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In the figure to the right, ∆PST is an isosceles right triangle, and P [#permalink]
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Updated on: 27 May 2017, 08:53
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In the figure to the right, ∆PST is an isosceles right triangle, and PS = 2. What is the area of the shaded region URST? Attachment:
vbvbbvbv.jpg [ 7.46 KiB  Viewed 1089 times ]
(A) 4 (B) 2 (C) 5/4 (D) 5/6 (E) 1/2
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Originally posted by SajjadAhmad on 27 May 2017, 08:42.
Last edited by Bunuel on 27 May 2017, 08:53, edited 1 time in total.
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Re: In the figure to the right, ∆PST is an isosceles right triangle, and P [#permalink]
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27 May 2017, 17:23
SajjadAhmad wrote: In the figure to the right, ∆PST is an isosceles right triangle, and PS = 2. What is the area of the shaded region URST? Attachment: vbvbbvbv.jpg (A) 4 (B) 2 (C) 5/4 (D) 5/6 (E) 1/2 Area of the shaded region URST = Area of \(\triangle\)PST  Area of \(\triangle\)PRU
Given; \(\triangle\)PST is an isosceles right triangle. PS = 2 \(\angle\)PTS is right angle. ie; \(90^{\circ}\). Therefore sides PT = TS Sides of isosceles right triangle are in ratio = 1:1:\(\sqrt{2}\) PS = 2 Therefore sides PT = TS = \(\frac{2}{\sqrt{2}}\) Area of Area of \(\triangle\)PST =\(\frac{1}{2}\) b x h = \(\frac{1}{2}\) x PT x TS = \(\frac{1}{2}\) x \(\frac{2}{\sqrt{2}}\) x \(\frac{2}{\sqrt{2}}\) = 1Area of \(\triangle\)PRU Given from the diagram; QR = PU = 1 \(\angle\)PUR is right angle. ie; \(90^{\circ}\). PU = UR = 1 Area of \(\triangle\)PRU = \(\frac{1}{2}\) b x h = \(\frac{1}{2}\) x PU x UR = \(\frac{1}{2}\) x 1 x 1 = \(\frac{1}{2}\)
Area of the shaded region URST = Area of \(\triangle\)PST  Area of \(\triangle\)PRU Area of the shaded region URST = 1  \(\frac{1}{2}\) = \(\frac{1}{2}\) Answer E...



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Re: In the figure to the right, ∆PST is an isosceles right triangle, and P [#permalink]
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27 May 2017, 20:17
Triangle PTS is isosceles So PT = TS = { PS / √2} = √2 So UT = √2  1 In trapezoid URST Area URST = { 1/2* ( UR+ST)* UT} = { 1/2 * ( √2 + 1 ) * ( √2 1) } = 1/2 Option E Sent from my ONE A2003 using GMAT Club Forum mobile app



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Re: In the figure to the right, ∆PST is an isosceles right triangle, and P [#permalink]
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27 May 2017, 22:00
Area of square = 1 Area of large triangle = 1/2*2=1 Area of Required region = Area of triangle  1/2 area of square 11/2=1/2
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In the figure to the right, ∆PST is an isosceles right triangle, and P [#permalink]
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28 May 2017, 01:41
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From the square, we can figure out that the triangle PUR is an isosceles triangle(454590)Also, the side PR is \(\sqrt{2}\)(diagonal of square with side 1) Triangle PUR and PST are similar(by AA similarity  Angle P is common and both angle U and S are 90 degree each) Also given PS = 2, hence RS = \(2  \sqrt{2}\) In similar triangles, corresponding sides are proportional. \(\frac{PR}{PS} = \frac{PU}{PT}\) Since PR = \(\sqrt{2}\), PS = 2 and PT = 1(given from the question stem) Hence PT = \(\sqrt{2}\) Triangle PST is also an isosceles triangle(454590), Area(Triangle PST) = \(\frac{1}{2} * \sqrt{2} * \sqrt{2}\) = 1 Area of shaded portion = Area(Triangle PST)  Area(Triangle PUS) = 1  \(\frac{1}{2} * 1 * 1\) = \(\frac{1}{2}\) (Option E)
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In the figure to the right, ∆PST is an isosceles right triangle, and P [#permalink]
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Updated on: 10 Apr 2018, 12:10
pushpitkc wrote: From the square, we can figure out that the triangle PUR is an isosceles triangle(454590) Also, the side PR is \(\sqrt{2}\)(diagonal of square with side 1) How do you know it's a square? Am I missing something here? pushpitkc Bunuel VeritasPrepKarishma Please help! Quote: Triangle PUR and PST are similar(by AA similarity  Angle P is common and both angle U and S are 90 degree each) Also given PS = 2, hence RS = \(2  \sqrt{2}\) In similar triangles, corresponding sides are proportional.
\(\frac{PR}{PS} = \frac{PU}{PT}\)
Since PR = \(\sqrt{2}\), PS = 2 and PT = 1(given from the question stem) Hence PT = \(\sqrt{2}\)
Triangle PST is also an isosceles triangle(454590), Area(Triangle PST) = \(\frac{1}{2} * \sqrt{2} * \sqrt{2}\) = 1 Area of shaded portion = Area(Triangle PST)  Area(Triangle PUS) = 1  \(\frac{1}{2} * 1 * 1\) = \(\frac{1}{2}\) (Option E)
Originally posted by rever08 on 08 Apr 2018, 09:35.
Last edited by rever08 on 10 Apr 2018, 12:10, edited 2 times in total.



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Re: In the figure to the right, ∆PST is an isosceles right triangle, and P [#permalink]
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08 Apr 2018, 18:44
I'm really confused by this question. How can you assume that RU is parallel to TS? How can you assume anything about triangle PQR?



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Re: In the figure to the right, ∆PST is an isosceles right triangle, and P [#permalink]
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11 Apr 2018, 03:04
rever08 wrote: pushpitkc wrote: From the square, we can figure out that the triangle PUR is an isosceles triangle(454590) Also, the side PR is \(\sqrt{2}\)(diagonal of square with side 1) How do you know it's a square? Am I missing something here? pushpitkc Bunuel VeritasPrepKarishma Please help! Quote: Triangle PUR and PST are similar(by AA similarity  Angle P is common and both angle U and S are 90 degree each) Also given PS = 2, hence RS = \(2  \sqrt{2}\) In similar triangles, corresponding sides are proportional.
\(\frac{PR}{PS} = \frac{PU}{PT}\)
Since PR = \(\sqrt{2}\), PS = 2 and PT = 1(given from the question stem) Hence PT = \(\sqrt{2}\)
Triangle PST is also an isosceles triangle(454590), Area(Triangle PST) = \(\frac{1}{2} * \sqrt{2} * \sqrt{2}\) = 1 Area of shaded portion = Area(Triangle PST)  Area(Triangle PUS) = 1  \(\frac{1}{2} * 1 * 1\) = \(\frac{1}{2}\) (Option E) Ideally, they should have given the 90 degree angles but even if it is not, it is still apparent. Since it is a PS question, the figure is what it looks like. You cannot solve it without assuming the 90 degree angles of the figure. You may not need to assume that it is a square but since PST is isosceles and triangles PUR and PTS are similar by AA, PU/PT = UR/TS PU/UR = PT/TS = 1/1 PU = UR
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