The number of triangles that can be formed by 5 points in a line and 3 points on a parallel line is
Answers
Answer: 45 triangles
Step-by-step explanation:
Let’s assume that,
The line formed with 5 points on it be denoted as line “L1”
Another line parallel to the previous line with 3 points on it be denoted as line “L2”
We know that triangles cannot be formed when the three points are colinear, so we cannot consider points alone from line L1 and alone from line L2 to form triangles as the area of the triangle will be zero.
So to form triangles we will form some possible combination of points i.e.,
No. of ways to choose a combination of 3 points from 8 points
= ⁸C₃
= 8! / (5! * 3!)
= (8*7*6) / (3*2*1)
= 56
No. of ways to choose a combination of 3 points from 5 points on line L1
= ⁵C₃
= 5! / (2! * 3!)
= (5 * 4) / (2 * 1)
= 10
No. of ways to choose a combination of 3 points from 3 points on line L2 = ³C₃ = 1
Thus,
The possible no. of triangles that can be formed with proper combinations is,
= 56 - 10 - 1
= 45
Answer:
45 Triangles can be Formed
Step-by-step explanation:
5 Points in a Line & 3 Points in a line
As Triangle has Three points which are not colinear then there can be triangles
Either With two points on Line With 5 Points Then 1 point is to be selected from line with 3 points
or With two points on Line with 3 points then 1 points is to be selected from line with 5 points
So total Trianges
⁵C₂ * ³C₁ + ³C₂ * ⁵C₁
= 10 * 3 + 3 * 5
= 30 + 15
= 45
45 Triangles can be Formed