Question

the number of points required to form 36 line segments no 3 points are collinear is i will mark brainliest who give correct answer

Answer 1

Correct question= There are 6 points, no three of which are collinear. How many straight lines and triangles can be formed by joining the points?

Solution:

There are 6 points, no three of which are collinear. How many straight lines and triangles can be formed by joining the points?

There are 6 points and since no 3 points are collinear, for simplicity, we can imagine the points to be vertices of a regular hexagon (A non-regular one also would do, as log as it’s a hexagon).

Let’s count the Lines:

Firstly there are those 6 lines for forming a hexagon. Then choose any point and we can draw 3 more lines. Choose the next (adjacent) point, going in anti clockwise direction and we can draw another 3 lines. with the next anti clockwise point, we can draw 2 lines and with the next anti clockwise point we can draw 1 more line. There are two (2) more points but the lines are already drawn (from those points) and thus we have 6 + 3 + 3 + 2 + 1 = 15 Lines, which could be summed up as 6C2 = (6*5) / (2*1) = 30 / 2 = 15 Lines.

Let’s go for the Triangles:

Let’s have 6 distinct points for the vertices of a Regular Hexagon. Let’s call the points as A, B, C, D, E and F in anti clockwise fashion.

Let’s take the base as AB and draw Distinct Triangles; we can draw 4 Triangles.

Now, let’s shift the base to BC and draw Distinct Triangles; we can draw 3 Triangles.

Now, let’s shift the base to CD and draw Distinct Triangles; we can draw 2 Triangles.

Now, let’s shift the base to DE and draw Distinct Triangles; we can draw 1 Triangle.

With the remaining 2 bases Viz. EF and FA, no more Distinct Triangles are possible. We can draw 2 more Distinct Triangles (1 each) with bases AC and BD and no more Distinct Triangles are possible with the remaining bases as all Triangles are now covered. So, here we have, 4 + 3 + 2 + 1 + 1 + 1 = 12. There must be some formula to arrive at it quickly but I am not able to figure it out, at the moment.

Thus, we have 15 Distinct Lines and 12 Distinct Triangles out of 6 points such that no 3 points are collinear.