Manhattan GMAT CAT Test maths question
A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?
4
6
8
10
12
Can you please provide some explanation for this questions.
The explanation given in the online exam is not very intuitive.
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- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
This is a tough one - and not a common question type for the test, so don't stress too much about it.
First, draw a coordinate plane. Place a dot on the origin. If it's a square and has to have an area of 100, the sides have to be 10.
The first, most obvious squares to draw are along the x and y axes - I can draw four squares, one for each quadrant. So that's at least 4. Look at your answers. 4 is on there and all of the other numbers are higher, so we can't eliminate anything yet. Keep going.
How else can I draw lines from the origin with a length 10 where the coordinates must be integers? This is a Pythagorean problem in disguise: a line radiating out from the origin that is NOT along the axes will be the hypotenuse of a triangle created by the x and y values of the point.
Illustration: put a point anywhere, as long as it's not on one of the axes. Draw a line from the point straight down to the x axis and over to the origin. Label the two legs of the resulting triangle - see how they're just the x and y values of the point? And then the length of that line to the origin is the hypotenuse of the triangle. Since it's a right triangle, I can use a^2 + b^2 = c^2 to find the length of the hypotenuse.
In this case, we want the length of the hypotenuse to equal 10 and we want the two legs to be integers (since we're told the coordinate points must use only integers). What integers would work?
Here's where you need to have your common right triangles memorized. The most common is 3-4-5 and the second most common is 6-8-10. That last is the useful one here. The two legs should be 6 and 8. Does the point (6,8) work? Yes. What about the point (8,6)? Yes again. Those are both in quadrant 1 - draw a clean coordinate plane and put both of those on. Add the points (10,0) and (0,10) - those work too, as we discussed above.
Take a look at what you have and look for a pattern. Basically, there will be one point on each axis that will work, for a total of 4 "starting points" (think of the starting point as the first side you draw - each one results in a new square). We also have two points in the 1st quadrant. I should expect equivalent points in each of the other three quadrants (eg, the pairs -6,8 and -8,6 in quadrant 2). That's two pairs per quadrant times four quadrants = 8 "starting points." I now have a total of 12. Oh - and that's the largest number in my answer choices, so I don't need to keep looking. I'm done! (Don't forget to check your answers periodically - they can really help to keep your time down!)
Hope this makes it a bit easier!
First, draw a coordinate plane. Place a dot on the origin. If it's a square and has to have an area of 100, the sides have to be 10.
The first, most obvious squares to draw are along the x and y axes - I can draw four squares, one for each quadrant. So that's at least 4. Look at your answers. 4 is on there and all of the other numbers are higher, so we can't eliminate anything yet. Keep going.
How else can I draw lines from the origin with a length 10 where the coordinates must be integers? This is a Pythagorean problem in disguise: a line radiating out from the origin that is NOT along the axes will be the hypotenuse of a triangle created by the x and y values of the point.
Illustration: put a point anywhere, as long as it's not on one of the axes. Draw a line from the point straight down to the x axis and over to the origin. Label the two legs of the resulting triangle - see how they're just the x and y values of the point? And then the length of that line to the origin is the hypotenuse of the triangle. Since it's a right triangle, I can use a^2 + b^2 = c^2 to find the length of the hypotenuse.
In this case, we want the length of the hypotenuse to equal 10 and we want the two legs to be integers (since we're told the coordinate points must use only integers). What integers would work?
Here's where you need to have your common right triangles memorized. The most common is 3-4-5 and the second most common is 6-8-10. That last is the useful one here. The two legs should be 6 and 8. Does the point (6,8) work? Yes. What about the point (8,6)? Yes again. Those are both in quadrant 1 - draw a clean coordinate plane and put both of those on. Add the points (10,0) and (0,10) - those work too, as we discussed above.
Take a look at what you have and look for a pattern. Basically, there will be one point on each axis that will work, for a total of 4 "starting points" (think of the starting point as the first side you draw - each one results in a new square). We also have two points in the 1st quadrant. I should expect equivalent points in each of the other three quadrants (eg, the pairs -6,8 and -8,6 in quadrant 2). That's two pairs per quadrant times four quadrants = 8 "starting points." I now have a total of 12. Oh - and that's the largest number in my answer choices, so I don't need to keep looking. I'm done! (Don't forget to check your answers periodically - they can really help to keep your time down!)
Hope this makes it a bit easier!
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
Instructor
Director, Content & Curriculum
ManhattanPrep
2 posts Page 1 of 1