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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Helpful Geometry formula sheet: http://gmatclub.com/forum/best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Last edited by Bunuel on 21 Jul 2014, 11:25, edited 1 time in total.

In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None B. Q Only C. Q and R D. P, Q and R only E. P, Q, R and S

The area of a rectangle equals to \(area=width*length\);

The area of a triangle equals to \(area=\frac{1}{2}*base*height\) (also the area of a square=diagonal^2/2);

Now it's easy to calculate the areas of given figure: \(area_P=\frac{1}{2}*4*1=2\); \(area_Q=1*2=2\); \(area_R=\frac{2^2}{2}=2\); \(area_S=\frac{1}{2}*3*1=1.5\).

Re: In the rectangular coordinate system above, for which of the [#permalink]

Show Tags

20 Oct 2011, 22:26

1

This post received KUDOS

1

This post was BOOKMARKED

Area of a triangle = 1/2 * Base * Height, from the graph we can find out length of base and height of the triangles and rectangles. P: Base = 4 Height = 1 Area = 1/2 (4)(1) = 2 Q: Base = 1 Height = 2 Area = Base * Height = 1 * 2 = 2 R: For a rhombus area can be calculated using the value of the diagonals ( D1 * D2 ) / 2 i:e (2*2)/2 = 2 S: Base = 3 Height = 1 Area = 1/2 (3)(1) = 1.5

Re: In the rectangular coordinate system above, for which of the [#permalink]

Show Tags

04 May 2014, 03:05

Bunuel wrote:

In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None B. Q Only C. Q and R D. P, Q and R E. P, Q, R and S

The area of a rectangle equals to \(area=width*length\);

The area of a triangle equals to \(area=\frac{1}{2}*base*height\) (also the area of a square=diagonal^2/2);

Now it's easy to calculate the areas of given figure: \(area_P=\frac{1}{2}*4*1=2\); \(area_Q=1*2=2\); \(area_R=\frac{2^2}{2}=2\); \(area_S=\frac{1}{2}*3*1=1.5\).

Answer: D.

Hey Bunuel, How did you find the area of R? Did you divide it into two triangles? And just for curiosity is that figure a square?

I got the right answer as I figured out the area of other three and the only option suitable was D

In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None B. Q Only C. Q and R D. P, Q and R E. P, Q, R and S

The area of a rectangle equals to \(area=width*length\);

The area of a triangle equals to \(area=\frac{1}{2}*base*height\) (also the area of a square=diagonal^2/2);

Now it's easy to calculate the areas of given figure: \(area_P=\frac{1}{2}*4*1=2\); \(area_Q=1*2=2\); \(area_R=\frac{2^2}{2}=2\); \(area_S=\frac{1}{2}*3*1=1.5\).

Answer: D.

Hey Bunuel, How did you find the area of R? Did you divide it into two triangles? And just for curiosity is that figure a square?

I got the right answer as I figured out the area of other three and the only option suitable was D

Yes, it's a square because its diagonals are equal and perpendicular bisectors of each other. The area of a square=diagonal^2/2.

Does this make sense?

P.S. \(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).
_________________

Re: In the rectangular coordinate system above, for which of the [#permalink]

Show Tags

21 Jul 2014, 10:54

Bunuel wrote:

b2bt wrote:

Bunuel wrote:

In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None B. Q Only C. Q and R D. P, Q and R E. P, Q, R and S

The area of a rectangle equals to \(area=width*length\);

The area of a triangle equals to \(area=\frac{1}{2}*base*height\) (also the area of a square=diagonal^2/2);

Now it's easy to calculate the areas of given figure: \(area_P=\frac{1}{2}*4*1=2\); \(area_Q=1*2=2\); \(area_R=\frac{2^2}{2}=2\); \(area_S=\frac{1}{2}*3*1=1.5\).

Answer: D.

Hey Bunuel, How did you find the area of R? Did you divide it into two triangles? And just for curiosity is that figure a square?

I got the right answer as I figured out the area of other three and the only option suitable was D

Yes, it's a square because its diagonals are equal and perpendicular bisectors of each other. The area of a square=diagonal^2/2.

Does this make sense?

P.S. \(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

Bunuel for calculating area of R : how did you get 2^2 /2 , where is 2 coming from? Can you explain the counting of boxes?

In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None B. Q Only C. Q and R D. P, Q and R E. P, Q, R and S

The area of a rectangle equals to \(area=width*length\);

The area of a triangle equals to \(area=\frac{1}{2}*base*height\) (also the area of a square=diagonal^2/2);

Now it's easy to calculate the areas of given figure: \(area_P=\frac{1}{2}*4*1=2\); \(area_Q=1*2=2\); \(area_R=\frac{2^2}{2}=2\); \(area_S=\frac{1}{2}*3*1=1.5\).

Answer: D.

Yes, it's a square because its diagonals are equal and perpendicular bisectors of each other. The area of a square=diagonal^2/2.

Does this make sense?

P.S. \(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

Bunuel for calculating area of R : how did you get 2^2 /2 , where is 2 coming from? Can you explain the counting of boxes?

same question for S and Q

R is a square --> the area of a square=diagonal^2/2 --> diagonal of R = 2 --> area = 2^2/2.

Q is a rectangle --> the area = length*width = 1*2.

S is a triangle --> the area = 1/2*base*height = 1/2*3*1 (consider vertical side as base).

Re: In the rectangular coordinate system above, for which of the [#permalink]

Show Tags

21 Jul 2014, 13:07

Great explanation, but how can I see that diagonal of R is 2? (like can you possibly label this on the diagram or give coordinates) also, I don't see how the width is 2 for the rectangle.

Great explanation, but how can I see that diagonal of R is 2? (like can you possibly label this on the diagram or give coordinates) also, I don't see how the width is 2 for the rectangle.

Sorry for the bother.

Ask yourself how many units are there in the diagonal of R, in the width of the rectangle, ...

Sorry, I don't know how to explain it better.
_________________

Re: In the rectangular coordinate system above, for which of the [#permalink]

Show Tags

11 Aug 2015, 05:34

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: In the rectangular coordinate system above, for which of the [#permalink]

Show Tags

10 Oct 2016, 08:32

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...