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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.

Re: Help me with this triangle geometry DS [#permalink]
05 Nov 2010, 17:32

1

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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

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. _________________

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. _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

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?

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! _________________

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". _________________

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.

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 _________________

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

So the answer is C.

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?