Triangle Friangle

Let us name the 2 base points P and Q

There are 5 lines generating from point P : { $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ }

Similarly, there are 5 lines generating from point Q : { $Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}$ }

Now, out of these two sets of lines one line is common to both the sets. Let us say, they are { $P_{5}$ or, $Q_{5}$ }

If we select any 2 lines from the set of { $P_{k}$ } and 1 line from set of { $Q_{k}$ }, we will form a triangle

This can be done in $5C2 \times 4C1 = 40 ways$

Also, If we select any 2 lines from the set of { $Q_{k}$ } and 1 line from set of { $P_{k}$ }, we will form a triangle

This can also be done in $5C2 \times 4C1 = 40 ways$

But, there are few triangles which are counted double.

As, $P_{5}$ and $Q_{5}$ are essentially the same line, so when we select

• $P_{5}$ or $Q_{5}$
• 1 line from the other 4 lines from the set of { $P_{k}$ }
• 1 line from the other 4 lines from the set of { $Q_{k}$ }

These triangles have been counted in either of the 40 cases

Such triangles can be formed in $4C1 \times 4C1 = 16 ways$

Total ways = $40 + 40 - 16 = \boxed{64}$

Let the triangle be ABC with the vertex being A

First take BC and see How many triangles are formed. You will get the no. to be 16

Then form the points on AB and vertex at C, 6 triangles will be formed. Go on moving inside and you will get 24 triangles in total. Similarly 24 for the other side (AC)

Therefore total no. of triangles 16+24+24=64.

there are two types of lines in the diagram . one originating from left-down vertex B and other originating from right-down vertex C. Base line BC is common to the both types. We can take it to be in any type. Now , a triangle is formed by taking either two lines from left side and one from right side or one from left side and two from right side. But we cannot make any triangle by taking single BC line from left side. So number of triangles with single line from left side and any combination of two lines out of 4 lines from right side is equal to 4 * (4c2) .= 24 Number of triangles with any combination of two lines out of 5 lines from left and one line from right is (5c2)*4= 40 So total triangles is 64

64 is the ans

Total number of triangles in this picture=9C3-5c3-5c3 =84-10-10=64