E, F, G, and H are the vertices of a polygon. Is polygon EFG

[Luci]

I know I´m loosing something here, it is just I cannot see it right now. Can you enlighten me please?


E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square?

(1) EFGH is a parallelogram.

(2) The diagonals of EFGH are perpendicular bisectors of one another.

Correct answer is E) Neither statment is sufficient.

Here is the explanation:

To prove that a quadrilateral is a square, you must prove that it is both a rhombus (all sides are equal) and a rectangle (all angles are equal).

(1) INSUFFICIENT: Not all parallelograms are squares (however all squares are parallelograms).

(2) INSUFFICIENT: If a quadrilateral has diagonals that are perpendicular bisectors of one another, that quadrilateral is a rhombus. Not all rhombuses are squares (however all squares are rhombuses).

If we look at the two statements together, they are still insufficient. Statement (2) tells us that ABCD is a rhombus, so statement one adds no more information (all rhombuses are parallelograms). To prove that a rhombus is a square, you need to know that one of its angles is a right angle or that its diagonals are equal (i.e. that it is also a rectangle).

The correct answer is E.


Well obviously each statement by itself is not sufficient, but, if the definition of a parallelogram is: In geometry, a parallelogram is a quadrilateral with two sets of parallel sides. The opposite sides of a parallelogram are of equal length, and the opposite angles of a parallelogram are congruent.

Knowing this, if we know that the diagonals are perpendicular and bisectors, shouldnt the figure has to be a square?
I cannot draw it here, but if we draw a rhombus with diagonals perpendicular bisectors (statement 2), if the sides of this rhombus have to be parallel (statement 1) then it has to be square, hasn´t it?

I imagine I´m missing something, and I know is hard to explain without drawing, but can you provide some help here?

Thanks

[Luci]

Thanks anyway and sorry for the post

[givemeanid]

Knowing this, if we know that the diagonals are perpendicular and bisectors, shouldnt the figure has to be a square?
I cannot draw it here, but if we draw a rhombus with diagonals perpendicular bisectors (statement 2), if the sides of this rhombus have to be parallel (statement 1) then it has to be square, hasn´t it?

Diagonals of a rhombus are perpendicular bisectors of each other. Sqaure is a special case but a rhombus DOES NOT have to be a square for the above mentioned diagonal property to be true.

[dbernst]

That is correct. For example, a diamond is an equaliteral parallelogram (rhombus) in which the diagonals are perpendicular bisectors of one another, but it is not a square.