Last visit was: 19 Nov 2025, 20:14 It is currently 19 Nov 2025, 20:14
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
655-705 Level|   Geometry|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,384
 [3]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,384
 [3]
The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT 05 May 2020, 10:22
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

51% (02:26) correct 49% (02:49) wrong based on 70 sessions

History

Date
Time
Result
Not Attempted Yet

The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT = RT, what is the area of ∆PST?

A. 3 cm^2
B. 2√2 cm^2
C. 2 cm^2
D. √3 cm^2
E. √2 cm^2


Are You Up For the Challenge: 700 Level Questions

Show Spoiler
Attachment:
mensuration-5-1.png
mensuration-5-1.png [ 9.61 KiB | Viewed 4873 times ]
ShowHide Answer
Official Answer
E
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 19 Nov 2025
Posts: 5,794
Own Kudos:
5,514
 [2]
Given Kudos: 161
Location: India
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
My Reviews
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
Posts: 5,794
Kudos: 5,514
 [2]
The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT 05 May 2020, 11:34
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel

The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT = RT, what is the area of ∆PST?

A. 3 cm^2
B. 2√2 cm^2
C. 2 cm^2
D. √3 cm^2
E. √2 cm^2


Are You Up For the Challenge: 700 Level Questions

Show Spoiler
Attachment:
The attachment mensuration-5-1.png is no longer available
Show more

Given:
1. The length of a side of square PQRS above is 2 cm.
2. IST = 2 cm
3. PT = RT
Asked: What is the area of ∆PST?

Attachment:
Screenshot 2020-05-06 at 12.03.13 AM.png
Screenshot 2020-05-06 at 12.03.13 AM.png [ 24.98 KiB | Viewed 4392 times ]

Area of triangle PSO = (√2*√2)/2 = 1 cm^2
Area of triangle PTO = (2+√2)*√2/2 = √2+1 cm^2
Area of triangle PST = Area of triangle PTO - Area of triangle PSO = √2+1 - 1 =  √2 cm^2

IMO E

Bunuel
Why some unwanted characters are shown in my answer?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,384
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,384
The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT 05 May 2020, 11:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kinshook
Bunuel

The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT = RT, what is the area of ∆PST?

A. 3 cm^2
B. 2√2 cm^2
C. 2 cm^2
D. √3 cm^2
E. √2 cm^2


Are You Up For the Challenge: 700 Level Questions

Show Spoiler
Attachment:
mensuration-5-1.png

Given:
1. The length of a side of square PQRS above is 2 cm.
2. IST = 2 cm
3. PT = RT
Asked: What is the area of ∆PST?

Attachment:
Screenshot 2020-05-06 at 12.03.13 AM.png

Area of triangle PSO = (√2*√2)/2 = 1 cm^2
Area of triangle PTO = (2+√2)*√2/2 = √2+1 cm^2
Area of triangle PST = Area of triangle PTO - Area of triangle PSO = √2+1 - 1 =  √2 cm^2

IMO E

Bunuel
Why some unwanted characters are shown in my answer?
Show more

Yes, I see those too. We'll look into it. Thank you for reporting.

P.S. Please do not delete them.
User avatar
AnirudhaS
User avatar
LBS Moderator
Joined: 30 Oct 2019
Last visit: 25 Jun 2024
Posts: 811
Own Kudos:
872
Given Kudos: 1,575
Posts: 811
Kudos: 872
The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT 05 May 2020, 12:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PST=PTO - SOP

now, PTO = \(\frac{1}{2}*\sqrt{2}*(2+\sqrt{2}) = \sqrt{2} + 1\)
And, SOP = \(\frac{1}{2}*\sqrt{2}*\sqrt{2} = 1\)

Therefore, PST = \(\sqrt{2} + 1 -1 = \sqrt{2}\)

Answer: E
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 19 Nov 2025
Posts: 21,716
Own Kudos:
26,998
 [2]
Given Kudos: 300
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,716
Kudos: 26,998
 [2]
The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT 09 May 2020, 06:06
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel

The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT = RT, what is the area of ∆PST?

A. 3 cm^2
B. 2√2 cm^2
C. 2 cm^2
D. √3 cm^2
E. √2 cm^2


Show more

We see that the area of triangle PST is equal to the area of triangle POT minus the area of triangle POS.

Notice that both triangles POT and POS are right triangles. The base of both triangles is PO, which has a length of ½ x 2√2 = √2. The height of triangle POS is OS, which also has a length of √2. The height of triangle POT is OT, which has a length of 2 + √2. So, the area of triangle POT is ½ x √2(2 + √2) and that of triangle POS is ½ x √2(√2). Therefore, the area of triangle PST is:

½ x √2(2 + √2) - ½ x √2(√2) = ½ x √2(2 + √2 - √2) = ½ x √2(2) = √2

Answer: E
User avatar
PolarisTestPrep
User avatar
GMAT Instructor
Joined: 30 Jun 2021
Last visit: 24 Apr 2022
Posts: 57
Own Kudos:
21
Given Kudos: 10
Expert
Expert reply
Posts: 57
Kudos: 21
The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT 15 Jul 2021, 11:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello!

Hopefully you see that the area of triangle PST can be found by subtracting the area of triangle POS from the area of triangle OTP.

To solve the problem, you have to find the area of triangle OTP and of triangle POS.

In a problem like this, you want to start with the information given to you.

Knowing that the sides of the square are 2, we can derive the length of the diagonal PR and of OP and OS.

This will also give us the length of OT because we know that ST is 2. This will allow us to calculate the area of triangle OTP.

We can also use that information to calculate the area of triangle POS and subtract it from OTP to get the area of triangle PST.

The key here is knowing some rules about squares.

If a square has side length x, the diagonal length will always be x\(\sqrt{2}\)

You can also derive the length of the diagonal PR using the Pythagorean theorem

\(a^2 + b^2 = c^2\)

\(2^2 + 2^2 = c^2\)

\(4 + 4 = c^2\)

\(8 = c^2\\
\)

\(c = \sqrt{8} = 2\sqrt{2}\) because \(\sqrt{8} = \sqrt{4}\) x \(\sqrt{2} = 2\sqrt{2}\)

The diagonals of a square bisect each other, meaning each diagonal divides the other into two equal halves.

That means that PO = OR = OS = \(\frac{2\sqrt{2}}{2}\) = \( \sqrt{2}\)

If OS = \(\sqrt{2}\) and we are given that ST = 2 cm, OT = 2 + \(\sqrt{2}\)

If OP = \(\sqrt{2}\), use the formula for the area of a triangle \(\frac{1}{2}\)(base)(height) to find the area:

\(\frac{1}{2}\)(\(\sqrt{2}\))(2 + \(\sqrt{2}\)) = \(\frac{1}{2}(2\sqrt{2} + 2) = \sqrt{2} + 1\)

The area of triangle OTP is \(\sqrt{2} + 1\)

If we can find the area of triangle POS, we can subtract it from the area of OTP to find the area of triangle PST

Not only do the diagonals of a square bisect each other, they are perpendicular to each other and create four right angles at the center. This means that angle POS is a right angle and triangle POS is a right triangle.

We know that OS = OP = \(\sqrt{2}\)

The area of POS is therefore \(\frac{1}{2}(\sqrt{2})(\sqrt{2})\)\( = \frac{1}{2}(2) = 1\)

Area of PST = Area of OTP - Area of POS

Area of PST = \(\sqrt{2} + 1 - 1 = \sqrt{2}\)

The answer is (E) \(\sqrt{2}\) cm\(^2\)
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts