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In the figure above, if MNOP is a trapezoid and NOPR is a pa

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In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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In the figure above, if MNOP is a trapezoid and NOPR is a parallelogram, what is the area of triangular region MNR?

(1) The ares of NOPR is 30
(2) The area of the shaded region is 5

[Reveal] Spoiler:
Attachment:
Capture.JPG
Capture.JPG [ 12.08 KiB | Viewed 28998 times ]
[Reveal] Spoiler: OA

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Last edited by Bunuel on 16 Aug 2015, 11:47, edited 2 times in total.
Renamed the topic and edited the question.

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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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roygush wrote:
in the figure above, if MNOPis a trapzoid and NOPRis a parallelogram, what is the area of triangular regoin MNR?

(1) The are of NOPR is 30
(2) The are of OQP is 5

If someone can please explain.
Thanks.

Dear roygush,
I'm happy to help. :-)

First of all, here's a important geometry theorem to know:
http://magoosh.com/gmat/2012/isosceles- ... -the-gmat/
Triangle MNR has equal base angles --- therefore, it's isosceles, which means MN = NR. Because NOPR is a parallelogram, NR = OP, so that MN = NR = OP.

Here's a version of the graph with a new perpendicular line drawn:
Attachment:
parallelogram with isosceles triangle.JPG
parallelogram with isosceles triangle.JPG [ 20.79 KiB | Viewed 38639 times ]

Because the bases are parallel, OQ = NT --- the distance between two parallel lines is always the same.

Statement #1: The area of NOPR is 30
From that, there's no way to determine the areas inside the triangle MNR. That statement, alone and by itself, is insufficient.

Statement #2: The area of OQP is 5
Well, if the shaded region is 5, we could figure out the pieces of the isosceles triangle. Still, we wouldn't know anything about that central part, NOQR. That would remain unknown. Therefore, this statement, alone and by itself, is insufficient.

Combined statements
Now, we know the area of NOPR is 30. We know that the two little triangles, MNT and TNR, are congruent to QOP, so each has an area of 5. That gives a total area of 30 + 5 + 5 = 40 to the whole trapezoid. With both statements, we can calculate the area, so combined, the statements are sufficient.

Answer = (C)

Does all this make sense?
Mike :-)
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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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IMO B

The question asks only about area of triangle MNR

In the figure given by mick..consider the two triangles NTR and OQP

angle NRT = angle OPQ ( as NR and OP are parallel and equal also)
angle NTR = angle OQP = 90 (altitudes)
=> angle TNR = angle QOP

As two angles and the corresponding side (NR and OP) are equal the two triangles, as per ASA rule,are congruent. Hence both will have same area.

Similarly the triangles MNT and NTR are congruent (ASA rule again). Hence the area of triangle MNR = area of MNT + area of NTR = 5+5 = 10

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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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adityapagadala wrote:
IMO B
The question asks only about area of triangle MNR

Dear adityapagadala,
You are totally 100% correct. I am embarrassed to say --- I made the most boneheaded of all GMAT mistakes: was so concerned with analyzing the diagram, I lost track of exactly what the prompt was asking and assumed it was asking for the area of the whole. You are perfectly correct --- the answer has to be (B).
Mike :-)
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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 12 Jan 2014, 15:36
mikemcgarry wrote:
adityapagadala wrote:
IMO B
The question asks only about area of triangle MNR

Dear adityapagadala,
You are totally 100% correct. I am embarrassed to say --- I made the most boneheaded of all GMAT mistakes: was so concerned with analyzing the diagram, I lost track of exactly what the prompt was asking and assumed it was asking for the area of the whole. You are perfectly correct --- the answer has to be (B).
Mike :-)


Hi Mike,

Would you mind explaining in detail why statement 1 is insufficient. I lost a lot of time on this part of the problem.

Thanks
-Chris

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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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CVarga wrote:
Hi Mike,
Would you mind explaining in detail why statement 1 is insufficient. I lost a lot of time on this part of the problem.
Thanks
-Chris

Dear Chris,
I'm happy to help. :-)

The area of a parallelogram is (base)*(height). We know nothing about the slant angle x, so we have no idea how much of the parallelogram the triangle occupies. First of all, we know b*h = 30 from Statement #1, but we don't know if we have a base of 1 and a height of 30, or a base of 30 and a height of 1, or something else. Furthermore, we don't know the slant angle x ----- perhaps x = 89°, nearly vertical, so that the side triangle is just a tall skinny sliver, or maybe x is something like 20° or 30°, so that the base of the triangle (QP) is most of the length of the base of the parallelogram (RP). Even knowing the parallelogram has an area of 30, we could design things consistent with this statement & the diagram so that the triangle OQP has an area of nearly zero or nearly half the parallelogram or anything in between.

Does all this make sense?
Mike :-)
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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 03 Jul 2014, 12:16
adityapagadala wrote:
IMO B

The question asks only about area of triangle MNR

In the figure given by mick..consider the two triangles NTR and OQP

angle NRT = angle OPQ ( as NR and OP are parallel and equal also)
angle NTR = angle OQP = 90 (altitudes)
=> angle TNR = angle QOP

As two angles and the corresponding side (NR and OP) are equal the two triangles, as per ASA rule,are congruent. Hence both will have same area.

Similarly the triangles MNT and NTR are congruent (ASA rule again). Hence the area of triangle MNR = area of MNT + area of NTR = 5+5 = 10


hey nice solution.

can you please explain the colored part it seems that NRT is exterior angle of given parallelogram, while OPQ is interior angle of the parallellogram.
how are they equal? i lack basics in these geometrical concepts.
thanks in advance.

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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 04 Jul 2014, 18:06
saggii27 wrote:
hey nice solution.

can you please explain the colored part it seems that NRT is exterior angle of given parallelogram, while OPQ is interior angle of the parallellogram.
how are they equal? i lack basics in these geometrical concepts.
thanks in advance.

Dear Saggii27,
I'm happy to respond. :-)

Because NOPR is a parallelogram, this means NR is parallel to OP. This means that RP is a line that cuts across two parallel lines --- such a line is known as a "transversal." See this blog for all the angles that must be equal in this situation:
http://magoosh.com/gmat/2013/angles-and ... -the-gmat/

Does all this make sense?
Mike :-)
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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 25 Feb 2016, 16:18
The official answer is B but I'm not quite understanding why we can't use A to divide the area by 3 and get 10. You can tell from the picture that it fits 3 times....

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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 04 Mar 2016, 15:51
DJ1986 wrote:
The official answer is B but I'm not quite understanding why we can't use A to divide the area by 3 and get 10. You can tell from the picture that it fits 3 times....


Never make any assumptions about figures - especially on DS

They draw them out of scale on purpose to catch people in this very trap.

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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 15 Jul 2016, 07:22
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In the figure above, if MNOP is a trapezoid and NOPR is a parallelogram, what is the area of triangular region MNR?

(1) The ares of NOPR is 30
(2) The area of the shaded region is 5

A) area of NOPR is 30, this is a parallelogram. It consists of two congruent triangle.

A right proved by others : triangle OPR is congruent to triangle NTR (or half of triangle NMR)

so basically triangle OPR is one of the two congruent triangles which makes up the parallelogram. Since OPR = NTR.

hence half the area of the parallelogram is the area of the req triangle MNR -----> sufficient

Im sure Im going wrong somewhere but this is what i figured in the test. Can someone please point out where am i going wrong in my logic?

Geometry is my weakest area. Thanks in advance
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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 15 Jul 2016, 08:39
aniketm.87@gmail.com wrote:
Image
In the figure above, if MNOP is a trapezoid and NOPR is a parallelogram, what is the area of triangular region MNR?

(1) The ares of NOPR is 30
(2) The area of the shaded region is 5

A) area of NOPR is 30, this is a parallelogram. It consists of two congruent triangle.

A right proved by others : triangle OPR is congruent to triangle NTR (or half of triangle NMR)

so basically triangle OPR is one of the two congruent triangles which makes up the parallelogram. Since OPR = NTR.

hence half the area of the parallelogram is the area of the req triangle MNR -----> sufficient

Im sure Im going wrong somewhere but this is what i figured in the test. Can someone please point out where am i going wrong in my logic?

Geometry is my weakest area. Thanks in advance

Dear aniketm.87,

I'm happy to respond. :-)

My friend, either diagonal of a parallelogram will indeed divide the parallelogram into two congruent triangles. Do you understand exactly what the the word "congruent" means? It means totally matching in every way: every side in one triangle is exactly equal to the corresponding side in the other, and every angle in one triangle is exactly equal to the corresponding angle in the other. It's absolutely impossible for two triangles to be congruent if one has a right angle and the other doesn't.

In parallelogram NOPR, it's true that triangle OPR is congruent to triangle NOR, but that doesn't help us answer the prompt question at all.

Triangle NTR, the right triangle that is half of triangle NMR, is not even pretending to be congruent to either OPR or NOR. Triangle NTR has a right angle, and neither OPR or NOR has a right angle.

Does all this make sense?
Mike :-)
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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 09 Sep 2017, 20:58
- Shape NOPR is a parallelogram so that means lines NR and OP are parallel.
- If those lines are parallel, then angle R and angle P are equal. So angle P = x.
- If angle R and angle P are equal then triangle QOP is half of triangle MNR, because triangle MNR base angles are both x.
- You can imagine a line straight down the middle of triangle MNR to create a perpendicular angle, and since the two base angles are both x then triangle MNR is basically just two of triangle QOP.

1. Not sufficient because have no base or height information.
2. Sufficient because we know triangle QOP is half or triangle MNR. Since triangle QOP is area 5, then triangle MNR must be area 10.

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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 03 Oct 2017, 08:32
"As two angles and the corresponding side (NR and OP) are equal the two triangles, as per ASA rule,are congruent. Hence both will have same area."

Is this theory true ? can someone please confirm this.

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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]

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New post 03 Oct 2017, 09:26
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pra1785 wrote:
"As two angles and the corresponding side (NR and OP) are equal the two triangles, as per ASA rule,are congruent. Hence both will have same area."

Is this theory true ? can someone please confirm this.

Dear pra1785,

I'm happy to respond. :-)

Yes, that's 100% correct. :-) For more detail, see:
GMAT Data Sufficiency: Congruence Rules

Let me know if you have further questions after reading that blog.

Mike :-)
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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa   [#permalink] 03 Oct 2017, 09:26
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