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

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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

Thanks.

Dear roygush,
I'm happy to help.

First of all, here's a important geometry theorem to know:
https://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 [ 20.79 KiB | Viewed 147209 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.

Does all this make sense?
Mike
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IMO B

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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IMO B

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).
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mikemcgarry wrote:
IMO B

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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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]
IMO B

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.
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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.

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:
https://magoosh.com/gmat/2013/angles-and ... -the-gmat/

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

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

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

How triangles MNT and NTR are congruent?
You are saying that as per ASA rule they are congruent, but how angle TNM and TNR are equal?
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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]
Why can just assume that the line that seemingly creates a triangle from Q to O actually meets in point O ? I understand how to solve the problem based on the assumption but would it not be wrong to just assume that it creates a triangle?
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Re: In the figure above, if MNOP is a trapezoid and NOPR is a pa [#permalink]
mikemcgarry wrote:
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

Thanks.

Dear roygush,
I'm happy to help.

First of all, here's a important geometry theorem to know:
https://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

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.

Does all this make sense?
Mike

Thanks Mike - this makes sense, but how can you see that TR = QP? I can see how the rest of the angles/sides are congruent but not that particular piece.
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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 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

Attachment:
The attachment Capture.JPG is no longer available

Let x=40.
Since triangle MNR is isosceles, MNOP is a trapezoid, and NOPR is a parallelogram, the following figure is yielded:
Attachment:

MNR and NOPR.png [ 166.33 KiB | Viewed 52085 times ]

The case above illustrates that triangle MNS = triangle NRS = triangle OPQ.

Statement 1; The area of region NOPR is 30
It's possible that NOQR = 25 and that triangle OPQ = 5.
In this case, since triangle MNS = triangle NRS = triangle OPQ = 5, triangle MNR = triangle MNS + triangle NRS = 5+5 = 10.

It's possible that NOQR = 26 and that triangle OPQ = 4.
In this case, since triangle MNS = triangle NRS = triangle OPQ = 4, triangle MNR = triangle MNS + triangle NRS = 4+4 = 8.

Since triangle MNR can be different values, INSUFFICIENT.

Statement 2: The area of the shaded region is 5.
Since triangle MNS = triangle NRS = triangle OPQ = 5, triangle MNR = triangle MNS + triangle NRS = 5+5 = 10.
SUFFICIENT.

.

Originally posted by GMATGuruNY on 11 Sep 2020, 04:44.
Last edited by GMATGuruNY on 08 Oct 2020, 11:58, edited 1 time in total.
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