If four vertices of a regular octagon are cho | vedclass

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(D) regular octagon has $8$ vertices. The total number of ways to choose $4$ vertices out of $8$ is given by the combination formula $^nC_r = \frac{n!

Vedclass · Accepted Answer

$\frac{1}{35}$

Vedclass · Answer

(D) regular octagon has $8$ vertices. The total number of ways to choose $4$ vertices out of $8$ is given by the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$.Total number of cases $= ^8C_4 = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$.In a regular octagon,a rectangle is formed by choosing $4$ vertices such that the diagonals of the rectangle are diameters of the circumscribed circle. There are only $2$ such rectangles possible in a regular octagon (as shown in the figure: $ADEH$ and $GFCB$).Number of favourable cases $= 2$.Therefore,the required probability $= \frac{	ext{Number of favourable cases}}{	ext{Total number of cases}} = \frac{2}{70} = \frac{1}{35}$.

If four vertices of a regular octagon are chosen at random,then the probability that the quadrilateral formed by them is a rectangle is

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