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Coordinate Geometry

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by kuiper » Fri Dec 03, 2010 5:06 pm
https://www.postimage.org/image.php?v=PqPnlgJ

Question on this image along with answer choices. I actually answered [spoiler]C - 10sqrt(2)[/spoiler]


I am using the approach - Area of Triangle - 1/2 * b * h

The base of the triangle can be figured out as 5sqrt(2) given the two coordinates of the base and using the distance formula.

The height of the triangle is - 4? (Drawn from point R to x-axis)

And therefore - [spoiler]10sqrt(2) [/spoiler]- which is incorrect

Seems like I might be miscalculating the height but I am not sure why that is incorrect?

I appreciate any help.
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by Night reader » Fri Dec 03, 2010 5:45 pm
kuiper wrote:https://www.postimage.org/image.php?v=PqPnlgJ

Question on this image along with answer choices. I actually answered [spoiler]C - 10sqrt(2)[/spoiler]


I am using the approach - Area of Triangle - 1/2 * b * h

The base of the triangle can be figured out as 5sqrt(2) given the two coordinates of the base and using the distance formula.

The height of the triangle is - 4? (Drawn from point R to x-axis)

And therefore - [spoiler]10sqrt(2) [/spoiler]- which is incorrect

Seems like I might be miscalculating the height but I am not sure why that is incorrect?

I appreciate any help.
Unfortunately your answer is not correct; even if the figures are not drawn to the scale, the answer is not correct.

Look you don't need tedious calculations here besides finding the area of three triangles and one quadrilateral.

Area triangle EFB = (4*3)/2 = 6
Area triangle DFC = (3*4)/2 = 6
Area triangle ADE = (7*1)/2 = 3.5

Area qdrl. ABCD = 7*4 = 28

Area of triangle PQR = Area ABCD - Areas (EFB, DFC, ADE) or 28-6-6-3.5 = 12.5

Answer A.
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by kuiper » Fri Dec 03, 2010 5:53 pm
No, you are absolutely correct. The answer in the sample image was not correct either.

All I am wondering is - why is the height of the triangle in the image not 4? - (per my post above)
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by Night reader » Fri Dec 03, 2010 5:56 pm
kuiper wrote:No, you are absolutely correct. The answer in the sample image was not correct either.

All I am wondering is - why is the height of the triangle in the image not 4? - (per my post above)
Why you are supplying us with the incorrect answer selected on the GMAT Prep then?!

you mentioned different answer of yours...

please quit your question, for you were demonstrated the easiest way how to solve this problem
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by Night reader » Fri Dec 03, 2010 6:04 pm
kuiper wrote:https://www.postimage.org/image.php?v=PqPnlgJ

Question on this image along with answer choices. I actually answered [spoiler]C - 10sqrt(2)[/spoiler]


I am using the approach - Area of Triangle - 1/2 * b * h

The base of the triangle can be figured out as 5sqrt(2) given the two coordinates of the base and using the distance formula.

The height of the triangle is - 4? (Drawn from point R to x-axis)

And therefore - [spoiler]10sqrt(2) [/spoiler]- which is incorrect

Seems like I might be miscalculating the height but I am not sure why that is incorrect?

I appreciate any help.
I Don't understand 5sqrt(2) ?

we may not assume the PQR is a right triangle, also we are not given any angle at the base (but we can calculate these).

sqrt(2) coefficient is usual per the right triangles of which the non-hypotenuse sides are equal; this speeds up calculations.
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by beat_gmat_09 » Fri Dec 03, 2010 7:44 pm
OPQ is right angled at origin with sides 3,4 hypotenuse PQ=5 from pythagorean triplets.
PR is a hypotenuse formed if you drop a perpendicular from R to x-axis
This forms another pythagorean triplet, with PR=5
Similarly RQ is one side of triangle formed when perpendicular from R is dropped on Q (y-axis)
RQ=7 and in this triangle another length is 1
Hypotenuse RQ^2 = 7^2 + 1^2 = 50
RQ = Sqrt(50)
PQ= 5, RP=5 from RQ it can be found out that RPQ is right angled at P.
Area of PQR = 1/2(5)(5) = 12.5
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by Night reader » Fri Dec 03, 2010 7:58 pm
beat_gmat_09 wrote: RQ = Sqrt(50)
PQ= 5, RP=5 from RQ it can be found out that RPQ is right angled at P.
the coordinates of the point R could be skewed to new coordinates (4<x<7; 0<y<4) this will invalidate your solution;
that 5^2 and 5^2 should make Sqrt(50) coincidence as long as your assumption of a right triangle - the values could not make up hypotenuse...
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by beat_gmat_09 » Fri Dec 03, 2010 8:10 pm
Image

x+y=90 in triangle OPE 3-4-5 triplet
x+y=90 in triangle PCR 3-4-5 triplet
Angle F = 180 - (x+y) = 180 - 90 = 90
Triangle PQR is right angled at P.
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by Night reader » Fri Dec 03, 2010 8:25 pm
You may have skewed R, different sides (not 3 and 4) x and x are not equal, y and y are not equal, in your example what makes them relative is 3 and 4 in two triangles {3;4}, {4;3}

this solution is unique and works only for this problem...
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by beat_gmat_09 » Fri Dec 03, 2010 8:32 pm
If you cut triangle RPC rotate it 90 degrees clockwise and place it on triangle POQ you'll
see the triangles are exactly same, their sides and their angles (common angle 90 degree and angles x and y).
Do you still disagree ?
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by Night reader » Fri Dec 03, 2010 8:36 pm
beat_gmat_09 wrote:If you cut triangle RPC rotate it 90 degrees clockwise and place it on triangle POQ you'll
see the triangles are exactly same, their sides and their angles (common angle 90 degree and angles x and y).
Do you still disagree ?
I don't disagree, your solution is unique to this problem with the given set of conditions. If an aim is to solve the problem - you nailed it; while this solution would not be useful if R is skewed.

By principle, the first solution is more handy; not because it's mine - but for its universality
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by beat_gmat_09 » Fri Dec 03, 2010 8:41 pm
Either way.
Whichever works in 2 minutes ;)
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