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

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HideShow timer Statistics 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

Attachment: Triangle.GIF [ 2.17 KiB | Viewed 221357 times ]

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134 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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7
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

So the answer is C.
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Re: In the figure shown, what is the value of x?  [#permalink]

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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.

My answer would be C.
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1
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]

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2
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
Considering the quadrilateral now: 90+(180-z)+x+(180-y)=360
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]

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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]

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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]

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Ok. Here is what I don't get.
To solve this question it is assumed that the two triangles RQS and SUT are isosceles triangles. This assumption is based on the condition QS=QR. My problem however is that an equilateral triangle QS=QR=SR would also satisfy the condition QS=QR. In that way the angles could be either 45-45-90 or 60-60-60. So what am I missing that gives proof that these triangles are Isosceles? As far as I know one should never make assumptions based on the graphs in the GMAT.

thanks for quick response
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Re: In the figure shown, what is the value of x?  [#permalink]

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Petermeterdeter wrote:
Ok. Here is what I don't get.
To solve this question it is assumed that the two triangles RQS and SUT are isosceles triangles. This assumption is based on the condition QS=QR. My problem however is that an equilateral triangle QS=QR=SR would also satisfy the condition QS=QR. In that way the angles could be either 45-45-90 or 60-60-60. So what am I missing that gives proof that these triangles are Isosceles? As far as I know one should never make assumptions based on the graphs in the GMAT.

thanks for quick response Two notes on this:
1. An isosceles triangle has at least two sides of the same length, which means that if a triangle is equilateral it could still be considered as isosceles.

2. Triangle QSR is not necessarily 45-45-90 or 60-60-60. Why cannot it be say, 15-15-150?
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Re: In the figure shown, what is the value of x?  [#permalink]

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Hi

for geom it is often more convenient to name angles your own way to avoid excessive abstraction: a, b, c, d (please the attached file) - because many of the angles get cancelled out during calculations and overall the less astractive the more comprehensible.

We are given that angle RPT=90. We derive that QSTU is a rectangle where one of the angles = x = 360 - 90 - c - d. We also understand that x = 180 - a - b.

ST 1: QR=RS hence a=a=angle RQS=angle RSQ - insufficient, because we do not know how other angles relate to each other and to x. At this point we understand that knowing b would suffice the question.

ST 2: ST=UT hence b=b - insufficient for the same reasons stated above. Clearly this statement gives just the same piece of information as the statement 1 does. We also understand it is a good thing to combine both statements.

ST 2 + ST 1: at this point we can stop and choose C because the stimulus gives you a bold hint that one of the angles of the triangle equals 90, x is within the rectangle where one of the angles is 90, 2 of the smaller triangles are isosceles, those base angles in the triangles are adjacent to x - having taken a short glimpse of all those factors you could figure out halfway that this is pretty much enough for a geom question like that to avoid answer E and choose C.

However, this is the solution for C:

1. x=180-(a+b) and x=360-(90+c+d)
2. a=180-c and b=180-d
3. combine (1) and (2): 360-90-c-d = 180 - (180-c) - (180-d)
4. c+d=225
5. x=360-(90+225)=45
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File comment: visualization Capture.JPG [ 20.69 KiB | Viewed 44036 times ]

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Originally posted by shasadou on 01 Sep 2015, 03:13.
Last edited by shasadou on 22 Oct 2015, 08:12, edited 1 time in total.
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Re: In the figure shown, what is the value of x?  [#permalink]

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Bunuel wrote: 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: In the figure shown, what is the value of x?  [#permalink]

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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

So the answer is C.

Hi subhashghosh,

Let's break your statement into two parts.

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

1) Xº + 90º + 180º = 360º ---> They make up the sum of all the angles of the quadrilateral PQSU.

2) 180 - RSQ - TSU = Xº

Hence, you are saying that 360 º + X = 360º, which is incorrect.
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Re: In the figure shown, what is the value of x?  [#permalink]

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blazov wrote:
In the figure shown , what is the value of x?

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

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

Statement 1: Gives us nothing , since we don't know any other angles or property to compute other angles

Statement 2: Gives us nothing , since we don't know any other angles or property to compute other angles

Combined: We know that $$\angle$$RQS and $$\angle$$RSQ are same, since have also equal sides RS=RQ. Also angles $$\angle$$TSU and $$\angle$$TUS are same since ST=TU

We will name angles as below for simplicity;

$$\angle$$ RQS=Z, $$\angle$$RSQ=Z, $$\angle$$QRS=B

$$\angle$$TSU=W, $$\angle$$TUS=W, $$\angle$$STU=C

So, we know that: $$90^{\circ}$$+B+C=180 meaning, B+C=$$90^{\circ}$$

I we add up angles of both triangles of QRS and STU: $$\angle$$B+2 *$$\angle$$Z+$$\angle$$C+2 *$$\angle$$W=$$360^{\circ}$$

Since we know $$\angle$$B+$$\angle$$C=$$90^{\circ}$$, we subtract it: 2 *$$\angle$$Z+2 *$$\angle$$W=$$270^{\circ}$$

We can divide both sides by 2 and get : $$\angle$$Z+$$\angle$$W=$$135^{\circ}$$

To find $$\angle$$SQP and $$\angle$$SUP
360-$$\angle$$(Z+W)=$$225^{\circ}$$

$$\angle$$SQPU=$$360^{\circ}$$
90+225+x=$$360^{\circ}$$

X=$$45^{\circ}$$

Both statements together are enough. Answer C

+1 if this helped, I put a lot effort Attachments

File comment: Explanation GMAT triangle.png [ 87.14 KiB | Viewed 43306 times ]

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Re: In the figure shown, what is the value of x?  [#permalink]

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Prompt analysis

Let angle PRS be a, angle PTR be b.

Super set
The angle x will be in the range of 0 to 180.

Translation
In order to find x, we need
1# the exact value of all the angles
2# the exact value of all the sides
3# some information/ property that will lead us to find the value of x

Statement analysis

St 1: Since the triangle QRS is an isosceles triangle we can say that angle RSQ = (180 - a)/2. But we don't have any other information about the rest of the angles. INSUFFICIENT. Hence option A. D eliminated.

St 2: Since the triangle STU is an isosceles triangle we can say that angle TSU = (180 - b)/2. But we don't have any other information about the rest of the angles. INSUFFICIENT. HEnce option A,B, D eliminated.

St 1 & St 2: angle RSQ + angle QSU + angle TSU = 180. Or (180 - a)/2 + x+ (180 - b)/2 = 180; 180 -(a+b)/2 +x =180; x = (a+b)/2 as it is a straight line. Also if we look at the triangle PRT we can say that a +b + 90 = 180 or a +b =90. Putting the value the equation x = (a+b)/2, we get x =45. ANSWER

Option C
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Re: In the figure shown, what is the value of x?  [#permalink]

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Bunuel wrote: 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.

Hi Bunuel, I'm trying to understand the working after $$x+\angle{QSR}+\angle{UST}=180$$ but I can't see how you reach the answer. Can you help clarify for me?
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Re: In the figure shown, what is the value of x?  [#permalink]

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Cez005 wrote:
Bunuel wrote: 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.

Hi Bunuel, I'm trying to understand the working after $$x+\angle{QSR}+\angle{UST}=180$$ but I can't see how you reach the answer. Can you help clarify for me?

For (1)+(2) we have $$x+\angle{QSR}+\angle{UST}=180$$.

From (1): $$\angle{QSR}=\frac{180-\angle{R}}{2}$$.

From (2): $$\angle{UST}=\frac{180-\angle{T}}{2}$$.

Substitute these into $$x+\angle{QSR}+\angle{UST}=180$$ to get $$x+\frac{180-\angle{R}}{2}+\frac{180-\angle{T}}{2}=180$$. This leads to $$x+\frac{360-(\angle{R}+\angle{T})}{2}=180$$. Since $$\angle{R}+\angle{T}=90$$, then $$x+\frac{360-90}{2}=180$$ --> $$x=45$$.

Hope it helps.
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In the figure shown, what is the value of x?  [#permalink]

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Here is a picture which should help. I used fake number "7's" to show it was isosceles. Also the 7's could have been changed to any number and the 2 triangles didn't have to both have 7's. It was just an easier way to remind myself that the triangle was isosceles.
Attachments answer.jpg [ 32.29 KiB | Viewed 42432 times ]

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Re: In the figure shown, what is the value of x?  [#permalink]

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I'd like to get a suggestion as to whether my approach is valid or coincidental. Unlike an algebra combination problem (eg. solve for x+y), this is a geometry problem solving for x. Hence the x variables and x equations rule should be valid.

One can notice that there are 9 angles - including x and hence 9 variables. One can also see that there are 7 equations that can be formed that are unique: Triangle QRS, Triangle SUT, Big Triangle PRT, Quadrilateral PQSU, Line RST, Line RQP, and Line PUT. Hence in order to solve for x I need to reduce it down to 7 variables.

S1 reduces it from 9 variables/angles to 8 since one is the same. S2 essentially does the same thing. S1+S2 reduces it from 9 now to 7, which makes this solvable.

The key thing- I believe - is to ensure that we are not solving for combinations as everyone may know from algebra problems that although something may not look solvable it can be because something cancels out.

Feedback would be much appreciated whether this approach is valid or not. If it is, then this monstrosity of a problem can be solved simply in less than a minute.
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In the figure shown, what is the value of x?  [#permalink]

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Bunuel wrote: 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.

Thanks Bunuel, perfect solution. For more practice, can you please provide links for similar problems, thanks.
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