Find the probability of getting an equilateral triangle from the vertices of a regular hexagon

Text Solution

Solution : ABCDEF is a regular hexagon. Total number of triangles `6_(C_3) = 20. ` <br> . Of these only `triangleACE;`<br> ` triangle BDF` are equilateral triangles.<br> Therefore, required probability `=2/20 = 1/10`<br>

Answer

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Hint:First of all, find the total number of triangles that is possible by taking 3 points of a regular hexagon which has 6 vertices. Total number of triangles is \[^{6}{{C}_{3}}\] . \[\Delta DFB\] and \[\Delta AEC\]are those triangles which have all of three sides equal to each other. So, there are two equilateral triangles possible in a regular hexagon. Probability can be calculated using the formula , \[\text{Probability}=\dfrac{\text{total number of equilateral triangles}}{\text{total number of triangles possible}}\].

Complete step-by-step answer:

We have connected the vertex A,E and C and can see that we got an equilateral triangle.Similarly, We have connected the vertex D,F and B and can see that we got an equilateral triangle.Suppose if we connect the vertex A,D and E, we don’t get an equilateral triangle. Because, according to the diagram we can see that all three sides are not equal to each other.We have only two equilateral triangles to be formed using a regular hexagon. Out of six points in a hexagon, we have to select only three points at a time.The total number of triangles to be formed using a regular hexagon=${}^{6}{{C}_{3}}\ ways$\[\begin{align}  & =\dfrac{6\times5\times4}{1\times2\times3} \\  & =\dfrac{120}{6} \\  & =20 \\ \end{align}\]Out of 20 triangles, there are only two equilateral triangles that are \[\Delta \]DFB and $\Delta$ AEC.Probability of choosing equilateral triangle \[=\dfrac{2}{20}=\dfrac{1}{10}\] .Note: In this question, one can make mistakes in taking the number of equilateral triangles. One can think that there can be six equilateral triangles that are \[\Delta DEF\], \[\Delta DCB\] , \[\Delta EFA\] , \[\Delta FAB\] , \[\Delta CBA\] and \[\Delta DEC\]. But in these triangles the third side is not equal to the remaining two sides.

Option 1 : \(\dfrac{1}{10}\)

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Concept:

• The probability of occurrence of an event A out of a total possible outcomes N, is given by \(\rm P(A)=\dfrac{n(A)}{N}\), where n(A) is the number of ways in which event A can occur.
• The total number of triangles that can be drawn using three points from a set of n non-collinear points, is given by nC3.

Calculation:

A hexagon has 6 vertices. The total number of triangles that can be drawn using these 6 vertices will be:

\(\rm N={^6C_{3}}=\dfrac{6!}{3!(6-3)!}=20\).

An equilateral triangle can be drawn from these six vertices if we select three vertices alternately.

This leaves us with three more alternately spaced vertices, which can form another equilateral triangle.

These two triangles together cover all six vertices.

Therefore, the total number of equilateral triangles that can be drawn = 2.

Required probability = \(\dfrac{2}{20}=\dfrac{1}{10}\).

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