mau5 wrote:
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)