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

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18 Feb 2012, 16:44

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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?

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 8863 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]

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18 Feb 2012, 17:19

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.

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]

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22 Feb 2012, 09:20

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

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]

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30 Sep 2012, 04:59

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]

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24 Mar 2013, 12:39

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]

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21 Oct 2013, 09:34

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, 09:49, edited 2 times in total.

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]

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21 Oct 2013, 11:27

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]

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21 Oct 2013, 11:55

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]

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22 Oct 2013, 00:28

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]

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29 Oct 2014, 07:39

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

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15 Nov 2014, 03:16

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=]

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

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05 Dec 2015, 09:47

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).

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