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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
6
the fastest approach is by 30:60:90 angle formed by BAE
do it quick by 1:root3:2
you get root3/2 in quantity A as well as in quantity B
so c is the answer
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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bim1946 wrote:
the fastest approach is by 30:60:90 angle formed by BAE
do it quick by 1:root3:2
you get root3/2 in quantity A as well as in quantity B
so c is the answer



is it 30-60-90 or 45-90-45?

Can you show in details
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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Thank you.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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Can Someone Pls explain how we know that BA = BD...or just how BDAE is a square.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
4
1
pranab01 wrote:
pclawong wrote:
explain please


This is a tough ques, let me try

consider AB=DE =x

now join Band D

in the quadrilateral ABDE we have angle A =90 degree

angle E = 90 degree

and AB=DE = x

So quadrilateral ABDE is a square.

and Area of square ABDE = x^2

Now we need the area of triangle ABE = Area of square ABDE - area of triangle BDE - area of triangle DEF

Area of triangle BDF = 1/2 * base * altitude =\(\frac{1}{2}\) \(* x *\)\(\frac{1}{2}*x\)(since side BD =x and 1/2 * total distance = altitude of triangle BDF, as BDEA is square)

and similarly Area of Triangle DEF =\(\frac{1}{2}\) \(*x *\) \(\frac{1}{2}*x\)

Therefore Area of Triangle ABE = \(x^2\) - (\(\frac{1}{4}*x^2\) + \(\frac{1}{4}*x^2\))

=\(\frac{1}{2} *x^2\)

Now in quadrilaterl BCDF we have
BC=CD , BC parallel to FD and BE parallel to CD and angle F =90degree (since the diagonal cuts at 90 degree )

Therefore we can consider BCDF is a square

Area of square BCDF=\(\frac{diagonal ^2}{2}\)

= \(\frac{1}{2} *x^2\)

Hence option C.

All queries are welcome!!!


and AB=DE = x

So quadrilateral ABDE is a square.

To be a square AB = AE has to be given. Not AB=DE.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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LethalMonkey wrote:

and AB=DE = x

So quadrilateral ABDE is a square.

To be a square AB = AE has to be given. Not AB=DE.


Plz see the attached diagram,


In the quadrilateral ABDE,

we have AB = DE and angle A = angle E = 90 degree and the diagonal bisect each other at equal length . SO angle b = angle d = 90

Now the triangle ABF and triangle BDF are equal, since they both have the common same height

i.e AB = BD.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
bim1946 wrote:
the fastest approach is by 30:60:90 angle formed by BAE
do it quick by 1:root3:2
you get root3/2 in quantity A as well as in quantity B
so c is the answer


Dear friend,
how do you get 30:60:90, please explain, really need your help.
thanks
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



Attachment:
box.jpg



AB = DE, BC = CD, BE is parallel to CD, and BC is parallel to DF.

Quantity A
Quantity B
The area of triangle ABE
The area of quadrilateral BCDF


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

I seem to be the only one who thinks B is greater. My reasoning is thus:

They both have same lengths for Base and Height. However, Area of a Triangle is 1/2bh which makes it smaller than bh (Area of a quadrilateral).

@poster can you provide the solution for this question?
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
Expert Reply
Please, see the explanation above.

Regards
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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pranab01 wrote:
LethalMonkey wrote:

and AB=DE = x

So quadrilateral ABDE is a square.

To be a square AB = AE has to be given. Not AB=DE.


Plz see the attached diagram,


In the quadrilateral ABDE,

we have AB = DE and angle A = angle E = 90 degree and the diagonal bisect each other at equal length . SO angle b = angle d = 90

Now the triangle ABF and triangle BDF are equal, since they both have the common same height

i.e AB = BD.


I still do not understand how is ABDE a square. How do we know AB=AE or AB = BD? All angles are 90 degrees, which means it can be a rectangle.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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We have no way of knowing whether quadrilateral BCDF is a square or not, we also have no way of knowing whether ABE is a 30-60-90 triangle. Fortunately, these details don't matter. Observe:

The area of triangle ABE is \(\frac{AB*AE}{2}\). Now, note that AB=DE and since CD || BE and BC || DF, we must have BC = BF = DF = CD. We can then infer that CF projects orthogonally down onto AE, meaning its length is the same as DE which forms a right angle with AE. Thus, CF = DE = AB. We then can subdivide the quadrilateral BCDF into four equal sized right triangles each with base \(\frac{AE}{2}\) and height \(\frac{AB}{2}\), giving each an area of \(\frac{AE*AB}{8}\). We then multiply this quantity by four to get \(\frac{AB*AE}{2}\), the same as the area of triangle ABE.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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garaidh wrote:
We have no way of knowing whether quadrilateral BCDF is a square or not, we also have no way of knowing whether ABE is a 30-60-90 triangle. Fortunately, these details don't matter. Observe:

The area of triangle ABE is \(\frac{AB*AE}{2}\). Now, note that AB=DE and since CD || BE and BC || DF, we must have BC = BF = DF = CD. We can then infer that CF projects orthogonally down onto AE, meaning its length is the same as DE which forms a right angle with AE. Thus, CF = DE = AB. We then can subdivide the quadrilateral BCDF into four equal sized right triangles each with base \(\frac{AE}{2}\) and height \(\frac{AB}{2}\), giving each an area of \(\frac{AE*AB}{8}\). We then multiply this quantity by four to get \(\frac{AB*AE}{2}\), the same as the area of triangle ABE.


Hi,

Could you plz let me know how you have proved BC = BF = DF = CD as only AB=DE and since CD || BE and BC || DF is mentioned?


Moreover can we really infer " that CF projects orthogonally down onto AE "? unless BCDF is a rhombus,square,parallelogram
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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pranab01 wrote:
garaidh wrote:
We have no way of knowing whether quadrilateral BCDF is a square or not, we also have no way of knowing whether ABE is a 30-60-90 triangle. Fortunately, these details don't matter. Observe:

The area of triangle ABE is \(\frac{AB*AE}{2}\). Now, note that AB=DE and since CD || BE and BC || DF, we must have BC = BF = DF = CD. We can then infer that CF projects orthogonally down onto AE, meaning its length is the same as DE which forms a right angle with AE. Thus, CF = DE = AB. We then can subdivide the quadrilateral BCDF into four equal sized right triangles each with base \(\frac{AE}{2}\) and height \(\frac{AB}{2}\), giving each an area of \(\frac{AE*AB}{8}\). We then multiply this quantity by four to get \(\frac{AB*AE}{2}\), the same as the area of triangle ABE.


Hi,

Could you plz let me know how you have proved BC = BF = DF = CD as only AB=DE and since CD || BE and BC || DF is mentioned?


Moreover can we really infer " that CF projects orthogonally down onto AE "? unless BCDF is a rhombus,square,parallelogram


Certainly. Since AB=DE, CD || BE, and BC || DF, we have BC=CF=DF=CD due to the statement "If two parallel lines intersect two parallel lines, each of the resulting segments will be equal to their opposite." This is equivalent to the Euclidean parallel postulate (which we are allowed to assume on all GRE questions). It should not be two hard to prove this or find a proof of this if you are unconvinced, but it should make sense intuitively because if it were the case that BF > DF, we would necessarily have BF (BE) intersecting CD, which is impossible by assumption CD || BE. Thus it must be the case that BC = DF and BF = CD and since we know BC = CD, we have by transitivity of equality that BF = DF (that is, they are all equal to each other).

As for your second question, we do know that BCDF is a parallelogram because CD || BF and BC || DF. Furthermore, since we know by the above reasoning that CD=DF, if we were to draw a circle of radius CD centered at D, it would intersect both C and F. We can then draw another circle of the same radius centered at B, which will also intersect C and F. You might recall that this is the construction of a perpendicular to BD. Now, we know that AB || DE and that AB=DE. We then have that BD || AE (again by statement equivalent to Euclid's Parallel Postulate), thus CF must be perpendicular with AE. We have CF || DE because both make right angles with AE and both intersect the parallel lines BE and CD. Then by Euclid's Fifth, we have CF=DE.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
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garaidh wrote:

Certainly. Since AB=DE, CD || BE, and BC || DF, we have BC=CF=DF=CD due to the statement "If two parallel lines intersect two parallel lines, each of the resulting segments will be equal to their opposite." This is equivalent to the Euclidean parallel postulate (which we are allowed to assume on all GRE questions). It should not be two hard to prove this or find a proof of this if you are unconvinced, but it should make sense intuitively because if it were the case that BF > DF, we would necessarily have BF (BE) intersecting CD, which is impossible by assumption CD || BE. Thus it must be the case that BC = DF and BF = CD and since we know BC = CD, we have by transitivity of equality that BF = DF (that is, they are all equal to each other).

As for your second question, we do know that BCDF is a parallelogram because CD || BF and BC || DF. Furthermore, since we know by the above reasoning that CD=DF, if we were to draw a circle of radius CD centered at D, it would intersect both C and F. We can then draw another circle of the same radius centered at B, which will also intersect C and F. You might recall that this is the construction of a perpendicular to BD. Now, we know that AB || DE and that AB=DE. We then have that BD || AE (again by statement equivalent to Euclid's Parallel Postulate), thus CF must be perpendicular with AE. We have CF || DE because both make right angles with AE and both intersect the parallel lines BE and CD. Then by Euclid's Fifth, we have CF=DE.


HI,
As per Euclidean parallel postulate, the distance between two parallel lines at point will always be equal, so yes we can concur that CD =BF and BC = FD, but since BC = CD, we can write as BC = CD = DF = FB,

This led us that BCDF is a rhombus, moreover the diagonal bisect each perpendicularly so CF and BD will be perpendicular and if we consider two triangles BCD and BDF, they will always have equal area.

The only confusion with the statement CF || DE,

However I approached a different way,

SInce AB || DE and BD || AE also AB= DE and BD = AE (Euclidean parallel postulate), so ABDE is a parallelogram and the diagonals of the parallelogram divides it into 4 triangles of equal area.
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
Critical and C
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
:lol: :lol:
imagine it like a letter symmetric envelope. very fast method
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Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]
7
It's not given that it is a square. May be a square, maybe not. But it's DEFINITELY a RECTANGLE. Join BD to complete the rectangle.

Now, area of triangle is width*length / 2.

To find the area of the rhombus, use BD dissector and calculate the area of top and bottom triangles. The length is l for both. And the height is half of the width because it's a rectangle and the diagonal intersects at the center, dividing the sides into equal proportions. So height = width/2.

So rhombus area = 2 *[(w/2)*(l)]/2 = wl/2.

Hence both areas are equal.


(Note: Ultimately being a square or rectangle does not matter in the end since the diagram is symmetric and can be stretched to form a square OR a rectangle but the PROPORTION of the areas of the 2 quantities will not change.)
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