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Q and T are the midpoints of opposite sides of square PRSU [#permalink]
10 Jun 2018, 09:04
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#GREpracticequestion Q and T are the midpoints of opposite.jpg [ 13.89 KiB  Viewed 532 times ]
Q and T are the midpoints of opposite sides of square \(PRSU\)
Quantity A 
Quantity B 
The area of region \(PQST\) 
\(\frac{3}{2}\) 
A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The two quantities are equal. D)The relationship cannot be determined from the information given.
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
11 Jun 2018, 03:42
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Carcass wrote: Attachment: The attachment square.jpg is no longer available Q and T are the midpoints of opposite sides of square \(PRSU\)
Quantity A 
Quantity B 
The area of region \(PQST\) 
\(\frac{3}{2}\) 
A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The two quantities are equal. D)The relationship cannot be determined from the information given. Join QT and say this side as x. Now say ST as y we have right angled triangle QTS = ( \(\sqrt{5}\))^2 = x^2+ y^2. Trick here is that x and y have to be only positive can't be negative. x^2+ y^2 = 5. Let's take the combination value. when x = 2 and y = 1 , this satisfies the equation. when y = 2 and x = 1 , this satisfies the equation and I don't think any values are possible apart from this. To get area of QSTP = 1/2 ( xy ) + 1/2 ( xy) = xy. What ever the value of xy from above we finally get as 2. This value is greater than 1.5 When we consider x and y as \(\sqrt{2}\) and \(\sqrt{3}\) or vice versa, xy value is greater than 1.5 So answer A. Hope this is clear.
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Last edited by msk0657 on 24 Jun 2018, 06:34, edited 1 time in total.



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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
24 Jun 2018, 04:47
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does someone have a easy and official explanation to this question? I got the answer right but I took like 5 mins to solve it



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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
24 Jun 2018, 06:28
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msk0657 wrote: To get area of QSTP = 1/2 ( xy ) + 1/2 ( xy) = xy. and if we consider x and y to be \(\sqrt{3}\) and \(\sqrt{2}\)then will RSUP be a square?
Hi can you explain how you jump to the above conclusion?
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
24 Jun 2018, 06:35
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Carcass wrote: Attachment: square.jpg Q and T are the midpoints of opposite sides of square \(PRSU\)
Quantity A 
Quantity B 
The area of region \(PQST\) 
\(\frac{3}{2}\) 
A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The two quantities are equal. D)The relationship cannot be determined from the information given. Let us take the right angled triangle RSQ,as PRSU is square By pythogorus theorem, \(RS^2 + RQ^2 = QS^2\) \(RS^2 + RQ^2 =5\) (since QS = \(\sqrt{5}\)) i.e. RS will be equal to =2 and RQ will be equal to =1 (since RSUP is a square and q is the midpoint of RP so RQ=RS/2) Therefore the Area of the triangle RSQ = \(\frac{1}{2} * 2 * 1 = 1\) Similarly area of triangle PTU = \(\frac{1}{2} *2 *1 = 1\) and Area of the square = \(side^2\) = \(2^2\) =4 Therefore the area of PQST = Area of the square  area of triangle PTU  Area of the triangle RSQ = 4  1  1 =2 Hence statement 1 is greater.
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
24 Jun 2018, 06:37
pranab01 wrote: msk0657 wrote: To get area of QSTP = 1/2 ( xy ) + 1/2 ( xy) = xy.
Hi can you explain how you jump to the above conclusion? You can refer the diagram and we are clearly given that figure is a square. QSTP is a parallelogram i.e. QST is one triangle and QPT is another triangle as highlighted in the diagram as T1 and T2. QSTP = T1 + T2 = two triangles = 1/2 ( xy ) + 1/2 ( xy) = xy. Hope this clears.
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
24 Jun 2018, 06:46
pranab01 wrote: msk0657 wrote: To get area of QSTP = 1/2 ( xy ) + 1/2 ( xy) = xy. and if we consider x and y to be \(\sqrt{3}\) and \(\sqrt{2}\)then will RSUP be a square?
Hi can you explain how you jump to the above conclusion? Note here, we have \(\sqrt{5}\) only for QST triangle as it is right angled triangle. We get X^2 + Y^2 = (\(\sqrt{5}\))^2 X^2 + Y^2 = 5. Now take combination, x = 2 and y = 1 satisfies and also vice versa. Now take x = \(\sqrt{2}\) and y = \(\sqrt{3}\) or vice versa , satifies the above equation. You are considering whole square and this is a deviation.
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
24 Jun 2018, 09:57
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Your graph is fascinating in its reasoning. However, the best and easier way to solve this question is, as almost always, an evaluation upfront and observation. Attachment:
square.jpg [ 14.32 KiB  Viewed 1132 times ]
Notice that we do have the triplet \(x:x:x \sqrt2\) So, \(x \sqrt{2} = \sqrt{5} = \frac{\sqrt5}{sqrt2} = 1.5\) \(x =\sqrt{5} = 2.2\) Area of the rectangle \(QS * ST= b*h = 1.5 * 2.2 = 3.3\) this is quantity A Quantity B, which of course, is 1.5. A is the answer
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
24 Jun 2018, 19:33
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Carcass wrote: Your graph is fascinating in its reasoning. However, the best and easier way to solve this question is, as almost always, an evaluation upfront and observation.
Notice that we do have the triplet \(x:x:x \sqrt2\)
So, \(x \sqrt{2} = \sqrt{5} = \frac{\sqrt5}{sqrt2} = 1.5\)
\(x =\sqrt{5} = 2.2\)
Area of the rectangle \(QS * ST= b*h = 1.5 * 2.2 = 3.3\) this is quantity A
Quantity B, which of course, is 1.5.
A is the answer Great explanation, I just missed this approach I just want to add that we donot actually need to know the value of \sqrt{5} SInce \(\sqrt{4}\) = 2 so \(\sqrt{5}\) should be more than 2 let us take = 2.05 and we know \(\sqrt{2}\) = 1.4 so if we take \(\sqrt{5}\) = 2.05 then \(\frac{\sqrt5}{sqrt2}\) = \(\frac{2.05}{1.4}\) = \(\frac{21}{28}\) But if we know consider the area PQST = length * breadth = \(\frac{21}{28}\) * \(\sqrt{5}\) which will always be greater than statement B.
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
24 Jun 2018, 20:04
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msk0657 wrote: pranab01 wrote: msk0657 wrote: To get area of QSTP = 1/2 ( xy ) + 1/2 ( xy) = xy. and if we consider x and y to be \(\sqrt{3}\) and \(\sqrt{2}\)then will RSUP be a square?
Hi can you explain how you jump to the above conclusion? Now take combination, x = 2 and y = 1 satisfies and also vice versa. Now take x = \(\sqrt{2}\) and y = \(\sqrt{3}\) or vice versa , satifies the above equation. You are considering whole square and this is a deviation. Hi, I doubt if you can use y=1 and x= 2 satisfies PRSU as square or either taking as \(\sqrt{2}\)and \(\sqrt{3}\). Moreover QSTP is a rectangle and we can easily find out the area as = Length * Breadth However I agree with Caracass with approach
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
19 Jan 2019, 07:13
Carcass wrote: Notice that we do have the triplet \(x:x:x \sqrt2\)
So, \(x \sqrt{2} = \sqrt{5} = \frac{\sqrt5}{sqrt2} = 1.5\)
I am confused. Please explain how is possible x:x:xsqrt2 because that is not 45:45: 90 but 30: 60: 90



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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
20 Jan 2019, 05:55
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AE wrote: Carcass wrote: Notice that we do have the triplet \(x:x:x \sqrt2\)
So, \(x \sqrt{2} = \sqrt{5} = \frac{\sqrt5}{sqrt2} = 1.5\)
I am confused. Please explain how is possible x:x:xsqrt2 because that is not 45:45: 90 but 30: 60: 90 Since PRSU is square and line is meeting the midpoint of the opposite side
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
23 Jan 2019, 20:33
Answer is A
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
29 Jan 2019, 19:17
since the shape is sqaure, x^2+1/2*x^2=5 then pqst=5/3



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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
03 Mar 2019, 08:50
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this is how I calculated
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Re: Q and T are the midpoints of opposite sides of square PRSU [#permalink]
04 Mar 2019, 11:17
thanks




Re: Q and T are the midpoints of opposite sides of square PRSU
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