6 points are place on the line a, 4 points are placed on the line b. How many triangles is it possible to form such that their verticies will be the given points, if a ∥b?

96

Step-by-step explanation:

Ok, lines a and b are parallel.

We can separate this problem in two cases:

Case 1: 2 vertex in line a, and one vertex in line b.

Here we use the relation:

"In a group of N elements, the total combinations of sets of K elements is given by"

Here, the total number of points in the line is N, and K is the ones that we select to make the vertices of the triangle.

Then if we have two vertices in line a, we have:

N = 6, K = 2

And the other vertex can be on any of the four points on the line b, so the total number of triangles is:

C = 15*4 = 60.

But we still have the case 2, where we have 2 vertices on line b, and one on line a.

First, the combination for the two vertices in line b is:

We use N = 4 and K = 2.

And the other vertice of the triangle can be on any of the 6 points in line a, so the total number of triangles that we can make in this case is:

C = 6*6 = 36

Then, putting together the two cases, we have a total of:

60 + 36 = 96 different triangles

step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.