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555-605 (Medium)|   Geometry|      
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ShahadatHJ
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB Updated on: 26 Sep 2021, 04:42
Originally posted by ShahadatHJ on 26 Sep 2021, 02:36.
Last edited by Bunuel on 26 Sep 2021, 04:42, edited 1 time in total.
Renamed the topic and edited the question.
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69% (02:20) correct 31% (02:43) wrong based on 77 sessions

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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB. If the area of ABCD is 36 sq. cm. and EX = 6. What is the area of the triangle ABX in sq. cm? (Picture is not drawn to scale)

A. 16
B. 12
C. 15
D. 09
E. None

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D
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 26 Sep 2021, 11:53
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ShahadatHJ

In the Parallelogram, E is the midpoint of AC and EF is parallel to AB. If the area of ABCD is 36 sq. cm. and EX = 6. What is the area of the triangle ABX in sq. cm? (Picture is not drawn to scale)

A. 16
B. 12
C. 15
D. 09
E. None

Show Spoiler
Attachment:
Picture.jpg
Show more


Of course you can solve it otherwise, but easiest would be to take ABCD as a square.
So Each side is 6.
AC=6, and AE = 6/3 = 2.

Hence area of ABX=\(\frac{1}{2}*6*3=9\)


OR

EF divides the parallelogram in two equal parts, so area of each is 18 sqm.

Now area of ABX = area of AEX + area of BFX, as the height is same being between two same parallel lines and base too is equal.
AB = EF = EX+XF

So area of ABX = \(\frac{18}{2}=9\)

D
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 27 Sep 2021, 01:05
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ShahadatHJ

In the Parallelogram, E is the midpoint of AC and EF is parallel to AB. If the area of ABCD is 36 sq. cm. and EX = 6. What is the area of the triangle ABX in sq. cm? (Picture is not drawn to scale)

A. 16
B. 12
C. 15
D. 09
E. None

Show Spoiler
Attachment:
Picture.jpg
Show more
The area of the first half = 36/2 = 18

Let the individual section of the second make up of area x , 2x and x respectively
Therefore 4x =18

We need 2x area which equals = 18/2
Therefore IMO D
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 27 Sep 2021, 01:51
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ShahadatHJ

In the Parallelogram, E is the midpoint of AC and EF is parallel to AB. If the area of ABCD is 36 sq. cm. and EX = 6. What is the area of the triangle ABX in sq. cm? (Picture is not drawn to scale)

A. 16
B. 12
C. 15
D. 09
E. None

Show Spoiler
Attachment:
Picture.jpg


Of course you can solve it otherwise, but easiest would be to take ABCD as a square.
So Each side is 6.
AC=6, and AE = 6/3 = 2.

Hence area of ABX=\(\frac{1}{2}*6*3=9\)


OR

EF divides the parallelogram in two equal parts, so area of each is 18 sqm.

Now area of ABX = area of AEX + area of AFX, as the height is same being between two same parallel lines and base too is equal.
AB = EF = EX+XF

So area of ABX = \(\frac{18}{2}=9\)

D
Show more
Hi chetan2u, 2 quick questions:

Now area of ABX = area of AEX + area of AFX, as the height is same being between two same parallel lines and base too is equal.

Shouldn't it be BFX instead of AFX? Are we considering AB the base for these 2 triangles or AE and BF as bases?

Thank you!
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 27 Sep 2021, 01:56
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ShahadatHJ

In the Parallelogram, E is the midpoint of AC and EF is parallel to AB. If the area of ABCD is 36 sq. cm. and EX = 6. What is the area of the triangle ABX in sq. cm? (Picture is not drawn to scale)

A. 16
B. 12
C. 15
D. 09
E. None

Show Spoiler
Attachment:
Picture.jpg
The area of the first half = 36/2 = 18

Let the individual section of the second make up of area x , 2x and x respectively
Therefore 4x =18

We need 2x area which equals = 18/2
Therefore IMO D
Show more
Hi Crytiocanalyst, question;
how do I know these areas are divided into X, X, 2X and 2X is equal to AEX + BFX? Since No value is given other than EX=6.
Thank You!
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 27 Sep 2021, 02:05
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ShahadatHJ
chetan2u
ShahadatHJ

In the Parallelogram, E is the midpoint of AC and EF is parallel to AB. If the area of ABCD is 36 sq. cm. and EX = 6. What is the area of the triangle ABX in sq. cm? (Picture is not drawn to scale)

A. 16
B. 12
C. 15
D. 09
E. None

Show Spoiler
Attachment:
Picture.jpg


Of course you can solve it otherwise, but easiest would be to take ABCD as a square.
So Each side is 6.
AC=6, and AE = 6/3 = 2.

Hence area of ABX=\(\frac{1}{2}*6*3=9\)


OR

EF divides the parallelogram in two equal parts, so area of each is 18 sqm.

Now area of ABX = area of AEX + area of AFX, as the height is same being between two same parallel lines and base too is equal.
AB = EF = EX+XF

So area of ABX = \(\frac{18}{2}=9\)

D
Hi chetan2u, 2 quick questions:

Now area of ABX = area of AEX + area of AFX, as the height is same being between two same parallel lines and base too is equal.

Shouldn't it be BFX instead of AFX? Are we considering AB the base for these 2 triangles or AE and BF as bases?

Thank you!
Show more


Yes, it is BFX. Corrected the typo. Thanks
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 27 Sep 2021, 07:14
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ShahadatHJ
chetan2u
ShahadatHJ

In the Parallelogram, E is the midpoint of AC and EF is parallel to AB. If the area of ABCD is 36 sq. cm. and EX = 6. What is the area of the triangle ABX in sq. cm? (Picture is not drawn to scale)

A. 16
B. 12
C. 15
D. 09
E. None

Show Spoiler
Attachment:
Picture.jpg


Of course you can solve it otherwise, but easiest would be to take ABCD as a square.
So Each side is 6.
AC=6, and AE = 6/3 = 2.

Hence area of ABX=\(\frac{1}{2}*6*3=9\)


OR

EF divides the parallelogram in two equal parts, so area of each is 18 sqm.

Now area of ABX = area of AEX + area of AFX, as the height is same being between two same parallel lines and base too is equal.
AB = EF = EX+XF

So area of ABX = \(\frac{18}{2}=9\)

D
Hi chetan2u, 2 quick questions:

Now area of ABX = area of AEX + area of AFX, as the height is same being between two same parallel lines and base too is equal.

Shouldn't it be BFX instead of AFX? Are we considering AB the base for these 2 triangles or AE and BF as bases?

Thank you!
Show more

The ABX region can be further divide into 2 similar areas by drawing a perpendicular and we can make the resultant triangle and take their individual areas as X hence when we add we get 4x hope it clarifies
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 27 Sep 2021, 08:23
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The ABX region can be further divide into 2 similar areas by drawing a perpendicular and we can make the resultant triangle and take their individual areas as X hence when we add we get 4x hope it clarifies
Show more
Sir, to divide the triangle into 2 equal parts we have to consider that X is in the middle of EF, isn't it? However, the question does not mention that. Is there something that I am missing?
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 27 Sep 2021, 10:02
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ShahadatHJ
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The ABX region can be further divide into 2 similar areas by drawing a perpendicular and we can make the resultant triangle and take their individual areas as X hence when we add we get 4x hope it clarifies
Sir, to divide the triangle into 2 equal parts we have to consider that X is in the middle of EF, isn't it? However, the question does not mention that. Is there something that I am missing?
Show more


Triangle ABX can be divided into 2 equal half and not the outer triangle the trianglr which is situated in the middle is divided , yes youre right for the triangles to be divide the the outer triangles it needs to be in between , however i never touched the outside triangle in the first place
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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 16 Oct 2021, 01:51
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The key to answering the question is to notice that the entire figure is a parallelogram and that EF is the parallel mid segment that divides the parallelogram into 2 symmetric/congruent areas.


Thus, we have parallelogram ABEF = Area of (1/2) (36) = 18


Rule: between 2 parallel lines, the perpendicular height will always be the same.

Call the area of triangle AEX = [AEX]

Area of triangle ABX = [ABX]

Area of triangle BFX = [BFX]


[AEX] + [ABX] + [BFX] = 18 (equation 1)


Choose base sides for each triangle:

EX = 6

Let XF = b

Since the opposite sides of a parallelogram are equal, AB = 6 + b


And all 3 triangles will have an equivalent perpendicular height drawn to the respective bases above ——> call it H = perpendicular distance between the 2 parallel lines AB and EF

[AEX] = (1/2) (6) (H) = 6H / 2

[BFX] = (1/2) (b) (H) = bH / 2


[ABX] = (1/2) (b + 6) (H) = (6H + bH) / 2


As you can see, the Area of triangle AEX + the Area of triangle BFX = the target triangle’s Area, triangle ABX


[AEX] + [BFX] = [ABX] (equation 2)

Combining this with (equation 1) above

[AEX] + [BFX] + [ABX] = 18


Substitute (equation 2) into (equation 1)


[ABX] + [ABX] = 18

(2) * [ABX] = 18

Area of triangle ABX = [ABX] = 9

D

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In the Parallelogram, E is the midpoint of AC and EF is parallel to AB 17 Oct 2022, 09:09
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