GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 20 May 2019, 00:29

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Senior RC Moderator
User avatar
V
Status: Preparing GMAT
Joined: 02 Nov 2016
Posts: 2741
Location: Pakistan
GPA: 3.39
In the figure to the right, ∆PST is an isosceles right triangle, and P  [#permalink]

Show Tags

New post Updated on: 27 May 2017, 08:53
3
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

77% (02:35) correct 23% (02:48) wrong based on 87 sessions

HideShow timer Statistics

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
vbvbbvbv.jpg [ 7.46 KiB | Viewed 1921 times ]



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

_________________
New Project RC Butler 2019 - Practice 2 RC Passages Everyday
Final days of the GMAT Exam? => All GMAT Flashcards.
This Post Helps = Press +1 Kudos
Best of Luck on the GMAT!!

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.
Edited the question.
Director
Director
User avatar
V
Joined: 04 Dec 2015
Posts: 750
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)
GMAT ToolKit User
Re: In the figure to the right, ∆PST is an isosceles right triangle, and P  [#permalink]

Show Tags

New post 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 \(\frac{2}{\sqrt{2}}\) = 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...
Intern
Intern
avatar
B
Joined: 28 May 2015
Posts: 36
Location: India
GMAT 1: 710 Q51 V34
GPA: 4
Reviews Badge
Re: In the figure to the right, ∆PST is an isosceles right triangle, and P  [#permalink]

Show Tags

New post 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
Senior Manager
Senior Manager
User avatar
G
Joined: 02 Mar 2017
Posts: 256
Location: India
Concentration: Finance, Marketing
Re: In the figure to the right, ∆PST is an isosceles right triangle, and P  [#permalink]

Show Tags

New post 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
1-1/2=1/2
_________________
Kudos-----> If my post was Helpful
Senior PS Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3386
Location: India
GPA: 3.12
In the figure to the right, ∆PST is an isosceles right triangle, and P  [#permalink]

Show Tags

New post 28 May 2017, 01:41
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)
_________________
You've got what it takes, but it will take everything you've got
Manager
Manager
User avatar
S
Joined: 21 Jul 2017
Posts: 191
Location: India
Concentration: Social Entrepreneurship, Leadership
GMAT 1: 660 Q47 V34
GPA: 4
WE: Project Management (Education)
Premium Member Reviews Badge
In the figure to the right, ∆PST is an isosceles right triangle, and P  [#permalink]

Show Tags

New post Updated on: 10 Apr 2018, 12:10
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, 09:35.
Last edited by rever08 on 10 Apr 2018, 12:10, edited 2 times in total.
Intern
Intern
avatar
B
Joined: 30 Jan 2018
Posts: 2
Re: In the figure to the right, ∆PST is an isosceles right triangle, and P  [#permalink]

Show Tags

New post 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?
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 9224
Location: Pune, India
Re: In the figure to the right, ∆PST is an isosceles right triangle, and P  [#permalink]

Show Tags

New post 11 Apr 2018, 03:04
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
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT Club Bot
Re: In the figure to the right, ∆PST is an isosceles right triangle, and P   [#permalink] 11 Apr 2018, 03:04
Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.