GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 02 Jul 2020, 01:51

### 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
Your Progress

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

# In the figure above, ABCD is a parallelogram and E is the midpoint of

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4296
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

21 Apr 2020, 09:27
30
00:00

Difficulty:

75% (hard)

Question Stats:

52% (01:46) correct 48% (01:34) wrong based on 447 sessions

### HideShow timer Statistics

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:

Screenshot 2020-04-21 at 10.55.06 PM.png [ 52.76 KiB | Viewed 3279 times ]

Attachment:

2020-04-28_1844.png [ 11.28 KiB | Viewed 2854 times ]

_________________
Prepare with PERFECTION to claim Q≥50 and V≥40 !!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

Most affordable l Comprehensive l 2000+ Qn ALL with Video explanations l LINK: Courses and Pricing
Click for FREE Demo on VERBAL & QUANT
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
FREE GMAT Resource: 22 FREE (FULL LENGTH) GMAT CATs LINKS l NEW OG QUANT 50 Qn+VIDEO Sol.
##### Most Helpful Expert Reply
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4947
Location: Canada
GMAT 1: 770 Q49 V46
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

03 May 2020, 12:10
6
Top Contributor
4
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png

Attachment:
2020-04-28_1844.png

Since we're asked to find a certain fraction, we can assign some nice values to the diagram (values that satisfy the given information!)

E is the midpoint of side AD
This means AE = ED
So, let's let AE = ED = 1
We get:

ABCD is a parallelogram
Property: Opposite sides in a parallelogram have equal lengths
Since AD = 2, it must also be the case that CB = 2

To find the areas of triangle ABE and trapezoid BCDE, we need the height of both shapes.
So, let's say the height of both shapes is 1

Area of triangle = (base)(height)/2

So, the area of ABE = (1)(1)/2 = 0.5

Area of trapezoid = (base1 + base2)(height)/2

So, the area of trapezoid BCDE = (1 + 2)(1)/2 = 3/2 = 1.5

The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?
(area of triangular region ABE)/(area of the quadrilateral region BCDE) = 0.5/1.5 = 1/3

Answer: B

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
##### General Discussion
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 6411
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

21 Apr 2020, 09:38
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png

let BC = x and AB =y
so area of ∆ ABE = 1/2 * x/2 * y ; 1/4 * x*y
and area of ll ogram ABCD ; x*y
so area of side BCDE
xy-1/4 * xy ; 3/4 x*y
hence The area of triangular region ABE is what fraction of the area of the region BCDE
(1/4) * x*y / (3/ 4) * (x*y) =
1/3
OPTION B; 1/3
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 1968
Location: India
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

21 Apr 2020, 09:48
Area (ABE)

= 1/2 * AB *AE * sin(A)

= 1/2 * AB *(AD/2) * sin(A)

= 1/4 * AB *AD * sin(A)

= 1/4 * (Area of parallelogram ABCD)

(Area of ABCD) = Area of (ABE) + Area (BCDE)

Area (BCDE) = 3/4 * (Area of parallelogram ABCD)

area of triangular region ABE is (1/3) rd of the area of the quadrilateral region BCDE.

GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png
Intern
Joined: 13 Apr 2020
Posts: 24
Location: United Kingdom
Schools: LBS '16 (A\$)
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

21 Apr 2020, 09:48
2
A quick visual way to solve this is to add point F as the midpoint of BC and draw the triangle EFD.

We have therefore split the parallelogram into 4 triangles that can be shown to be all equal (due to symmetries)

BCDE makes up 3 of those equal triangles and ABE makes up the other one
Hence the ratio is 1/3 ==> Answer (B)
_________________
Graeme
Founder of MBA Backstage
http://www.mbabackstage.com/guides/
All you need to know about your MBA & GMAT journey
GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4296
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

21 Apr 2020, 22:37
1
1
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png

Draw a line BD

∆ ABD = (1/2)*ABCD

Area of triangle = (1/2)*Base*Height

Now, BE divides the basis of triangle ABD in two equal parts

i.e. Area of ∆BAE = Area of ∆BDE =
(1/2)*ABD = (1/4)*ABCD

i.e. Area BCDE = ABCD - (1/4)*ABCD = (3/4)*ABCD

Now, ABE / BCDE = (1/4)*ABCD / (3/4)*ABCD = 1/3

Answer: Option B
_________________
Prepare with PERFECTION to claim Q≥50 and V≥40 !!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

Most affordable l Comprehensive l 2000+ Qn ALL with Video explanations l LINK: Courses and Pricing
Click for FREE Demo on VERBAL & QUANT
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
FREE GMAT Resource: 22 FREE (FULL LENGTH) GMAT CATs LINKS l NEW OG QUANT 50 Qn+VIDEO Sol.
Intern
Joined: 29 Apr 2018
Posts: 2
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

25 Apr 2020, 13:24
1
1
Since its given that ABCD is a parallelogram, and not other constraints given, assume it as a square a solve it
GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4296
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

26 Apr 2020, 00:21
rahulraj1988 wrote:
Since its given that ABCD is a parallelogram, and not other constraints given, assume it as a square a solve it

Hi rahulraj1988

That's really a smart thinking...
_________________
Prepare with PERFECTION to claim Q≥50 and V≥40 !!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

Most affordable l Comprehensive l 2000+ Qn ALL with Video explanations l LINK: Courses and Pricing
Click for FREE Demo on VERBAL & QUANT
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
FREE GMAT Resource: 22 FREE (FULL LENGTH) GMAT CATs LINKS l NEW OG QUANT 50 Qn+VIDEO Sol.
Intern
Joined: 04 Jan 2016
Posts: 30
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

29 Apr 2020, 12:07
1
The simplest way i could think of is:

Draw a line connecting E and C

Now we have two equal triangles ABE and CDE (since the base and height of the two triangles are the same)

This leads to the third triangle BEC
Area of BEC = 1/2 * BC * height of the parallelogram

If we notice the area of the parallelogram could also be BC * h

This means that the triangle BEC is 1/2 of the Parallelogram which means the rest of the two triangles ABE and CDE combined are the other half of the parallelogram
and since ABE and CDE triangles are equal it would mean that each triangle is 1/4th of the Parallelogram

Hope this helps!
Board of Directors
Joined: 01 Sep 2010
Posts: 3496
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

03 May 2020, 12:29
Top Contributor
BrentGMATPrepNow wrote:
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png

Attachment:
2020-04-28_1844.png

Since we're asked to find a certain fraction, we can assign some nice values to the diagram (values that satisfy the given information!)

E is the midpoint of side AD
This means AE = ED
So, let's let AE = ED = 1
We get:

ABCD is a parallelogram
Property: Opposite sides in a parallelogram have equal lengths
Since AD = 2, it must also be the case that CB = 2

To find the areas of triangle ABE and trapezoid BCDE, we need the height of both shapes.
So, let's say the height of both shapes is 1

Area of triangle = (base)(height)/2

So, the area of ABE = (1)(1)/2 = 0.5

Area of trapezoid = (base1 + base2)(height)/2

So, the area of trapezoid BCDE = (1 + 2)(1)/2 = 3/2 = 1.5

The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?
(area of triangular region ABE)/(area of the quadrilateral region BCDE) = 0.5/1.5 = 1/3

Answer: B

Cheers,
Brent

Outstanding. very easy to grasp this difficult problem !!
_________________
Sloan MIT School Moderator
Joined: 07 Mar 2019
Posts: 1277
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

07 May 2020, 04:39
1
1
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02
Attachment:
The attachment Screenshot 2020-04-21 at 10.55.06 PM.png is no longer available

Attachment:
The attachment 2020-04-28_1844.png is no longer available

The best part about the question is that it asks for a ratio and since no dimensions are given we are free to take any value. Refer figure (a) and figure (b):
Attachment:

Area parallelogram ABCD.png [ 56.25 KiB | Viewed 2196 times ]

Consider figure (a). Visual Solution:
Know that if we draw a parallel EF line to AB, the line we are dividing the area of parallelogram ABCD into two equal parts. Similarly, area of parallelogram AEFB is divided by AE into two equal parts.
Thus, if area of parallelogram ABCD = a THEN
Area of parallelogram AEFB = $$\frac{a}{2}$$
And Area of triangle AEB = 1/2 * $$\frac{a}{2}$$ = $$\frac{a}{4}$$

Area of quadrilateral BCDE = a - $$\frac{a}{4}$$ = $$\frac{3a}{4}$$
Hence, area of triangle ABE / area of quadrilateral BCDE = $$\frac{a}{4}$$/$$\frac{3a}{4}$$ = $$\frac{1}{3}$$

OR

Consider figure (b). Parallelogram ABCD is remolded into rectangle ABCD to make things better understandable.
Let dimensions are AB = 3 and AD = 4. AE = 2.
Area of triangle ABE = $$\frac{1}{2} * 3 * 2 = 3$$
Area of quadrilateral BCDE = Area of rectangle ABCD - Area of triangle ABE
Area of quadrilateral BCDE = 4*3 - 3 = 9

Hence, , area of triangle ABE / area of quadrilateral BCDE = $$\frac{3}{9}$$ = $$\frac{1}{3}$$

Answer B.
Note: A rectangle satisfies a parallelogram's conditions.
_________________
Manager
Joined: 12 Nov 2018
Posts: 98
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

11 May 2020, 08:01
Archit3110 wrote:
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png

let BC = x and AB =y
so area of ∆ ABE = 1/2 * x/2 * y ; 1/4 * x*y
and area of ll ogram ABCD ; x*y
so area of side BCDE
xy-1/4 * xy ; 3/4 x*y
hence The area of triangular region ABE is what fraction of the area of the region BCDE
(1/4) * x*y / (3/ 4) * (x*y) =
1/3
OPTION B; 1/3

how can the height of triangle ABE be y?

It isn't the altitude to x/2
GMAT Club Legend
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4296
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

11 May 2020, 08:31
prototypevenom wrote:
Archit3110 wrote:
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png

let BC = x and AB =y
so area of ∆ ABE = 1/2 * x/2 * y ; 1/4 * x*y
and area of ll ogram ABCD ; x*y
so area of side BCDE
xy-1/4 * xy ; 3/4 x*y
hence The area of triangular region ABE is what fraction of the area of the region BCDE
(1/4) * x*y / (3/ 4) * (x*y) =
1/3
OPTION B; 1/3

how can the height of triangle ABE be y?

It isn't the altitude to x/2

prototypevenom

DIstance between any two parallel lines remain constant so at every point on one of the parallel lines the perpendicular distance will be y

I hope this help!
_________________
Prepare with PERFECTION to claim Q≥50 and V≥40 !!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

Most affordable l Comprehensive l 2000+ Qn ALL with Video explanations l LINK: Courses and Pricing
Click for FREE Demo on VERBAL & QUANT
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
FREE GMAT Resource: 22 FREE (FULL LENGTH) GMAT CATs LINKS l NEW OG QUANT 50 Qn+VIDEO Sol.
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4947
Location: Canada
GMAT 1: 770 Q49 V46
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

11 May 2020, 08:38
2
Top Contributor
GMATinsight wrote:
prototypevenom wrote:
Archit3110 wrote:
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png

let BC = x and AB =y
so area of ∆ ABE = 1/2 * x/2 * y ; 1/4 * x*y
and area of ll ogram ABCD ; x*y
so area of side BCDE
xy-1/4 * xy ; 3/4 x*y
hence The area of triangular region ABE is what fraction of the area of the region BCDE
(1/4) * x*y / (3/ 4) * (x*y) =
1/3
OPTION B; 1/3

how can the height of triangle ABE be y?

It isn't the altitude to x/2

prototypevenom

DIstance between any two parallel lines remain constant so at every point on one of the parallel lines the perpendicular distance will be y

I hope this help!

If ABCD were a rectangle, then the height = AB. However, since ABCD is a parallelogram, the height is not necessarily AB.
The fact of the matter is that the height a parallelogram ABCD can be ANY positive value.
As you can see from other solutions, the height of the triangle cancels out with the height of the trapezoid.

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Sloan MIT School Moderator
Joined: 07 Mar 2019
Posts: 1277
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

11 May 2020, 08:41
1
prototypevenom wrote:
Archit3110 wrote:
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png

let BC = x and AB =y
so area of ∆ ABE = 1/2 * x/2 * y ; 1/4 * x*y
and area of ll ogram ABCD ; x*y
so area of side BCDE
xy-1/4 * xy ; 3/4 x*y
hence The area of triangular region ABE is what fraction of the area of the region BCDE
(1/4) * x*y / (3/ 4) * (x*y) =
1/3
OPTION B; 1/3

how can the height of triangle ABE be y?

It isn't the altitude to x/2

Its a typo. Archit must have meant y as altitude(perpendicular distance between BC and AD).
_________________
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 6411
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

11 May 2020, 08:59
prototypevenom
given that lines BC AND AD are ll so distance b/w them would be same ; which I have taken as y to find area of ∆ ABE ..

prototypevenom wrote:
Archit3110 wrote:
GMATinsight wrote:

In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6

PS49220.02

Attachment:
Screenshot 2020-04-21 at 10.55.06 PM.png

let BC = x and AB =y
so area of ∆ ABE = 1/2 * x/2 * y ; 1/4 * x*y
and area of ll ogram ABCD ; x*y
so area of side BCDE
xy-1/4 * xy ; 3/4 x*y
hence The area of triangular region ABE is what fraction of the area of the region BCDE
(1/4) * x*y / (3/ 4) * (x*y) =
1/3
OPTION B; 1/3

how can the height of triangle ABE be y?

It isn't the altitude to x/2
Intern
Joined: 22 Mar 2020
Posts: 13
Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

22 May 2020, 06:53
Whats wrong with following approach?

1) asume distance AB to be 10
2) Perpendicular distance from base to point b is 5

Therefore are of parallelogram is 5×10=50

Area of tringle will be half AB so AE which will be 5 times the heigt to B which is 5 times 1/2

5×5×0.5=12.5

12.5/50=1/4

what am I missing?

Posted from my mobile device
Intern
Joined: 02 Dec 2018
Posts: 2
In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

27 May 2020, 16:52
1
we are essentially solving for

$$\frac{area of triangle}{(area of parallelogram - area of triangle)}$$

let B = base of parallelogram
let b = base of triangle = $$\frac{B}{2}$$
let h = height

area of triangle = $$\frac{bh}{2}$$ = $$\frac{1}{2}*\frac{Bh}{2}$$=$$\frac{Bh}{4}$$
[/m]
area of parallelogram = Bh

so final equation =$$\frac{Bh}{4}/\frac{Bh-Bh}{4}$$= $$\frac{Bh}{4}/\frac{3Bh}{4}$$ = $$\frac{1}{3}$$
Intern
Joined: 30 May 2020
Posts: 19
In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

### Show Tags

31 May 2020, 05:15
Plot a midpoint on BC and name it as F.

Now connect EC and EF.

Now there will be three triangles in BEDC.

now the answer would be = $$\frac{No.of triangles in ABE}{No.of triangles in BEDC}$$

= $$\frac{1}{3}$$
_________________
Founder
Perfect Review
http://www.perfectreview.co.in
In the figure above, ABCD is a parallelogram and E is the midpoint of   [#permalink] 31 May 2020, 05:15

# In the figure above, ABCD is a parallelogram and E is the midpoint of

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne