Geometry question
Help in solving the following two questions Q1. Ten points are marked on a straight line and eleven points are marked on another straight line. How many triangles can be constructed with vertices from among the above points? 1. 495 2. 550 3. 1045 4. 2475 Q2. Ten points are marked on a straight line and eleven points are marked on another straight line which is parallel to the first line. How many rectangles can be constructed with vertices from among the above points?
for the first question i am
for the first question i am getting 1045.
the general formula for m points on one line and n points on other line
is m*nC2 + n*mC2 = 1045 in this case.
I suppose
for first question solution :
out of two parlell line we can take two points on one side and one on other side. But this can be done in two ways
Choosing two points from side from 10 points or choosing two points from 11 points so number of triangles may be (10C2 * 11C1) + (11C2 * 10C1) = 1045
for second question similarly:
for forming an rectangle when you pick two points other two points are fixed so 10C2 = 45 rectangles can be formed.(As only 10 parllel points are available)
correct me if iam wrong
Goofey
There is no way you can say
There is no way you can say that because we don’t know where the points are on the lines - to be rectangular the angle must be 90 degree
The answer I think should have the option - can’t be determined
what anita's saying dis
what anita's saying dis cross my mind. but i am hoping that the problem assumes that all points are such that they can form a rectangle, but then that could be taking it too far.
Though "cant be determined"
Though "cant be determined" should be the answer to the above question we can rephrase it as
Ten points are marked on a straight line of 10 units starting from one end and eleven points are marked on another straight line of 11 units with two points at two ends. The distance between two consecutive points on each line is same. If the two lines start from same y co-ordinate and are parallel to each other then how many rectangles can be constructed with vertices from among the above points?
And the answer is 9 + 8 + 7. . . . = 45
But then it does not matter what is the length of the second line what ever is the length number of rectangles‘ll be always same.
Only 1 condition is an exception else its fine
The answer will be 1045 clearly...the onlt point to be mentioned is that the lines dont intersect each other at a common point.
Sushil Singh
Software Engineer
eForce Global Inc.
KOlkata,India
solution
answer to the first question is 1045 (c),,i.e
21c3(total no of ways in which 3 points out of 21 (11+10) points can be selected)-11c3(no of ways in which selection of ALL 3 points on one line has been made)-10c3(no of ways in which selection of ALL 3 points on the other line has been made)
For the second question, rectangles can only be formed if some of the points on either line, lie on the common perpendicular(number of such points need to be specified). Secondly,location of points on the two lines are important in this question.