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:

The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

18 Feb 2012, 17:44

2

This post received KUDOS

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

63% (02:49) correct
37% (02:18) wrong based on 359 sessions

HideShow timer Statistics

Attachment:

Triangle.PNG [ 2.68 KiB | Viewed 7751 times ]

The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?

Length of PQ = 13 i.e. this is 5:12:13 Right angle triangle.

We are told that this length (13) is equal to the height of the equilateral triangle XYZ. An equilateral triangle can be cut into two 30-60-90 triangles, where the height of the equilateral triangle is equal to the long leg of each 30-60-90 triangle. We know that the height of XYZ is 13 so the long leg of each 30-60-90 triangle is equal to 13. So the sides of the 30:60:90 triangle are x:x\(\sqrt{3}\):2x.

Now here I am stuck. What will be the three sides of a triangle?

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

18 Feb 2012, 18:08

1

This post received KUDOS

Expert's post

7

This post was BOOKMARKED

The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ? A. 169/√3 B. 84.5 C. 75√3 D. 169√3 /4 E. 225√3 /4

Attachment:

Triangle.PNG [ 2.68 KiB | Viewed 7725 times ]

The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

So the distance between P and Q is \(d=\sqrt{(-2+7)^2+(9+3)^2}=13\). (Else you can find the length of PQ by realizing that PQ is a hypotenuse of 5:12:13 right triangle)

So we know that the height in equilateral triangle XYZ equals to 13: \(height=13\)

Now, since the height of the equilateral triangle divides it into two 30-60-90 right triangles with the ratio of the sides \(1:\sqrt{3}:2\) then height becomes the leg opposite 60 degrees angle and the hypotenuse, which is the side of an equilateral triangle can be found from the ratio: \(\frac{height}{side}=\frac{\sqrt{3}}{2}\) --> \(side=\frac{26}{\sqrt{3}}\).

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

18 Feb 2012, 18:19

This is where I always get confused:

Height becomes the leg opposite 60 degrees angle and the hypotenuse, which is the side of an equilateral triangle can be found from the ratio: \(\frac{height}{side}=\frac{\sqrt{3}}{2}\) --> \(side=\frac{26}{\sqrt{3}}\).

How come we decide that it will be the leg opposite to 60 degrees angle?

I am fine with the rest, but struggles in the above concept.

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

18 Feb 2012, 18:26

1

This post received KUDOS

Expert's post

enigma123 wrote:

This is where I always get confused:

Height becomes the leg opposite 60 degrees angle and the hypotenuse, which is the side of an equilateral triangle can be found from the ratio: \(\frac{height}{side}=\frac{\sqrt{3}}{2}\) --> \(side=\frac{26}{\sqrt{3}}\).

How come we decide that it will be the leg opposite to 60 degrees angle?

I am fine with the rest, but struggles in the above concept.

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

22 Feb 2012, 10:20

Bunuel i have small query since we have given coordinates of points P and Q but where its mentioned that these points are going to form the equlateral triangle , that triangle can be some other triangle also, whose height is the distance between points P and Q. and if we use two formulaes one Height for the equt triangle = rt3/side from there we can get side and from side we can find area of the equlateral triangle rt3/4 side sqr am i right , need ur feedback thanks

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

22 Feb 2012, 10:29

Expert's post

pbull78 wrote:

Bunuel i have small query since we have given coordinates of points P and Q but where its mentioned that these points are going to form the equlateral triangle , that triangle can be some other triangle also, whose height is the distance between points P and Q. and if we use two formulaes one Height for the equt triangle = rt3/side from there we can get side and from side we can find area of the equlateral triangle rt3/4 side sqr am i right , need ur feedback thanks

The red part: we are not told that. We are given two points P and Q and are told that the height of some equilateral triangle XYZ is the same as the length of line segment PQ. Next, we calculated the length of PQ, which tuned out to be 13 and found the area of an equilateral triangle which has the height of 13.

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

30 Sep 2012, 05:59

wondering the level of this problem

this prob wont take much time if you know the formulae for distance between 2 points and the one for height of a equilateral triangle _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

24 Mar 2013, 13:39

Bunuel wrote:

The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ? A. 169/√3 B. 84.5 C. 75√3 D. 169√3 /4 E. 225√3 /4

Attachment:

Triangle.PNG

The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).

So the distance between P and Q is \(d=\sqrt{(-2+7)^2+(9+3)^2}=13\). (Else you can find the length of PQ by realizing that PQ is a hypotenuse of 5:12:13 right triangle)

So we know that the height in equilateral triangle XYZ equals to 13: \(height=13\)

Slightly different approach that boils down to the same after finding the side of PQ = 13:

The height of an equilateral triangle equals: S\(\sqrt{3}\)/2, in which S represents the length of the side of the triangle. S is found when the height is multiplied by 2 and divided by \(\sqrt{3}\): 13*2 / \(\sqrt{3}\).

The area of a triangle is found with (height * side)/2, hence: (13* 26\(\sqrt{3}\) / 2) so: 338\(\sqrt{3}\)/2 which is 169/\(\sqrt{3}\).

Hope this helps for the people that got stuck after finding the height.

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

21 Oct 2013, 10:34

AccipiterQ wrote:

The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?

The formula for the distance between two points (x1, y1) and (x2, y2) is: .

One way to understand this formula is to understand that the distance between any two points on the coordinate plane is equal to the hypotenuse of a right triangle whose legs are the difference of the x-values and the difference of the y-values (see figure). The difference of the x-values of P and Q is 5 and the difference of the y-values is 12. The hypotenuse must be 13 because these leg values are part of the known right triangle triple: 5, 12, 13.

We are told that this length (13) is equal to the height of the equilateral triangle XYZ. An equilateral triangle can be cut into two 30-60-90 triangles, where the height of the equilateral triangle is equal to the long leg of each 30-60-90 triangle. We know that the height of XYZ is 13 so the long leg of each 30-60-90 triangle is equal to 13. Using the ratio of the sides of a 30-60-90 triangle (1:\(\sqrt{3}\): 2), we can determine that the length of the short leg of each 30-60-90 triangle is equal to 13/. The short leg of each 30-60-90 triangle is equal to half of the base of equilateral triangle XYZ. Thus the base of XYZ = 2(13/) = 26/.

The question asks for the area of XYZ, which is equal to 1/2 × base × height:

The correct answer is A.

For the LIFE of me I don't get why the OE is correct, and why my method is wrong, I'll post my method in the first reply

OK, so the OE states to figure out what PQ equals, which is very easy, just pythagoras it, and you get 13. It then says that this number is equal to the height of equilateral triangle XYZ. So my logic is this:

we know that the height of an equilateral triangle = \(\sqrt{3}/2\)*side. So since we already know the height, we can solve for any side:

13= \(\sqrt{3}/2\)*side

26=\(\sqrt{3}\)*side

\(26/\sqrt{3}\)=side

Great, now we have a side.

Area of an equilateral triangle? \((side^2*\sqrt{3})/4\)

so we have \((26/\sqrt{3}^2)*\sqrt{3}\))/4

that ends up with \(((676/3)*\sqrt{3}\))/4 which is obviously not the right answer. How on Earth does this not work??

edit: edited for formatting

Last edited by AccipiterQ on 21 Oct 2013, 10:49, edited 2 times in total.

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

21 Oct 2013, 10:42

Expert's post

AccipiterQ wrote:

AccipiterQ wrote:

The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?

The formula for the distance between two points (x1, y1) and (x2, y2) is: .

One way to understand this formula is to understand that the distance between any two points on the coordinate plane is equal to the hypotenuse of a right triangle whose legs are the difference of the x-values and the difference of the y-values (see figure). The difference of the x-values of P and Q is 5 and the difference of the y-values is 12. The hypotenuse must be 13 because these leg values are part of the known right triangle triple: 5, 12, 13.

We are told that this length (13) is equal to the height of the equilateral triangle XYZ. An equilateral triangle can be cut into two 30-60-90 triangles, where the height of the equilateral triangle is equal to the long leg of each 30-60-90 triangle. We know that the height of XYZ is 13 so the long leg of each 30-60-90 triangle is equal to 13. Using the ratio of the sides of a 30-60-90 triangle (1:\(\sqrt{3}\): 2), we can determine that the length of the short leg of each 30-60-90 triangle is equal to 13/. The short leg of each 30-60-90 triangle is equal to half of the base of equilateral triangle XYZ. Thus the base of XYZ = 2(13/) = 26/.

The question asks for the area of XYZ, which is equal to 1/2 × base × height:

The correct answer is A.

For the LIFE of me I don't get why the OE is correct, and why my method is wrong, I'll post my method in the first reply

OK, so the OE states to figure out what PQ equals, which is very easy, just pythagoras it, and you get 13. It then says that this number is equal to the height of equilateral triangle XYZ. So my logic is this:

we know that the height of an equilateral triangle = \(\sqrt{3}/2\)*side. So since we already know the height, we can solve for any side:

13= \(\sqrt{3}/2\)*side

26=\(\sqrt{3}\)*side

\(26/\sqrt{3}\)=side

Great, now we have a side.

Area of an equilateral triangle? \((side^2*\sqrt{3})/4\)

so we have \((26/\sqrt{3}^2)*\sqrt{3}\))/4

that ends up with \(676/3*\sqrt{3}\)/4 which is obviously not the right answer. How on Earth does this not work??

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

21 Oct 2013, 12:27

The trouble there is that you squared 26 and so end up in a foggy place where only calculators can safely travel. If you had not multiplied, you'd have gotten an answer with 2*2 in the numerator, and this would have been an easy cancelation with the 4 in the denominator. After all 676/4 is 169.

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

21 Oct 2013, 12:55

Gmatdojo wrote:

The trouble there is that you squared 26 and so end up in a foggy place where only calculators can safely travel. If you had not multiplied, you'd have gotten an answer with 2*2 in the numerator, and this would have been an easy cancelation with the 4 in the denominator. After all 676/4 is 169.

yeah I see that now, I actually just gave up after i squared 26, because I figured I was too far off track. Looks like I had it right after all haha

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

22 Oct 2013, 01:28

enigma123 wrote:

The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?

A. 169/√3 B. 84.5 C. 75√3 D. 169√3 /4 E. 225√3 /4

Attachment:

Triangle.PNG

This is is how I am trying to solve this:

Length of PQ = 13 i.e. this is 5:12:13 Right angle triangle.

We are told that this length (13) is equal to the height of the equilateral triangle XYZ. An equilateral triangle can be cut into two 30-60-90 triangles, where the height of the equilateral triangle is equal to the long leg of each 30-60-90 triangle. We know that the height of XYZ is 13 so the long leg of each 30-60-90 triangle is equal to 13. So the sides of the 30:60:90 triangle are x:x\(\sqrt{3}\):2x.

Now here I am stuck. What will be the three sides of a triangle?

I would give a simple solution which Bunuel can correct me if I am wrong....PQ=13 which is from distance formulae and its given that it is equals to height of any equilateral triangle. So my logic is as per 30-60-90 rule , altitute is rt3timesX (X is side opposite to 30degree). So now here rt3X=13. So X=13/rt3.. And area of triangle = 1/2*base(13/rt3)*13 =169/rt3.. Bunuel correct me if I am wrong please.

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

29 Oct 2014, 08:39

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: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

15 Nov 2014, 04:16

Hi Bunel,

I graphically find that the PQ=13, and the rest two sides are 5 and 12. Then how can this be an equilateral triangle. Shouldn't the sides be of same length then?

What am I missing???

TIA [quote="enigma123"]

Attachment:

Triangle.PNG

The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?

A. 169/√3 B. 84.5 C. 75√3 D. 169√3 /4 E. 225√3 /4[spoiler=]

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

15 Nov 2014, 05:12

Expert's post

anki2762 wrote:

Hi Bunel,

I graphically find that the PQ=13, and the rest two sides are 5 and 12. Then how can this be an equilateral triangle. Shouldn't the sides be of same length then?

What am I missing???

TIA

enigma123 wrote:

Attachment:

Triangle.PNG

The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?

A. 169/√3 B. 84.5 C. 75√3 D. 169√3 /4 E. 225√3 /4[spoiler=]

Attachment:

Triangle.PNG

You are not reading the question and the solution carefully.

XYZ is an equilateral triangle, not PQR. _________________

Re: The (x, y) coordinates of points P and Q are (-2, 9) and [#permalink]

Show Tags

05 Dec 2015, 10:47

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. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...