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

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

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

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

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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 03 May 2020, 12:10
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GMATinsight wrote:
Image
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: Image


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
Image


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
Image


Area of triangle = (base)(height)/2
Image
So, the area of ABE = (1)(1)/2 = 0.5


Area of trapezoid = (base1 + base2)(height)/2
Image
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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 21 Apr 2020, 09:38
GMATinsight wrote:
Image
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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 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:
Image
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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 21 Apr 2020, 09:48
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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)
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 21 Apr 2020, 22:37
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GMATinsight wrote:
Image
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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 25 Apr 2020, 13:24
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Since its given that ABCD is a parallelogram, and not other constraints given, assume it as a square a solve it
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 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... :thumbsup: :)
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 29 Apr 2020, 12:07
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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!
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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


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
Image


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
Image


Area of triangle = (base)(height)/2
Image
So, the area of ABE = (1)(1)/2 = 0.5


Area of trapezoid = (base1 + base2)(height)/2
Image
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 !!
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 07 May 2020, 04:39
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1
GMATinsight wrote:
Image
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
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.
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 11 May 2020, 08:01
Archit3110 wrote:
GMATinsight wrote:
Image
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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 11 May 2020, 08:31
prototypevenom wrote:
Archit3110 wrote:
GMATinsight wrote:
Image
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!
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 11 May 2020, 08:38
2
Top Contributor
GMATinsight wrote:
prototypevenom wrote:
Archit3110 wrote:
GMATinsight wrote:
Image
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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 11 May 2020, 08:41
1
prototypevenom wrote:
Archit3110 wrote:
GMATinsight wrote:
Image
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).
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 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:
Image
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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 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?

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

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New post 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 =\(\frac{Bh}{4}/\frac{Bh-Bh}{4}\)= \(\frac{Bh}{4}/\frac{3Bh}{4}\) = \(\frac{1}{3}\)
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In the figure above, ABCD is a parallelogram and E is the midpoint of  [#permalink]

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New post 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 = \(\frac{No.of triangles in ABE}{No.of triangles in BEDC}\)

= \(\frac{1}{3}\)
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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

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