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605-655 (Medium)|   Geometry|      
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In the figure to the right, ∆PST is an isosceles right triangle, and P Updated on: 05 Jul 2019, 05:03
Originally posted by Sajjad1994 on 27 May 2017, 08:42.
Last edited by Sajjad1994 on 05 Jul 2019, 05:03, edited 2 times in total.
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In the figure below, ∆PST is an isosceles right triangle, and PS = 2. What is the area of the shaded region URST?

Attachment:
vbvbbvbv.jpg
vbvbbvbv.jpg [ 7.46 KiB | Viewed 10408 times ]

(A) 4
(B) 2
(C) 5/4
(D) 5/6
(E) 1/2

Source: Nova GMAT
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E
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In the figure to the right, ∆PST is an isosceles right triangle, and P 27 May 2017, 17:23
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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


(A) 4
(B) 2
(C) 5/4
(D) 5/6
(E) 1/2
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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 \(\frac{2}{\sqrt{2}}\) = 1

Area 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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In the figure to the right, ∆PST is an isosceles right triangle, and P 27 May 2017, 20:17
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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

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In the figure to the right, ∆PST is an isosceles right triangle, and P 27 May 2017, 22:00
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Area of square = 1
Area of large triangle = 1/2*2=1
Area of Required region = Area of triangle - 1/2 area of square
1-1/2=1/2
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In the figure to the right, ∆PST is an isosceles right triangle, and P 28 May 2017, 01:41
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From the square, we can figure out that the triangle PUR is an isosceles triangle(45-45-90)
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(45-45-90), 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 Updated on: 10 Apr 2018, 12:10
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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pushpitkc
From the square, we can figure out that the triangle PUR is an isosceles triangle(45-45-90)
Also, the side PR is \(\sqrt{2}\)(diagonal of square with side 1)
Show more

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(45-45-90), 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 08 Apr 2018, 18:44
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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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In the figure to the right, ∆PST is an isosceles right triangle, and P 11 Apr 2018, 03:04
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pushpitkc
From the square, we can figure out that the triangle PUR is an isosceles triangle(45-45-90)
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(45-45-90), 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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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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In the figure to the right, ∆PST is an isosceles right triangle, and P 07 Apr 2021, 19:54
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