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# In the figure to the right, ∆PST is an isosceles right triangle, and P

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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, 07:53
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Question Stats:

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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 1558 times ]

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

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Last edited by Bunuel on 27 May 2017, 07:53, edited 1 time in total.
Edited the question.
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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, 16:23
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 $$\frac{2}{\sqrt{2}}$$ = 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 - $$\frac{1}{2}$$ = $$\frac{1}{2}$$
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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, 19:17
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

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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, 21:00
Area of square = 1
Area of large triangle = 1/2*2=1
Area of Required region = Area of triangle - 1/2 area of square
1-1/2=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, 00:41
1
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  [#permalink]

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Updated on: 10 Apr 2018, 11:10
pushpitkc wrote:
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)

Originally posted by rever08 on 08 Apr 2018, 08:35.
Last edited by rever08 on 10 Apr 2018, 11:10, 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, 17:44
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, 02:04
rever08 wrote:
pushpitkc wrote:
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)

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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Re: In the figure to the right, ∆PST is an isosceles right triangle, and P &nbs [#permalink] 11 Apr 2018, 02:04
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