teal wrote:

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.

For B:

Given one angle of the parallelogram, all its angles are determined as you mentioned.

So, the "shape" of the parallelogram is uniquely determined, meaning all of them are similar, they are just "smaller" or "larger".

Similarity also means the lengths of the sides and the diagonals (which can also be seen as sides in some triangles, like half of the parallelogram) increase or decrease proportionally, so their difference cannot remain a constant.

For C:

Please, refer to the attached drawing, where y in fact is \(2x+2-x\sqrt{3}\).

The diagonals then can be expressed in terms of x, but we don't have any specific value for x.

Hence, answer E.

Attachments

Parallelogram30.jpg [ 11.92 KiB | Viewed 1891 times ]

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