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Quadrilateral

Expert replies
by yellowho » Mon Mar 07, 2011 10:59 pm
1) What is the relationship between diagonals and the main quadrilateral? (parallelogram, rhombus, square, rectangle)

2) Namely this property: "the two diagonal are bisector of each other"

a) What can we conclude if given this statement?

b) If we are given one of the shape above what can we conclude about the diagonals?

c) Are these diagonals necessarily perpendicular?

d) What if the diagonals are NOT bisector? What can we conclude?


I keep getting these problems wrong and the exp. has not been helpful.
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Source: — Problem Solving |

by Geva@EconomistGMAT » Tue Mar 08, 2011 1:38 am
yellowho wrote:1) What is the relationship between diagonals and the main quadrilateral? (parallelogram, rhombus, square, rectangle)

2) Namely this property: "the two diagonal are bisector of each other"

a) What can we conclude if given this statement?
Just that - that the diagonals cut each other in half - the point of intersection between diagonals is the midpoint of each diagonal.

b) If we are given one of the shape above what can we conclude about the diagonals?
Square: diags are equal, perpendicular and bisect each other. Basically, diags in a square create four equal right isosceles triangles.
Rectangle: diagonals are not longer perpendicular, but still equal and bisect each other. the diagonals create four isosceles (no longer right) triangles.
Parallelogram: diags are no longer perpendicular or equal, but still bisect each other. We have one long and one short diagonal, but the point of intersection is the midpoint of both diagonals.
Rhombus: a specific case of a parallelogram with all sides equal. the diagonals of a Rhombus are not equal (still one long, one short), but they still bisect each other, and, as opposed to other parallelograms, they are also perpendicular.

c) Are these diagonals necessarily perpendicular?

d) What if the diagonals are NOT bisector? What can we conclude?
Not sure i understand the question.

I keep getting these problems wrong and the exp. has not been helpful.
Geva
Senior Instructor
Master GMAT
1-888-780-GMAT
https://www.mastergmat.com
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by yellowho » Tue Mar 08, 2011 6:28 am
Geva,

My question D: "If given a shape and we know that it does not have a bisecting diagonal what can we conclude about the figure?" However, it sounds like all of those quadrilateral has a bisector from your discussion.

Can you take a look at this problem then:

Is there a diagonal in quadrilateral ABCD that is also a bisector in ABCD?

(1) ABCD is a parallelogram.

(2) The diagonals of ABCD intersect at right angles.

Should stat 1 be sufficient then?

Stat 2: Can you go backward in the logic? If perpendicular bisector then squares or rhombus or trapezoid? Is the relationship only 1 way?



[quote="Geva@MasterGMAT"][quote="yellowho"]1) What is the relationship between diagonals and the main quadrilateral? (parallelogram, rhombus, square, rectangle)

2) Namely this property: "the two diagonal are bisector of each other"

a) What can we conclude if given this statement?
Just that - that the diagonals cut each other in half - the point of intersection between diagonals is the midpoint of each diagonal.

b) If we are given one of the shape above what can we conclude about the diagonals?
Square: diags are equal, perpendicular and bisect each other. Basically, diags in a square create four equal right isosceles triangles.
Rectangle: diagonals are not longer perpendicular, but still equal and bisect each other. the diagonals create four isosceles (no longer right) triangles.
Parallelogram: diags are no longer perpendicular or equal, but still bisect each other. We have one long and one short diagonal, but the point of intersection is the midpoint of both diagonals.
Rhombus: a specific case of a parallelogram with all sides equal. the diagonals of a Rhombus are not equal (still one long, one short), but they still bisect each other, and, as opposed to other parallelograms, they are also perpendicular.

c) Are these diagonals necessarily perpendicular?

d) What if the diagonals are NOT bisector? What can we conclude?
Not sure i understand the question.

I keep getting these problems wrong and the exp. has not been helpful.[/quote][/quote]
Last edited by yellowho on Tue Mar 08, 2011 6:45 am, edited 1 time in total.
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by Geva@EconomistGMAT » Tue Mar 08, 2011 6:33 am
A completely different matter. Bisector, by default, means "bisector of the angle" - i.e. a line that emerges from a vertice and bisects the angle created by the vertice. You asked about diagonals bisecting each other: if that is the meaning, the question will then explicitly use : is there a diagonal in ABCD that bisects the other diagonal? (or bisects the figure).

yellowho wrote:Geva,

My question D: "If given a shape and we know that it does not have a bisecting diagonal what can we conclude about the figure?" However, it sounds like all of those quadrilateral has a bisector from your discussion.

Can you take a look at this problem then:

Is there a diagonal in quadrilateral ABCD that is also a bisector in ABCD?

(1) ABCD is a parallelogram.

(2) The diagonals of ABCD intersect at right angles.

Should stat 1 be sufficient then and that stat 2 is also sufficient because info. in 2 suggests that the figure can be a rhombus or a square? (not sure about 2 but shouldn't stat 1 be suff.?)


Geva@MasterGMAT wrote:
yellowho wrote:1) What is the relationship between diagonals and the main quadrilateral? (parallelogram, rhombus, square, rectangle)

2) Namely this property: "the two diagonal are bisector of each other"

a) What can we conclude if given this statement?
Just that - that the diagonals cut each other in half - the point of intersection between diagonals is the midpoint of each diagonal.

b) If we are given one of the shape above what can we conclude about the diagonals?
Square: diags are equal, perpendicular and bisect each other. Basically, diags in a square create four equal right isosceles triangles.
Rectangle: diagonals are not longer perpendicular, but still equal and bisect each other. the diagonals create four isosceles (no longer right) triangles.
Parallelogram: diags are no longer perpendicular or equal, but still bisect each other. We have one long and one short diagonal, but the point of intersection is the midpoint of both diagonals.
Rhombus: a specific case of a parallelogram with all sides equal. the diagonals of a Rhombus are not equal (still one long, one short), but they still bisect each other, and, as opposed to other parallelograms, they are also perpendicular.

c) Are these diagonals necessarily perpendicular?

d) What if the diagonals are NOT bisector? What can we conclude?
Not sure i understand the question.

I keep getting these problems wrong and the exp. has not been helpful.
Geva
Senior Instructor
Master GMAT
1-888-780-GMAT
https://www.mastergmat.com
Join the discussion

by yellowho » Tue Mar 08, 2011 6:48 am
How do you determine the bisector angle then? (Although the other info. are useful too for other problems)

[quote="Geva@MasterGMAT"]A completely different matter. Bisector, by default, means "bisector of the [b]angle[/b]" - i.e. a line that emerges from a vertice and bisects the angle created by the vertice. You asked about diagonals bisecting each other: if that is the meaning, the question will then explicitly use : is there a diagonal in ABCD that bisects the other diagonal? (or bisects the figure).


[quote="yellowho"]Geva,

My question D: "If given a shape and we know that it does not have a bisecting diagonal what can we conclude about the figure?" However, it sounds like all of those quadrilateral has a bisector from your discussion.

Can you take a look at this problem then:

Is there a diagonal in quadrilateral ABCD that is also a bisector in ABCD?

(1) ABCD is a parallelogram.

(2) The diagonals of ABCD intersect at right angles.

Should stat 1 be sufficient then and that stat 2 is also sufficient because info. in 2 suggests that the figure can be a rhombus or a square? (not sure about 2 but shouldn't stat 1 be suff.?)



[quote="Geva@MasterGMAT"][quote="yellowho"]1) What is the relationship between diagonals and the main quadrilateral? (parallelogram, rhombus, square, rectangle)

2) Namely this property: "the two diagonal are bisector of each other"

a) What can we conclude if given this statement?
Just that - that the diagonals cut each other in half - the point of intersection between diagonals is the midpoint of each diagonal.

b) If we are given one of the shape above what can we conclude about the diagonals?
Square: diags are equal, perpendicular and bisect each other. Basically, diags in a square create four equal right isosceles triangles.
Rectangle: diagonals are not longer perpendicular, but still equal and bisect each other. the diagonals create four isosceles (no longer right) triangles.
Parallelogram: diags are no longer perpendicular or equal, but still bisect each other. We have one long and one short diagonal, but the point of intersection is the midpoint of both diagonals.
Rhombus: a specific case of a parallelogram with all sides equal. the diagonals of a Rhombus are not equal (still one long, one short), but they still bisect each other, and, as opposed to other parallelograms, they are also perpendicular.

c) Are these diagonals necessarily perpendicular?

d) What if the diagonals are NOT bisector? What can we conclude?
Not sure i understand the question.

I keep getting these problems wrong and the exp. has not been helpful.[/quote][/quote][/quote][/quote]
Join the discussion

by Geva@EconomistGMAT » Tue Mar 08, 2011 6:56 am
yellowho wrote:How do you determine the bisector angle then? (Although the other info. are useful too for other problems)

For the kind of problem you presented, just look at some figures.
For a square, the diagonals also bisect the angles: each diagonal bisects the right angles into two 45 degree angles.

For a rectangle and a parallelogram, the diagonals lose this feature as well - they are no longer bisectors.

For a Rhombus, the diagonals also bisect the angles as well.
Geva
Senior Instructor
Master GMAT
1-888-780-GMAT
https://www.mastergmat.com
Join the discussion