## GMAT Prep - Geometry (2)

Math questions from mba.com and GMAT Prep software
Guest

### GMAT Prep - Geometry (2)

rschunti

Drop perpendicular from P to the X axis and lets say the intersection point is L. Drop another perpendicular from Q to X axis and call it M.

Tan (PLO)=PL/LO=(1)/(root(3))

So angel PLO=60 Hence Angle QOM=30

So cos(30)=OM/2

Hence OM=root(3)

What is OA?
Guest

QA: B or 1

And still confused.
sheetal

From the figure, it is given that OQ = OP
=> (s)^2 + (t)^2 = 4 --- eqn (1)

Line OP & OQ are perpendiculars, so the product of their slopes = -1.

slope of OP = (0-1)/ (0 + sqrt(3)) ..eqn (2)
slope of OQ = t/s ...eqn (3)

Therefore

[ (-1)/ sqrt(3) ] * [t/s] = -1.
=> t = s * (sqrt(3)). Substituting this in eqn (1) will give s = 1 or -1

We have 1 in the ans choice. B is the answer.
tmmyc

### Re: GMAT Prep - Geometry (2)

First, see that after dropping perpendicular lines, we have two right triangles.

Detailed Explanation:
Let's begin with the triangle on the left.

We know the sides are 1 and (sqrt 3) from point P.
If you know your special right triangles, you will quickly see that this is a 30-60-90 right triangle.

The angle opposite '1' is 30 degrees.

Let's move on to the triangle on the right.

We know that a straight line has 180 degrees.

Since we know the lower angle of the triangle on the left is 30 degrees, and we also know the angle between the two line segments is 90 degrees, the lower angle of the triangle on the right must be 60 degrees in order to sum to 180 degrees. (30 + 90 + x = 180; x = 60)

This means the triangle on the right is also a 30-60-90 triangle. The hypotenuse of this triangle is the same as the other triangle's (which is '2' by the Pythagorean Theorem), since both are radii of the same circle.

Using the same properties of a 30-60-90 triangle, you can find the side lengths and finally the point (s,t) which gives the value for s.
Guest

Thanks, that definitely makes sense. After looking back at it, I was thinking there may be an easier way to approach it by getting the hypotenuse of the main 45/45/90 triangle, but for some reason it's not giving me exactly the right answer. Does this logic make sense?

1. The radius of 2 is easy enough to find with the given information, then
2. calculating the hypotenuse with two legs (radius) of 2 ---> 2*sqt2
3. subtract out the given Sqt3 --> 2*sqt2 - sqt3 = approximately 1.1 ((~2*1.41) - 1.73))

1 or "B" is the only one that is close to 1.1, but still you'd think that it would be exact.
Guest

Never mind... that would only be exactly equal if the Y data points were exactly the same height. I guess my approach would work in a rush as an estimation, or if I forget how to do it the right way. Thanks for your help.
RonPurewal
Students

Posts: 19747
Joined: Tue Aug 14, 2007 8:23 am

here's an awesome way to approach this.

see that 90 degree angle there? ok. that means that this question is really asking where the point (-rad3, 1) would go if the paper were rotated clockwise by 90 degrees.

so...

draw that point on your paper, and then physically rotate the paper by 90 degrees.

originally the x coordinate was negative rad 3 (to the left). when you rotate the paper, this is now upward, so it's positive rad 3 in the y direction.
originally the y coordinate was positive 1 (upward). when you rotate the paper, this is now to the right, so it becomes positive 1 in the x direction.

ergo, new coordinates = (1, rad3)

sweet
dslewis

Ron question goes to you: I have a method to solve this type of question, but not sure if its exactly valid.

For a question which asks what is the distance between 2 points on a coordinate plane this question is really asking you to find the hypotenuse. For example if you are given points (0,2) and (2,0) (both of these points lie on the hypotenuse) - in order to find the hypotenuse i prefer doing the rise over run method ( subtract the Y coordinates (rise) and subtract the X coordinates (run) ; Here I get 2 and 2 and using the pythag theorem the hypotenuse would be 2sqroot2.

If instead I was given only one point on the hypotenuse and the other point is the point not on the hypotenuse - ie (0,2) and (0,0). I figured out that if I subtracted the Y coordinates this value gives me the X coordinate for the other point that lies on the hypotenuse. And the Y coordinate is simply the Y coordinate for point not on the hypotenuse - giving me the third point of (2,0). Then I could figure out the hypotenuse from here.

Applying this method to the original question I subtract the Y coordinates P and O which from the second example above would give me the X value of point Q which would be 1 and our answer, but using the same method to find T the method doesnt work. Does this mean my method is wrong or that I can only use it to find the X corrdinate?
dslewis

to clarify point O lies on origin so its coordinates are (0,0)
dslewis

Correction my method would actually make T 1 so is Q(1,1)?
(the Y value should be the Y value for the point that lies on the same line - So for this question I use point P's Y value inorder to get Q's Y value - therefore using this method I get 1)

Is this correct? If not please let me know. THanks
RonPurewal
Students

Posts: 19747
Joined: Tue Aug 14, 2007 8:23 am

this method, as described, won't work. i'm not sure i understand 100%, but i'll attempt a response below.

dslewis wrote:For a question which asks what is the distance between 2 points on a coordinate plane this question is really asking you to find the hypotenuse. For example if you are given points (0,2) and (2,0) (both of these points lie on the hypotenuse) - in order to find the hypotenuse i prefer doing the rise over run method ( subtract the Y coordinates (rise) and subtract the X coordinates (run) ; Here I get 2 and 2 and using the pythag theorem the hypotenuse would be 2sqroot2.

so far so good.

dslewis wrote:If instead I was given only one point on the hypotenuse and the other point is the point not on the hypotenuse - ie (0,2) and (0,0). I figured out that if I subtracted the Y coordinates this value gives me the X coordinate for the other point that lies on the hypotenuse.

no way jose.

if you rephrase what you're saying here, it turns into this: 'if you tell me the length of one leg of a right triangle, then i automatically know the length of the other leg - and, moreover, it's the same length as the first leg.'
in other words, the logical conclusion of what you're saying here is that every right triangle drawn in the plane must be isosceles. i think we both know that isn't true.

check out the other solutions given on this page; there are lots of good insights.
GMAT Fever

RPurewal wrote:here's an awesome way to approach this.

see that 90 degree angle there? ok. that means that this question is really asking where the point (-rad3, 1) would go if the paper were rotated clockwise by 90 degrees.

so...

draw that point on your paper, and then physically rotate the paper by 90 degrees.

originally the x coordinate was negative rad 3 (to the left). when you rotate the paper, this is now upward, so it's positive rad 3 in the y direction.
originally the y coordinate was positive 1 (upward). when you rotate the paper, this is now to the right, so it becomes positive 1 in the x direction.

ergo, new coordinates = (1, rad3)

sweet

Whoa!! Nice solution!

Ron - Is there a method to your madness? Specifically, when do you advise we rotate our paper 90 degrees? Are there certain types of problems that we can implement this method or a rotation of X degrees depending on some specific problem types that you are aware of? Or are there certain elements we should look out for? I never would have thought of this approach.

If you have some specific cases or examples where we can use this method or other similar methods that would be fantastic!!
Dont want to go in the test flipping my scrap paper in every which way like a lunatic! :wink:
RonPurewal
Students

Posts: 19747
Joined: Tue Aug 14, 2007 8:23 am

GMAT Fever wrote:Whoa!! Nice solution!

thanks.

GMAT Fever wrote:Ron - Is there a method to your madness? Specifically, when do you advise we rotate our paper 90 degrees?

if the problem features a transformation involving a 90 degree rotation, then you should rotate the paper by 90 degrees to simulate that transformation. the same goes for a rotation through any other angle, though i doubt the test would ask you to rotate through any angle other than the orthogonal angles (90, 180, 270).

in general, you can and should use physical movements of your paper to simulate any transformation called for in a problem. the one exception to this rule is reflections; you can't simply flip the yellow plastic pad around, as it is completely opaque and the marker doesn't show through to the other side.

another nice feature of the yellow pad is that it features a graph-paper grid, which makes translations (i.e., shifting graphs up/down/left/right in their entirety) much easier.
Guest660

Confusion regarding this problem...

we know that each triangle is 30 60 90 ....

also that radius is 2... so if we draw a line joining both the points...we form a triangle... with sides equal(radius)...

each angle is 45 ..

now - is this correct.... that means my total sum of angels is not 90 for the angel p.. between a line joining pint p to q and a perpendicular to x axis...

am i missing something ??