Question

in a polygon no three diagonals are concurrent. if the total number of point of intersection of diagonals interior to the polygon be 210, then the number of diagonals of the polygon is ​

Answer 1

Answer:

Correct option is

A

20

A selection of four vertices of the polygon given an interior intersection.

∴ the number of sides =n

⇒

n

C

4

=70⇒n(n−1)(n−2)(n−3)

=24×70=8×7×6×5

∴n=8

∴ the number of diagonals =

8

C

2

−8=20

Answer 2

Answer:

The number of diagonals =  20

Step-by-step explanation:

A selection of four vertices of the polygon given an interior intersection.

The number of sides = n

$n_{C4}$=70

By expanding,

We get,

$\frac{n(n-1)(n-2)(n-3)}{4*3*2*1}$= 70

We need to find the value of n

n(n–1)(n–2)(n–3)24=70

n(n–1)(n–2)(n–3)=70×24

By multiplying,

We get,

$n^{4}$–6$n^{3}$+11–6n=1680

Using synthetic division, we can write polynomial completely as follows,

(n–8)(n+5)(n2–3n+42)=0

Here only n= 8

is applicable to this equation.

n=8

Thus the polygon has 8 sides.

The number of diagonals =$8_{C2}$−8

By expanding,

We get,

=$\frac{8!2!}{(8-2)!}$−8

=$\frac{8!}{2!6}$!−8

On further expanding,

=8×7×6×5×4×3×2×12×1×6×5×4×3×2×1−8

By simplifying,

We get,

=28−8=20

Hence, the number of diagonals =  20

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