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

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

ALTERNATE SOLUTION
If you're unsure how to solve the question (or you're quite far behind timing-wise), you can apply the following property:
The diagrams in GMAT Problem Solving questions are DRAWN TO SCALE unless stated otherwise.
So, we may be able to use this fact to solve the question (or at least eliminate some answer choices before guessing) by simply "eyeballing" the diagram.

If we mentally connect points E and D to the midpoint of side BC, we got something like this:

At this point, it certainly looks like region ABE is approximately 1/3 the area of the quadrilateral region BCDE.
Guess C.

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)

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

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.

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

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...

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!

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):

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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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 =\(\frac{Bh}{4}/Bh-\frac{Bh}{4}\)= \(\frac{Bh}{4}/\frac{3Bh}{4}\) = \(\frac{1}{3}\)

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.

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)

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

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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
Devmitra Sen(GMAT Quant Expert)
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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= \(1/3\)

Trapezoids area is minor base plus major base times height divided by 2

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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)

GIVEN:

• ABCD is a parallelogram.
• E is the mid-point of AD

TO FIND:

• Area of triangular region ABE as a fraction of the area of the quadrilateral region BCDE
  
  • = \(\frac{Area of ABE}{Area of BCDE}\) ------ (1)

SOLUTION:

Let’s first find both parts of the required fraction (1) – numerator and denominator - one by one. We will then write the fraction using what we get.

Area of triangle ABE:

• Area of a triangle = ½ × base × height
  
  • In triangle ABE, base = AE and height = BF. (where F lies on AE such that BF is perpendicular to AE)
  • Thus, area of ABE = ½ × AE × BF ---------- (2)

Area of quadrilateral BCDE:

• Since BC || DE, BCDE is a trapezoid. [Reason: In parallelogram ABCD, BC || AD and DE is a part of AD]
• Area of a trapezoid = ½ × (Sum of parallel sides) × height
  
  • In trapezoid BCDE, parallel sides are BC and DE and height = BF. [Since height of trapezoid = height of triangle]
  • Thus, area of BCDE = ½ × (BC + DE) × BF ---------- (3)

Required Fraction

• = \(\frac{Area of ABE}{Area of BCDE}\) -----------From (1)
• = (\(\frac{1}{2}\) × AE × BF)/(\(\frac{1}{2}\) × (BC + DE) × BF)------- From (2) and (3)
  
  • = \(\frac{AE}{(BC + DE)}\) [Cancelling out the common factors from numerator and denominator]
  • = \(\frac{AE}{(BC + AE)}\) [Since E is the mid-point of AD, AE = DE]
  • = \(\frac{AE}{(AD + AE)}\) [Since ABCD is a parallelogram, BC = AD]
  • = \(\frac{AE}{(2AE + AE)}\) [Since E is the mid-point of AD, AD = 2AE]
  • = \(\frac{AE}{3AE}\) = \(\frac{1}{3}\)

Correct Answer: Choice B



TAKEAWAYS:

1. Area of a triangle = ½ × base × height
2. Area of a trapezoid = ½ × (Sum of parallel sides) × height

Shweta Koshija
Quant Product Creator, e-GMAT
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