It is currently 30 May 2020, 20:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# AB = DE, BC = CD, BE is parallel to CD, and BC is par

Author Message
TAGS:
Founder
Joined: 18 Apr 2015
Posts: 11192
Followers: 239

Kudos [?]: 2809 [3] , given: 10653

AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  02 Aug 2017, 14:49
3
KUDOS
Expert's post
00:00

Question Stats:

51% (02:05) correct 48% (02:26) wrong based on 148 sessions

Attachment:

#GREpracticequestion AB = DE, BC = CD, BE is parallel to CD, and BC.jpg [ 18.52 KiB | Viewed 4170 times ]

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.
[Reveal] Spoiler: OA

_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

VP
Joined: 20 Apr 2016
Posts: 1278
WE: Engineering (Energy and Utilities)
Followers: 19

Kudos [?]: 1208 [2] , given: 241

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  26 Sep 2017, 22:20
2
KUDOS
pclawong wrote:

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!!!
Attachments

box (1).jpg [ 13.66 KiB | Viewed 7141 times ]

_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

VP
Joined: 20 Apr 2016
Posts: 1278
WE: Engineering (Energy and Utilities)
Followers: 19

Kudos [?]: 1208 [2] , given: 241

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  26 Sep 2017, 22:31
2
KUDOS
pranab01 wrote:
pclawong wrote:

This is a tough ques, let me try

All queries are welcome!!!

Based on the above solution if the reasoning is correct then we can also have the diagram (attached)

Now if we divide to 10 small triangles (considering all the assumption given and proven in previous response)

Then Area of Triangle = ABE = sum of 4 triangles

and Area of quadrialteral = sum of 4 triangles

This also led to option C.

Kindly let me know if my reasoning are correct.

Thanks in advance. and all queries are welcome
Attachments

box .jpg [ 16.57 KiB | Viewed 7139 times ]

_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

Intern
Joined: 26 Sep 2017
Posts: 31
Followers: 0

Kudos [?]: 4 [2] , given: 19

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  27 Sep 2017, 05:57
2
KUDOS
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
VP
Joined: 20 Apr 2016
Posts: 1278
WE: Engineering (Energy and Utilities)
Followers: 19

Kudos [?]: 1208 [2] , given: 241

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  27 Sep 2017, 09:17
2
KUDOS
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

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

Can you show in details
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

Intern
Joined: 28 Feb 2017
Posts: 6
Followers: 0

Kudos [?]: 2 [1] , given: 3

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  27 Sep 2017, 11:35
1
KUDOS
Thank you.
Intern
Joined: 21 Aug 2017
Posts: 2
Followers: 0

Kudos [?]: 1 [1] , given: 0

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  15 Nov 2017, 12:14
1
KUDOS
Can Someone Pls explain how we know that BA = BD...or just how BDAE is a square.
Intern
Joined: 14 Oct 2017
Posts: 5
Followers: 0

Kudos [?]: 4 [2] , given: 0

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  06 Dec 2017, 15:14
2
KUDOS
pranab01 wrote:
pclawong wrote:

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.
VP
Joined: 20 Apr 2016
Posts: 1278
WE: Engineering (Energy and Utilities)
Followers: 19

Kudos [?]: 1208 [1] , given: 241

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  06 Dec 2017, 22:37
1
KUDOS
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,

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.
Attachments

box %281%29.jpg [ 14.57 KiB | Viewed 6627 times ]

_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

Manager
Joined: 03 Dec 2017
Posts: 64
Followers: 0

Kudos [?]: 17 [0], given: 20

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  07 Dec 2017, 19:18
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

Dear friend,
thanks
Intern
Joined: 08 Apr 2018
Posts: 44
Followers: 0

Kudos [?]: 16 [2] , given: 15

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  10 Jun 2018, 18:53
2
KUDOS
Carcass wrote:

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?
Founder
Joined: 18 Apr 2015
Posts: 11192
Followers: 239

Kudos [?]: 2809 [0], given: 10653

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  20 Jun 2018, 14:42
Expert's post

Regards
_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

Manager
Joined: 27 Feb 2017
Posts: 188
Followers: 1

Kudos [?]: 73 [1] , given: 15

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  24 Jun 2018, 20:29
1
KUDOS
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,

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.
Intern
Joined: 16 Jul 2018
Posts: 4
Followers: 0

Kudos [?]: 5 [2] , given: 0

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  17 Jul 2018, 12:13
2
KUDOS
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.
VP
Joined: 20 Apr 2016
Posts: 1278
WE: Engineering (Energy and Utilities)
Followers: 19

Kudos [?]: 1208 [1] , given: 241

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  19 Jul 2018, 19:10
1
KUDOS
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
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

Intern
Joined: 16 Jul 2018
Posts: 4
Followers: 0

Kudos [?]: 5 [1] , given: 0

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  20 Jul 2018, 06:24
1
KUDOS
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.
VP
Joined: 20 Apr 2016
Posts: 1278
WE: Engineering (Energy and Utilities)
Followers: 19

Kudos [?]: 1208 [1] , given: 241

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  20 Jul 2018, 10:26
1
KUDOS
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.
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

Director
Joined: 09 Nov 2018
Posts: 506
Followers: 0

Kudos [?]: 54 [0], given: 1

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  18 Jan 2019, 22:26
Critical and C
Intern
Joined: 21 Apr 2019
Posts: 3
Followers: 0

Kudos [?]: 2 [0], given: 0

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  03 May 2019, 02:26

imagine it like a letter symmetric envelope. very fast method
Manager
Joined: 04 Apr 2020
Posts: 72
Followers: 0

Kudos [?]: 27 [1] , given: 16

Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par [#permalink]  03 May 2020, 10:25
1
KUDOS
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.)
Re: AB = DE, BC = CD, BE is parallel to CD, and BC is par   [#permalink] 03 May 2020, 10:25
Display posts from previous: Sort by