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In the figure above, ABCD is a parallelogram and E is the midpoint of
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21 Apr 2020, 09:27
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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:
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of
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03 May 2020, 12:10
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 20200421 at 10.55.06 PM.png Attachment: 20200428_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 ADThis means AE = ED So, let's let AE = ED = 1We get: ABCD is a parallelogramProperty: Opposite sides in a parallelogram have equal lengths Since AD = 2, it must also be the case that CB = 2To 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 1Area of triangle = (base)(height)/2 So, the area of ABE = ( 1)( 1)/2 = 0.5Area 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 = 1/3Answer: B Cheers, Brent
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of
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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 20200421 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 xy1/4 * 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
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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 20200421 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
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21 Apr 2020, 09:48
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
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21 Apr 2020, 22:37
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 20200421 at 10.55.06 PM.png Draw a line BD ∆ ABD = (1/2)*ABCDArea 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)*ABCDi.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
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25 Apr 2020, 13:24
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
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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 rahulraj1988That's really a smart thinking...
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Re: In the figure above, ABCD is a parallelogram and E is the midpoint of
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29 Apr 2020, 12:07
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
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03 May 2020, 12:29
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 20200421 at 10.55.06 PM.png Attachment: 20200428_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 ADThis means AE = ED So, let's let AE = ED = 1We get: ABCD is a parallelogramProperty: Opposite sides in a parallelogram have equal lengths Since AD = 2, it must also be the case that CB = 2To 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 1Area of triangle = (base)(height)/2 So, the area of ABE = ( 1)( 1)/2 = 0.5Area 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 = 1/3Answer: 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
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07 May 2020, 04:39
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 20200421 at 10.55.06 PM.png is no longer available Attachment: The attachment 20200428_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 = \(\frac{a}{4}\)/\(\frac{3a}{4}\) = \(\frac{1}{3}\) ORConsider 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
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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 20200421 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 xy1/4 * 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
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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 20200421 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 xy1/4 * 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 prototypevenomDIstance 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
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11 May 2020, 08:38
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 20200421 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 xy1/4 * 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 prototypevenomDIstance 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
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11 May 2020, 08:41
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 20200421 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 xy1/4 * 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
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11 May 2020, 08:59
prototypevenomgiven 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 20200421 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 xy1/4 * 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
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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
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27 May 2020, 16:52
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 =\(\frac{Bh}{4}/\frac{BhBh}{4}\)= \(\frac{Bh}{4}/\frac{3Bh}{4}\) = \(\frac{1}{3}\)



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