Last visit was: 18 Nov 2025, 19:53 It is currently 18 Nov 2025, 19:53
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
505-555 Level|   Geometry|                     
User avatar
Walkabout
Joined: 02 Dec 2012
Last visit: 30 Oct 2025
Posts: 172
Own Kudos:
28,184
 [64]
Given Kudos: 35
Products:
Posts: 172
Kudos: 28,184
 [64]
4
Kudos
Add Kudos
60
Bookmarks
Bookmark this Post
Most Helpful Reply
avatar
PareshGmat
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,534
Own Kudos:
8,099
 [23]
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,534
Kudos: 8,099
 [23]
14
Kudos
Add Kudos
7
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
Given Kudos: 99,964
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
 [13]
8
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
General Discussion
avatar
madn800
Joined: 07 May 2013
Last visit: 11 Aug 2014
Posts: 67
Own Kudos:
Given Kudos: 1
Posts: 67
Kudos: 66
Kudos
Add Kudos
Bookmarks
Bookmark this Post
What is the shaded part, the pentagon or the right angled triangle?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
Given Kudos: 99,964
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
Kudos
Add Kudos
Bookmarks
Bookmark this Post
madn800
What is the shaded part, the pentagon or the right angled triangle?

The triangle is shaded.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,086
 [1]
Given Kudos: 99,964
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bumping for review and further discussion.

GEOMETRY: Shaded Region Problems!
User avatar
Bigred2008
Joined: 10 Aug 2009
Last visit: 08 Mar 2017
Posts: 66
Own Kudos:
281
 [2]
Given Kudos: 45
Location: United States (VA)
Concentration: Entrepreneurship, Strategy
WE:Business Development (Other)
Products:
Posts: 66
Kudos: 281
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Walkabout
Attachment:
Shaded region.png
The shaded portion of the rectangular lot shown above represents a flower bed. If the area of the bed is 24 square yards and x = y + 2, then z equals

(A) \(\sqrt{13}\)
(B) \(2\sqrt{13}\)
(C) 6
(D) 8
(E) 10

I actually did this a different way.

For a triangle area=\(\frac{B*H}{2}\) so solving I get that \(B*H=48\) now I would solve by algebra, but the problem tells that the difference between x & y is two, so what two numbers multiply to 48 that separated by 2... 6 & 8. Since it doesn't matter which side is which length I saw this was a pythagorean triplet and z or the hypotenuse is 10.

Answer E

Just another way to think about this
User avatar
pacifist85
Joined: 07 Apr 2014
Last visit: 20 Sep 2015
Posts: 324
Own Kudos:
449
 [1]
Given Kudos: 169
Status:Math is psycho-logical
Location: Netherlands
GMAT Date: 02-11-2015
WE:Psychology and Counseling (Other)
Posts: 324
Kudos: 449
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi,

This one was tricky, not because the solution was difficult, but because of my mistakes in the calculations...

I also used Paresh's approach, but instead of solving for y and getting y=x-2, I just used y and y+2. But, I couldn't see the 6*4=48 ... so I couldn't find the possible solutions of the equation easily!

In fact I did this:
Area= (b*2) / 2
24=[y(y+z)] / 2
24= (y^2 + 2y) / 2
48 = y^2 + 2y
y^2 + 2y - 48 = 0.

I didn't see these solutions: -6, 8 and ended up using the quadratic formula, which is not handy when you are not quick with calculations! Haha! So, I passed the time...

Just as a small piece of advice, in similar situations, what makes it easier to find the roots is to make the prime factorization of the constant. Then you end up with this 2*2*2*2*3, so you have fewer values to test and find one that works.
avatar
kjax
Joined: 12 Apr 2015
Last visit: 03 Jul 2017
Posts: 5
Own Kudos:
Given Kudos: 88
Posts: 5
Kudos: 3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Great OG problem. Paresh's solution helped me see the light on this one. The others made sense, but I may not see that relationship on test day. Re-labeling the figure with the formulas will help add clarity. Great explanation!
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,785
Own Kudos:
12,804
 [5]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,785
Kudos: 12,804
 [5]
3
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Hi All,

On certain GMAT questions, you can "brute force" the work and quickly come up with the correct answer. Here's how:

From the prompt and the picture, we know that…
1) The triangle is a right triangle (because it's in the "corner" of a rectangle)
2) The two legs of the triangle differ by 2
3) The area of the triangle is 24 (A = (1/2)(B)(H))

Let's focus on the area = 24 for a moment. We know the two legs differ by 2, so we can probably "brute force" the possibilities and find the match:

If the legs are:
2 and 4, then the area = 4
4 and 6, then the area = 12
6 and 8, then the area = 24 ---> that's THE match

If the legs are 6 and 8, then we can use the Pythagorean Formula to figure out the value of Z. You might also recognize the "Pythagorean Triplet" and solve the problem that way.

Final Answer:
GMAT assassins aren't born, they're made,
Rich
User avatar
jaspreets
Joined: 10 Dec 2012
Last visit: 26 Jun 2024
Posts: 36
Own Kudos:
Given Kudos: 61
Posts: 36
Kudos: 57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I liked your approach, Simple and time saving, however not applicable in all cases.
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,785
Own Kudos:
12,804
 [1]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,785
Kudos: 12,804
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi jaspreets,

As you continue to study for the GMAT, you're going to find that 'shortcuts' vary from question to question, but there are almost always shortcuts to found (and in many cases there are alternative ways to approach the question besides 'just doing the math'). It's to your benefit to learn to spot the 'clues' that hint at these options - and then practice using those approaches. While not every shortcut will be available on every question, the list of possible shortcuts is definable, so you can train to use them again and again on Test Day.

GMAT assassins aren't born, they're made,
Rich
User avatar
minhaz3333
Joined: 11 Nov 2014
Last visit: 11 Mar 2020
Posts: 26
Own Kudos:
Given Kudos: 106
Concentration: Marketing, Finance
WE:Programming (Computer Software)
Posts: 26
Kudos: 114
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1/2(xy)= 24 ; => xy = 48 ;
x = y+2 ;
=> x-y = 2;
squaring both sides

=> x^2 + y^2 -2xy = 4
=> x^2 + y^2 = 4 + 2*48 = 100 (well we know x^2 + y^2 = z^2)

=> z^2 = 100
=> z = 10.
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 18 Nov 2025
Posts: 21,712
Own Kudos:
26,990
 [3]
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: 21,712
Kudos: 26,990
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Walkabout
Attachment:
Shaded region.png
The shaded portion of the rectangular lot shown above represents a flower bed. If the area of the bed is 24 square yards and x = y + 2, then z equals

(A) \(\sqrt{13}\)
(B) \(2\sqrt{13}\)
(C) 6
(D) 8
(E) 10

In solving this problem we first must recognize that the flower bed is the right triangle with sides of y yards, x yards, and z yards. We are given that the area of the bed (which is the right triangle) is 24 square yards. Since we know that area of a triangle is ½ Base x Height, we can say:

24 = ½(xy)

48 = xy

We also know that x = y + 2, so substituting in y + 2 for x in the area equation we have:

48 = (y+2)y

48 = y^2 + 2y

y^2 + 2y – 48 = 0

(y + 8)(y – 6) = 0

y = -8 or y = 6

Since we cannot have a negative length, y = 6.

We can use the value for y to calculate the value of x.

x = y + 2

x = 6 + 2

x = 8

We can see that 6 and 8 represent two legs of the right triangle, and now we need to determine the length of z, which is the hypotenuse. Knowing that the length of one leg is 6 and the other leg is 8, we know that we have a 6-8-10 right triangle. Thus, the length of z is 10 yards.

If you didn't recognize that 6, 8, and 10 are the sides and hypotenuse of a right triangle, you would have to use the Pythagorean to find the length of the hypotenuse: 6^2 + 8^2 = c^2 → 36 + 64 = c^2 → 100 = c^2. The positive square root of 100 is 10, and thus the value of z is 10.

Answer is E.
User avatar
Abhishek009
User avatar
Board of Directors
Joined: 11 Jun 2011
Last visit: 18 Jul 2025
Posts: 5,937
Own Kudos:
Given Kudos: 463
Status:QA & VA Forum Moderator
Location: India
GPA: 3.5
WE:Business Development (Commercial Banking)
Posts: 5,937
Kudos: 5,327
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Walkabout
Attachment:
Shaded region.png
The shaded portion of the rectangular lot shown above represents a flower bed. If the area of the bed is 24 square yards and x = y + 2, then z equals

(A) \(\sqrt{13}\)
(B) \(2\sqrt{13}\)
(C) 6
(D) 8
(E) 10

Quote:
If the area of the bed is 24 square yards

So, \(\frac{1}{2} * x * y\)= 24

Or, xy = 48

Given

Quote:
x = y + 2

Substitute the value in xy

(y+2) y = 48

So, y = 6

hence x = 8

So, \(z\) \(=\) \(\sqrt{8^2 - 6^2}\)

Hence z = 10 ; Answer will be (E)
User avatar
bimalr9
Joined: 20 Mar 2015
Last visit: 07 Nov 2017
Posts: 39
Own Kudos:
Given Kudos: 9
Posts: 39
Kudos: 327
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Walkabout
Attachment:
Shaded region.png
The shaded portion of the rectangular lot shown above represents a flower bed. If the area of the bed is 24 square yards and x = y + 2, then z equals

(A) \(\sqrt{13}\)
(B) \(2\sqrt{13}\)
(C) 6
(D) 8
(E) 10

I thought that the shaded part was isosceles right triangle and came with Z = \(4\sqrt{6}\). Is it because "\(x= y+2\)" that this is not an isos rt. triangle?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,086
 [1]
Given Kudos: 99,964
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,086
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
bimalr9
Walkabout
Attachment:
Shaded region.png
The shaded portion of the rectangular lot shown above represents a flower bed. If the area of the bed is 24 square yards and x = y + 2, then z equals

(A) \(\sqrt{13}\)
(B) \(2\sqrt{13}\)
(C) 6
(D) 8
(E) 10

I thought that the shaded part was isosceles right triangle and came with Z = \(4\sqrt{6}\). Is it because "\(x= y+2\)" that this is not an isos rt. triangle?

x is 2 units more than y, so x does not equal to y, and the triangle is not isosceles.
avatar
Ndkms
Joined: 09 Aug 2016
Last visit: 26 Jul 2017
Posts: 42
Own Kudos:
Given Kudos: 8
Posts: 42
Kudos: 71
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Answer E

The easy stuff: (x*y) / 2 = 24 therefore x*y = 48

So since x = y + 2 replace x to product x*y = 48. Hence (y+2) * y = 48 (equation 1) .

IMHO this IS the TRICKY part... and I think this is the moment where you can save so much time the only number that gives you 48 when is multiplied by itself +2 is 6! because 6 (6+2) = 48 HENCE y = 6 and x = 8 ... the rest is history as you apply Pythagoras theorem and you get z^2 = sqrt(100) ..... and so z = 10 BOOM!
User avatar
diegocml
Joined: 22 Apr 2016
Last visit: 22 Jun 2018
Posts: 122
Own Kudos:
Given Kudos: 392
Status:On a 600-long battle
Location: Hungary
Concentration: Accounting, Leadership
Schools: Erasmus '19
GMAT 1: 410 Q18 V27
GMAT 2: 490 Q35 V23
Schools: Erasmus '19
GMAT 2: 490 Q35 V23
Posts: 122
Kudos: 126
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pacifist85
Hi,

This one was tricky, not because the solution was difficult, but because of my mistakes in the calculations...

I also used Paresh's approach, but instead of solving for y and getting y=x-2, I just used y and y+2. But, I couldn't see the 6*4=48 ... so I couldn't find the possible solutions of the equation easily!

In fact I did this:
Area= (b*2) / 2
24=[y(y+z)] / 2
24= (y^2 + 2y) / 2
48 = y^2 + 2y
y^2 + 2y - 48 = 0.

I didn't see these solutions: -6, 8 and ended up using the quadratic formula, which is not handy when you are not quick with calculations! Haha! So, I passed the time...

Just as a small piece of advice, in similar situations, what makes it easier to find the roots is to make the prime factorization of the constant. Then you end up with this 2*2*2*2*3, so you have fewer values to test and find one that works.

pacifist85 I share the same problem as you do. Finding the roots can be tricky at times and time consuming too. I ended up prime factorizing the constant and saw the 6*8. I started to solve with the quadratic formula but said in my mind "Screw this - I'll be here till tomorrow solving this thing!"
avatar
Shiv2016
Joined: 02 Sep 2016
Last visit: 14 Aug 2024
Posts: 516
Own Kudos:
Given Kudos: 277
Posts: 516
Kudos: 211
Kudos
Add Kudos
Bookmarks
Bookmark this Post
There are two ways to solve it:

1) Factorization

Given y*x= 24*2=48
and x=y+2

(y+2)y=48
y^2+2y-48=0
y=6
Discard y= -8 because side can never be negative

Then
x^2 +y^2= z^2
z=10





2) We know xy= 48

and x-y=2
Square both the sides

(x-y)^2= 2^2
x^2 + y^2 -2xy= 4
x^2 +y^2 =4+ 2*48

x^2 +y^2= 100

Plug this value in the expression: x^2 +y^2= z^2
z=10
 1   2   
Moderators:
Math Expert
105355 posts
Tuck School Moderator
805 posts