Geometric Question........?

Question

Note tha an line (unlike a line segment) is considered to extend indefinitely in each direction.)

twelve points are aligned in two rows of 6, as shown.(let's say every "o" is a point)
o o o o o o
o o o o o o

a) the # of lines tha can be formed.
b) the # of triangles that can be formed.
c) the probability that three points chosen randomly will form a triangle.
d) the# of quadrilaterals that can be formed
e) the prabability that 4 points chosen randomly will form quadrilateral
will form a quadrilateral.
plz write the solution

Answer 1

Here is one answer for (a).
A line by definition is a "straight line."
Two points determine a straight line.
Number the points from left to right (Top)
1 2 3 4 5 6
Number the points from left to right (Bottom)
A B C D E F
Add these lines formed by two points:
5 - 1-2, 1-3, 1-4, 1-5, 1-6
5 - A-B, A-C, A-D, A-E, A-F
6 - A1...A6
6 - B1...B6
6 - C1...C6
6 - D1...D6
6 - E1...E6
6 - F1...F6
Total: 46 lines

Answer 2

a) the # of lines that can be formed.

You can form two horizontal lines plus a line from every point on the first row to every point on the second row.

# of possible lines = 2 + 6*6 = 38

b) the # of triangles that can be formed.

This is the number of combinations of two points on one line and one point on the other line.

# of triangles = (2C1) * (6C1) * (6C2) = 2*6*15 = 180

c) the probability that three points chosen randomly will form a triangle.

This is equal to the number of ways to form a triangle divided by the total number of ways to choose three points.

P( Triangle) = 180 / (12C3) = 180 / 220 = 9/11

d) the # of quadrilaterals that can be formed

This is the number of combinations of two points on one line and two points on the other line.

# of quadrilaterals = (6C2) * (6C2) = 15*15 = 225

e) the prabability that 4 points chosen randomly will form quadrilateral.

This is equal to the number of ways to form a quadrilateral divided by the total number of ways to choose four points.

P(Quadrilateral) = 225 / (12C4) = 225 / 495 = 5/11

Answer 3

a) The no. of lines that can be formed=6(6)+2=36+2=38
b) The no.of triangles that can be formed=2*(6c2)*(6c1)=2*15*6=180
c) The probabilty that 3 points chosen at random to form a triangle is (2*(6c2)*(6c1))/(12c3)=(180)/(220)=0.818
d) 6c2*6c2=225
e) ((6c2)*(6c2))/(12c4)=0.4545

Answer 4

i admire polygons because of the fact they are able to extremely be something and as a consequence are extra suitable to all shapes...they strike a cord in me of chameleons, different than circles, yet they only suck because of the fact they have infinite factors alongside them and that's purely too many...polygons FTW!