagonals of parallelogram \(ABCD\) ( \(BD \gt AC\) )?

\(AB - BC = 2\)
\(\angle ABC = 30\) degrees

OA:E

Here is my reasoning -

A) Difference between two diagonals does not restrict the shape of the quad. If it's shape is closer to rectangle then diff between diagonals is very less as opposed to if you change the quad's shape to move away from being a rectangle. ( Rectangle has equal diags.) - Insuff.

B) One angle =30. Due to the property of Parallelogram, this restricts the other angles to 30, 150, 30 (opp angles equal and adjacent angles supplementary). So we have a definite shape for parallelogram but NOT a fixed size so we can still vary the size. As size will vary, the difference between diagonals should also vary -

Here is my question for Forum moderators - as the size of parallelogram will increase will the difference between it's diagonals also increase , right?? vice versa for decrease???? Please confirm.

For C -

You can still vary the size of the quad, so the difference in diagonals should vary, hence E.

It says "sides" and not diagonals!!!

HI,

I have couple of queries here.

1. the question is talking about parallelogram. so it can be rectangle ( can not be square or rohumbus) or parallelogram ( oppsite sites and opposite angles are equal)

so what you mean by

"Difference between two diagonals does not restrict the shape of the quad. If it's shape is closer to rectangle then diff between diagonals is very less as opposed to if you change the quad's shape to move away from being a rectangle"

Thanks.

we can't fix the shape ( parallelogram or rhomus). hence E

BrentGMATPrepNow

Can you pls explain why C isn't correct

Aren't we locking the shape?

Same question as someone had a while ago…….


Statement 1 gives us the difference between adjacent sides of the parallelogram.

Statement 2 fixes the shape of this figure.

Thus:

-For a parallelogram in which opposite sides are equal and parallel

-With the angles locked in at 30, 150, 30, 150

-and with the difference between Adjacent Sides being 2

Isn’t this shape “fixed in place”? In other words, there is only 1 unique Parallelogram with the proper shape and size that fits these parameters. Thus, there must be only 1 unique value for the difference in diagonals.

Although we cannot find the Unique Value ourselves, there MUST be a Unique Value.

Can we therefore say C is the answer?

Posted from my mobile device

You are correct in saying that there exists one unique parallelogram with these conditions. However, for such DS questions that ask 'What is the value...?' we can say that A or B or whatever only if we get a unique value as an answer.

In this case we get a unique parallelogram but not a unique value as an answer to the question.

So E is the answer.

Great. Thank you!

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