Question

If a polygon of 'n' sides has 1/2 n (n-3) diagonals. How many sides are there in a polygon with 65 diagonals ? Is there
a polygon with 50 diagonals ? ​

Answer 1

Answer:

Step-by-step explanation:

The number of diagonals of a polygon is given by n(n-3)/2.

So if there are 65 diagonals then the number of sides of the polygon are

65*2 = 130 = n(n-3), or

n^2–3n-130 = 0, or

(n-13)(n+10) = 0, or

n = 13, -10 is not permissible.

Hence the polygon has 13 sides.

Again if the number of diagonals is

50 =n(n-3)/2, or

100 = n^2 - 3n, or

n^2–3n-100 = 0

It has no integral roots, so there is no polygon with 50 diagonals.

Answer 2

            ༒ Qυєѕтiσи ༒

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If a polygon of 'n' sides has 1/2 n (n-3) diagonals. How many sides are there in a polygon with 65 diagonals ? Is there a polygon with 50 diagonals ?

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            ★彡 Sσℓυтiσи 彡★

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Given :-

No. of diagonals of a polynomial with n-sides

= n(n - 3)/2

No. of diagonals of a given polygon = 65

i.e., n(n - 3)/2 = 65

Where n is number of sides of the polygon

⟹ n² - 3n = 2 × 65

⟹ n² - 3n - 130 = 0

⟹ n² - 13n + 10n - 130 = 0

⟹ n(n - 13) + 10(n - 13) = 0

⟹ (n - 13) (n + 10) = 0

⟹ n - 13 = 0 (or) n + 10 = 0

⟹ n = 13 (or) n = -10

But n can't be negative .

.°. n = 13 (i.e.) number of sides = 13

Also to check 50 as the number of diagonals of a polygon

.°. n(n - 3)/2 = 50

⟹ n² - 3n = 100

⟹ n² - 3n - 100 = 0

There is no real value of n for which the above equation is satisfied.

.°. There can't be a polygon with 50 diagonals.

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