How many triangles in a pentagon with diagonals

Answer

Verified

Hint: In geometry a pentagon is a polygon having 5 sides, and 5 internal angles, also triangles are polygons having 3 sides as well as internal angles. To solve the given question, we have to fix one of the vertices of a pentagon. Then, draw lines from the fixed vertex to the other four vertices, and count the number of triangles that can be formed. To make things simple, we will use a regular polygon for this. A regular polygon is a special polygon having all sides equal.

Complete step-by-step solution:

We are asked to count the number of triangles that can be found using the vertices of a pentagon. To solve the given question, we will have to fix one of the vertices of the pentagon. Then, draw lines from the fixed vertex to the other four vertices, and count the number of triangles that can be formed.We are given the pentagon,

We will fix the vertex A, and draw lines joining other vertices with it. By doing this, we get a figure as follows,

Thus, we are getting three triangles by drawing the lines that join the fixed vertex A with the vertices C, and D. The names of the triangles are \[\Delta ABC,\Delta ADC,\Delta AED\].

Note: Here, we join the vertex A with only two other vertices. The reason of this is as follows:

The pentagon has a total 5 vertices, we have fixed one vertex so there are four remaining. Out of the four vertices, two vertices B and E are adjacent to the fixed vertex. Hence, by joining them we will get a line. Thus, excluding these four, only two vertices remain.

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.

Break into cases:

all three vertices of the triangle are on the outer pentagon
two of the three vertices are adjacent on the outer pentagon while the third is in the inner pentagon
two of the three vertices are nonadjacent on the outer pentagon while the third is in the inner pentagon
only one of the three vertices are on the outer pentagon

In case 1, any choice of three distinct vertices will form a triangle, so case 1 contributes $\binom{5}{3}=10$ to the overall sum.

In case 2, notice that any choice of two distinct vertices from the outer pentagon will have a two choices of a third vertex from the inner pentagon such that the three form a triangle in the image. The number of adjacent pairs is $5$, each of which contributing three to the sum, for a total of $15$ being contributed.

In case 3, notice that any choice of two distinct vertices from the outer pentagon will have a unique choice of a third vertex from the inner pentagon such that the three form a triangle in the image. The number of nonadjacent pairs is $5$, each of which contributing one to the sum.

In case 4, notice that any choice of single vertex from the outer pentagon has a unique pair of vertices from the inner pentagon such that the three form a triangle in the image. Thus, case 3 contributes $\binom{5}{1}=5$ to the overall sum.

Finally, notice that no other triangles exist in the image.

This brings the overall sum to $10+15+5+5=35$