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# In the figure shown, what is the value of x?

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In the figure shown, what is the value of x? [#permalink]  25 Jul 2010, 19:24
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Triangle.GIF [ 2.17 KiB | Viewed 65210 times ]
In the figure shown, what is the value of x?

(1) The length of line segment QR is equal to the length of line segment RS

(2) The length of line segment ST is equal to the length of line segment TU
[Reveal] Spoiler: OA

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In the figure shown, what is the value of x? [#permalink]  26 Jul 2010, 13:56
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In the figure shown, what is the value of x?

$$x+\angle{QSR}+\angle{UST}=180$$ (straight line =180) and $$\angle{R}+\angle{T}=90$$ (as PRT is right angle)

(1) The length of line segment QR is equal to the length of line segment RS --> triangle QRS is isosceles --> $$\angle{RQS}=\angle{QSR}=\frac{180-\angle{R}}{2}$$ (as $$\angle{RQS}+\angle{QSR}+\angle{R}=180$$ --> $$2*\angle{QSR}+\angle{R}=180$$ --> $$\angle{QSR}=\frac{180-\angle{R}}{2}$$). Not sufficient.

(2) The length of line segment ST is equal to the length of line segment TU --> triangle UST is isosceles --> $$\angle{SUT}=\angle{UST}=\frac{180-\angle{T}}{2}$$. Not sufficient.

(1)+(2) $$x+\angle{QSR}+\angle{UST}=180$$ --> $$x+\frac{180-\angle{R}}{2}+\frac{180-\angle{T}}{2}=180$$ --> $$x+\frac{360-(\angle{R}+\angle{T})}{2}=180$$ --> since $$\angle{R}+\angle{T}=90$$ --> $$x+\frac{360-90}{2}=180$$ --> $$x=45$$. Sufficient.

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Re: Help me with this triangle geometry DS [#permalink]  05 Nov 2010, 17:32
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Well this is how I would solve this.

Original statement:

What is x?

Without jumping into statements, we can clearly see that x = 180 - ARSQ - ATSU.

Statement 1:

Statement 1 says that triangle QRS is isosceles. with ARSQ = ARQS

therefore we know that ARSQ = (180 - ASRQ)/2.

However, without information about triangle TSU cannot solve the above equation about x.

Insufficient

Statement 2:

Statement 2 says that triangle TSU is isosceles with ATSU = (180 - ASTU)/2.
Similar to statement 1, not enough information about triangle RSQ to find x.

Insufficient.

Statement 1 + 2:

From statement 1: we know that ARSQ = (180-ASRQ)/2
From statement 2: we know that ATSU = (180-ASTU)/2

Plugging these information into the x = 180 - ARSQ - ATSU we see that

x = 180 - (180-ASRQ)/2 - (180-ASTU)/2 = 180 - (360/2) + (ASRQ + ASTU)/2 = (ASRQ + ASTU)/2

Since we know points RTP makes a right triangle, we know that ASRQ + ASTU = 90.

x = 45.

Sufficient.

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Re: Help me with this triangle geometry DS [#permalink]  05 Nov 2010, 18:10
My answer would be C as well, from the

Isosceles triangle give us 45 degree angle.

Statement 1 : we know that angle RSQ = 45
Statement 2 : We know that angle UST = 90 (SUT = 45, UTS = 45, UST = 180 - (SUT + UTS)

Hence x equal with 180 - (45 +95) = 45
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GMAT Prep DS, geometry: Find angle X [#permalink]  20 Dec 2010, 13:57

Hey guys, I have another question regarding my reasoning in the solving of the problem.
[Reveal] Spoiler:
C

[Reveal] Spoiler:
E
and incorrectly so.

Now that I looked at it more, I came up with this solution, and I'd appreciate if you can tell me if my logic is correct:

Let's label angle QRS (or PRT) as w. Let's label angle STU as z. Since RPT is 90, we know that w+z=90
Now, RQS and RSQ are same angles. So, we can label them as j. At the same time, TSU and TUS are the same, so we can label them as k.
We have that 2j+w=180, that 2k+z=180, and that w+z=90. By substituting, we can see that w=180-2j and that z=180-2k.
Finally, we have that w+z=90 or that (180-2j)+(180-2k)=90 or that 2k+2j=250, or that j+k=125. Since j and k lie on a straight line, and combined with x produce 180, and j+k=125, we conclude that x=55.

Is my reasoning alright and is there a quicker way to solve this?

Much appreciated!
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Re: GMAT Prep DS, geometry: Find angle X [#permalink]  20 Dec 2010, 14:11
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MisterEko wrote:

Hey guys, I have another question regarding my reasoning in the solving of the problem.
[Reveal] Spoiler:
C

[Reveal] Spoiler:
E
and incorrectly so.

Now that I looked at it more, I came up with this solution, and I'd appreciate if you can tell me if my logic is correct:

Let's label angle QRS (or PRT) as w. Let's label angle STU as z. Since RPT is 90, we know that w+z=90
Now, RQS and RSQ are same angles. So, we can label them as j. At the same time, TSU and TUS are the same, so we can label them as k.
We have that 2j+w=180, that 2k+z=180, and that w+z=90. By substituting, we can see that w=180-2j and that z=180-2k.
Finally, we have that w+z=90 or that (180-2j)+(180-2k)=90 or that 2k+2j=250, or that j+k=125. Since j and k lie on a straight line, and combined with x produce 180, and j+k=125, we conclude that x=55.

Is my reasoning alright and is there a quicker way to solve this?

Much appreciated!

Except the calculation (the red part) seems that everything is OK. Check other solutions above for slightly different approaches.
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Re: GMAT Prep Question... plz explain??? [#permalink]  20 Dec 2010, 14:29
Thanks Bunnuel, I appreciate it. I see I made the wrong calculation, but the answer should be the same in this DS, which is C.

If you have any suggestions as to how I can search problems that are posted in a form of an image rather than words, it would be helpful so I don't have to repeat Qs... Thanks again!
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Re: GMAT Prep Question... plz explain??? [#permalink]  20 Dec 2010, 14:58
Expert's post
MisterEko wrote:
Thanks Bunnuel, I appreciate it. I see I made the wrong calculation, but the answer should be the same in this DS, which is C.

If you have any suggestions as to how I can search problems that are posted in a form of an image rather than words, it would be helpful so I don't have to repeat Qs... Thanks again!

Generally it's a good idea to search before posting. Though it's not a probelm to post a question that was posted before: if moderators find previous discussions they will merge the topics, copy the solution from there or give a link to it.

To make it easier to search I always copy a body text of every question to the post. So for example I'd search this question in DS subforum by the key words: "figure", "shown", "segment", "length".
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Re: GMAT Prep Question... plz explain??? [#permalink]  16 Mar 2011, 00:03
Bunuel wrote:
In the figure shown, what is the value of x?

x+<QSR+<UST=180 (straight line =180) and <R+<T=90 (as PRT is right angle)

(1) The length of line segment QR is equal to the length of line segment RS --> QRS is isosceles --> <PQS=<QSR=(180-R)/2 (as <PQS+<QSR+<R=180 --> 2<QSR+<R=180 --> <QSR=(180-R)/2). Not sufficient.

(2) The legnth of line segment ST is equal to the length of line segment TU --> UST is isosceles --> <SUT=<UST=(180-T)/2. Not sufficient.

(1)+(2) x+<QSR+<UST=180 --> x+(180-R)/2+(180-T)/2=180 --> x+(360-(R+T))/2=180 --> as R+T=90 --> x+(360-90)/2=180 --> x=45. Sufficient.

so kool
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Re: GMAT Prep Question... plz explain??? [#permalink]  16 Mar 2011, 20:19
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From (1) RSQ = RQS

From (2) SUT = TSU

Because these 3 angles are on a straight line:

x + RSQ + TSU = 180

As all the angles of a quadrilateral sum up to 360:

x+90 + 180 - RSQ + 180 - TSU = 360

=> x + 360 - (RSQ + TSU) + 90 = 360

=> x + 90 -(180-x) = 0

=> 2x - 90 = 0

=> x = 45

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Re: Geometry [#permalink]  08 Jan 2012, 05:17
kotela wrote:

stmnt 1. QR= RS then angle RQS = angle RSQ = y

We have no other detail and hence insuff

stmnt 2: like stmnt 1 this will also be insuff

taking together we have

QR= RS then angle RQS = angle RSQ = y

then angle QRS = 180 - 2y

PTR = 180 - ( 90 + 180 - 2y) = 2y - 90

we have TSU = TUS = [180 - (2y - 90)]/2 = 135 - y

RSQ + QSU + TUS = 180

y + x + 135 - y = 180

Hence C is Suff
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Re: Geometry [#permalink]  08 Jan 2012, 17:45
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C it is.

Two equations,
x = 180 - (a + b), where a and b are RQS and SUT.
x = 360 -[(180-a) + (180-b) + 90] = (a+b)-90.

Solve both, you get x.
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Re: In the figure shown, what is the value of x? [#permalink]  08 Aug 2012, 14:37
Hi Bunuel, I got the answer by plugging 2 diff values and I got the answer in a very short time. Am i doing anything wrong. Pls explain. My approach as follows:

I did this by plugging 2 different sets of values.
We know PRT is a right angle triangle. So let it be 90-60-30 or 90-45-45 triangle.
Now we can find out the value of X for both this situation. In both case the value of X comes as 45 degree.
So St 1+ St 2 Sufficient.
For 90-60-30: <RST + <TSU = 135 degree, so X =45 degree
For 90-45-45: <RST + <TSU = 135/2 + 135/2 = 135 Degree, so X =45 Degree
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Re: In the figure shown, what is the value of x? [#permalink]  17 Jul 2013, 12:50
shorteverything wrote:
Attachment:
Triangle.GIF
In the figure shown, what is the value of x?

(1) The length of line segment QR is equal to the length of line segment RS

(2) The length of line segment ST is equal to the length of line segment TU

hi bunuel,

just let me know if i go wrong.

x = angle QSU===>This angle is dependent on two things: 1)QS orientation 2) SU orientation.

statement 1 tell RS=RQ==>means orientation of QS is fixed===>but we dont know about SU==>INSUFFICIENT

STATEMENT 2==> TU=TS==>orientation of SU fixed ==>but we dont know about QS.==>INSUFFICIENT.

by combining both orientation of both QS and SU is fixed==>so will defenetely get fixed value.
hence C
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Re: GMAT Prep Question... plz explain??? [#permalink]  31 Aug 2013, 12:57
Hi Subash,

Could you please explain me how you got the eqn -> x+90 + 180 - RSQ + 180 - TSU = 360 ?

subhashghosh wrote:
From (1) RSQ = RQS

From (2) SUT = TSU

Because these 3 angles are on a straight line:

x + RSQ + TSU = 180

As all the angles of a quadrilateral sum up to 360:

x+90 + 180 - RSQ + 180 - TSU = 360

=> x + 360 - (RSQ + TSU) + 90 = 360

=> x + 90 -(180-x) = 0

=> 2x - 90 = 0

=> x = 45

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Re: GMAT Prep Question... plz explain??? [#permalink]  17 Jun 2014, 17:12
subhashghosh wrote:
From (1) RSQ = RQS

From (2) SUT = TSU

Because these 3 angles are on a straight line:

x + RSQ + TSU = 180

As all the angles of a quadrilateral sum up to 360:

x+90 + 180 - RSQ + 180 - TSU = 360

=> x + 360 - (RSQ + TSU) + 90 = 360

=> x + 90 -(180-x) = 0

=> 2x - 90 = 0

=> x = 45

Hi Subhash,

I got a little lost in the middle of this solution. Can you please explain how you're equating the angles to 360. I realize that you're equating the quadrilateral PQSU but i'm not sure where you're getting all the angles from?

Highlighted the above area in question. Isnt "180 - RSQ + 180 - TSU" actually giving you the the line "RT" if you add "x" in the mix of it?

Another question -- are we able to say that angle x is the sum of RQS and SRQ since they are opposite ends of the angle in question?

Thanks!
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Re: In the figure shown, what is the value of x? [#permalink]  20 Jun 2014, 02:30
shorteverything wrote:
Attachment:
Triangle.GIF
In the figure shown, what is the value of x?

(1) The length of line segment QR is equal to the length of line segment RS

(2) The length of line segment ST is equal to the length of line segment TU

Thanks for the share...
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Re: In the figure shown, what is the value of x? [#permalink]  24 Nov 2014, 08:55
The triangle in the figure has 2 triangles and a quadrilateral in it.
S1 and S2 insufficient
S1+S2
Considering the top-triangle: assign 'y' to equal angles
Considering the bottom triangle: assign 'z' to the equal angles
x+y+z= 180.............I
x-y-z=-90................II
Adding I and II gives me a value for x
(in the spirit of seeing things through- 2x=90 -----> x=90/2=45)
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Re: In the figure shown, what is the value of x? [#permalink]  16 Mar 2015, 03:40
How do I know it is not actually equilateral?
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Re: In the figure shown, what is the value of x? [#permalink]  16 Mar 2015, 04:20
Expert's post
Petermeterdeter wrote:
How do I know it is not actually equilateral?

All angles in equilateral triangle are 60 degrees, thus a right triangle (a triangle with one 90-degree angle) cannot be equilateral.

Does this make sense?
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Re: In the figure shown, what is the value of x?   [#permalink] 16 Mar 2015, 04:20

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