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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of [#permalink]
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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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of [#permalink]
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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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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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of [#permalink]
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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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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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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
Area parallelogram ABCD.png [ 56.25 KiB | Viewed 44508 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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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 =\(\frac{Bh}{4}/Bh-\frac{Bh}{4}\)= \(\frac{Bh}{4}/\frac{3Bh}{4}\) = \(\frac{1}{3}\)

Originally posted by BoscoVIX on 27 May 2020, 17:52.
Last edited by BoscoVIX on 18 Aug 2020, 18:39, edited 1 time in total.
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of [#permalink]
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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


The question is simple and needs no construction. Note that
Area of parallelogram ABCD = Base AD * Height
Area of triangle ABE = (1/2) * Base AE * Height
The height of both the triangle and the parallelogram is the same.

Since AE is half of AD, area of triangle ABE is (1/2)*(1/2) * area of parallelogram ABCD = 1/4 of area of parallelogram ABCD

Hence, area of quadrilateral BCDE is 3/4 area of parallelogram ABCD (after removing area of triangle ABE).

Area of triangle ABE / Area of quad BCDE = 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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According to me, we can look at this in another way. Please let me know if my thought process is wrong.

If I draw a similar line from D to BC to meet at the midpoint F, the two triangles would be identical.

Now the parallelogram BFDE would be 1/2 of parallelogram ABCD because the height is same but the base is half.

So, ABCD = T - ABE + P - BFDE + T - CDF
Talking in parts wrt to ABCD, if T - ABE = 1 part then P - BFDE = 2 parts (half) and T - CDF = 1 part.

Therefore, T-ABE/(P - BFDE + T - CDF) = 1/(2+1) = 1/3

Hence, the answer is (B).

Please find the image attached.
Attachments

parallel.png
parallel.png [ 11.56 KiB | Viewed 31307 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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Avoid calculating as much as possible when regular polygons are involved. Just take a point R as the midpoint of BC and draw the triangle.
Therefore we have 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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achloes wrote:
Bunuel KarishmaB

I solved the question correctly by applying a concept I know about inscribed triangles within rectangles (reasoned that parallelograms are also regular figures) but I'd appreciate your validation.

Concept: If a triangle inscribed in a rectangle shares a side with the rectangle, the area of the triangle is equal to half the area of the rectangle. In other words, the area of the two smaller triangles = area of main triangle.

Applying this rule to the parallelogram, I connected points E and C to create an inscribed triangle.

First small triangle / Main triangle + second small triangle = 1 / 3

Attachment:
Screenshot 2023-07-20 at 10.41.10 AM.png


You are absolutely correct. Why? The reasoning is given in my post here: https://gmatclub.com/forum/in-the-figur ... l#p3153698

Area of a rectangle = Base * Height
Area of triangle with same Base = (1/2) * Base * Height

Note that when the vertex of the triangle lies on the opposite side to the base, the height of both is the same.
Exactly the same concept is applicable to the parallelogram too. Area of parallelogram is also base * height.
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of [#permalink]
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
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of [#permalink]
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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



Another simple way to look at it is to make a line EF parallel to AB, and join AC.
Line EF will join opposite sides, and so would divide the llgm into two equal pieces.
AB further divides the half portion in half each.
So Area of ABE is 1/4 of total

C
Attachments

2020-04-28_1844.png
2020-04-28_1844.png [ 13.31 KiB | Viewed 30076 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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Solution:

Join EF where F is the midpoint of BC

In quadrilateral BFEA,

BF || AE (As BF is a part of the line segment BC and BC || AD in the ||gm ABCD)

AD =BC (opposite sides of a ||gm)

AE = ED (E is the midpoint of AD)

=>AE = BF (F is the midpoint of BC)

Thus opposite sides are parallel and equal.(AEFB is a ||gm).The same can be proved for EDCF)

Diagonals BE and DF divide the ||gm into 4 congruent triangles of which AEB is one.

Quadrilateral BCDE is made of 3 congruent triangles.

Thus area of triangular region ABE/of the area of the quadrilateral region BCDE

= 1 / 3 (option b)

Hope this helps :thumbsup:
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of [#permalink]
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 E is a mid-point, let's say it divides the line into two equal halves. K units on either side of the mid-point.

Let base of triangle ABE= k and Height of triangle ABE = h
so, the area of triangle is\( 1/2*k*h\) --- \(kh/2\)

For the quadrilateral, the height will be the same as the height of the triangle, which is = h. but since we have two different bases (k,2k). this is a trapezoid.

Area of BCDE=\( k+2k/2*h \)----> \(3k/2*h\)

Area of triangle/Area of Quadrilateral= \(1/3\)
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of [#permalink]
Expert Reply
Trapezoids area is minor base plus major base times height divided by 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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