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805+ (Hard)|   Geometry|      
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pitchern
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Right triangle ABO is drawn in the xy-plane, with OB as hypo 20 May 2020, 19:21
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I'm not sure if people still answer these questions on these old posts, but can I just clear one thing up.

When I read statement 1 the only useful information on get is that the values are integers.
Whether they are positive or negative is irrelevant for the area and the triangle can only possibility be in quadrant 1 anyway based on the angle ab being 90 degrees.

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Jezza
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Right triangle ABO is drawn in the xy-plane, with OB as hypo 11 Jun 2022, 01:25
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Certainly one of the most challenging one, took me 5 mins. I don't expect official questions to be this hard though, this is probably just on the outer limit of GMAT math.
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Engineer1
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Right triangle ABO is drawn in the xy-plane, with OB as hypo 28 Nov 2023, 11:57
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KarishmaB
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Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

Got it off Manhattan GMAT blog(Challenge of the week)

First off, draw the figure with whatever is given to you.
Attachment:
Ques3.jpg
Vertex A has the right angle and could be at many places - the triangle could be 45-45-90 in which case it will be in the middle of O and B, it could be 30-60-90 triangle or many other variations. The area of all these triangles will be different.

(1) The x- and y-coordinates of all three points are non-negative integers.

The co-ordinates are non negative integers so A lies in the first quadrant. Also, if the co-ordinates of A are (a, b), the two sides will be given by:
\(OA^2 = a^2 + b^2\)
\(AB^2 = (15 - a)^2 + b^2\)
(a and b are positive integers)

Since OAB is a right triangle at A, \(OA^2 + AB^2 = OB^2\)
\(a^2 + b^2 + (15 - a)^2 + b^2 = 15^2\)
\(b^2 = (15 - a)*a\)
Now, we can quickly find out the values of a and b by putting in values of a from 1 to 7. Only a = 3 (or 12) gives us a perfect square on the right hand side.
So we will get 2 triangles a = 3, b = 6 or a = 12, b = 6. But note that both triangles will have the same area given by (1/2)*15*6 = 45 (1/2*OB*Altitude)
Hence statement 1 alone is sufficient.

(2) No two sides of the triangle have the same length.
This tells us that the triangle is not 45-45-90. Still the triangle can have many other angle measures such as 30-60-90 or 40-50-90 etc. The area will be different in different cases.

Answer (A)
Show more

KarishmaB

Not sure why we have to do so much calculation. Wouldn't it follow Pythagorean theorem? If the hypotenuse is 15, the other sides can be found out by applying either 3:4:5 or 5:12:13. 3:4:5 would fit in. So the other two sides will be 12, 9 or 9, 12. Either way, the area will remain the same. Hence, A and B, each one is sufficient. Moreover, how does it matter the coordinates are positive or negative? The length of sides is what we are concerned about.

Please help explain, thank you.

Plus, I feel like this question is testing "mathematics" skills more than "business" decision skills. :/
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Right triangle ABO is drawn in the xy-plane, with OB as hypo 28 Nov 2023, 23:24
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Engineer1
KarishmaB
mau5
Right triangle ABO is drawn in the xy-plane, with OB as hypotenuse, where O is at the origin and B at (15, 0). What is the area of the triangle?

(1) The x- and y-coordinates of all three points are non-negative integers.

(2) No two sides of the triangle have the same length.

Got it off Manhattan GMAT blog(Challenge of the week)

First off, draw the figure with whatever is given to you.
Attachment:
Ques3.jpg
Vertex A has the right angle and could be at many places - the triangle could be 45-45-90 in which case it will be in the middle of O and B, it could be 30-60-90 triangle or many other variations. The area of all these triangles will be different.

(1) The x- and y-coordinates of all three points are non-negative integers.

The co-ordinates are non negative integers so A lies in the first quadrant. Also, if the co-ordinates of A are (a, b), the two sides will be given by:
\(OA^2 = a^2 + b^2\)
\(AB^2 = (15 - a)^2 + b^2\)
(a and b are positive integers)

Since OAB is a right triangle at A, \(OA^2 + AB^2 = OB^2\)
\(a^2 + b^2 + (15 - a)^2 + b^2 = 15^2\)
\(b^2 = (15 - a)*a\)
Now, we can quickly find out the values of a and b by putting in values of a from 1 to 7. Only a = 3 (or 12) gives us a perfect square on the right hand side.
So we will get 2 triangles a = 3, b = 6 or a = 12, b = 6. But note that both triangles will have the same area given by (1/2)*15*6 = 45 (1/2*OB*Altitude)
Hence statement 1 alone is sufficient.

(2) No two sides of the triangle have the same length.
This tells us that the triangle is not 45-45-90. Still the triangle can have many other angle measures such as 30-60-90 or 40-50-90 etc. The area will be different in different cases.

Answer (A)

KarishmaB

Not sure why we have to do so much calculation. Wouldn't it follow Pythagorean theorem? If the hypotenuse is 15, the other sides can be found out by applying either 3:4:5 or 5:12:13. 3:4:5 would fit in. So the other two sides will be 12, 9 or 9, 12. Either way, the area will remain the same. Hence, A and B, each one is sufficient. Moreover, how does it matter the coordinates are positive or negative? The length of sides is what we are concerned about.

Please help explain, thank you.

Plus, I feel like this question is testing "mathematics" skills more than "business" decision skills. :/
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Note that given the co-ordinate are integers doesn't necessarily mean that the sides will be integers too and hence only pythagorean triplets may not be in the play.
A right triangle with coordinates at (0,0), (3, 0) and (0, 2) doesn't give a pythagorean triplet.

Strong mathematical skills help you develop reasoning and business skills though the application of Geometry for managers is questionable. It is more relevant for technical roles. Perhaps that is why the switch away from much of Geometry.
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