Question

A polygon has 27 diagonals . How many sides does it have ? Answer me and I will mark you the brainliest.

Answer 1

Answer:

The number of diagonals in a polygon = n(n-3)/2, where n is the number of polygon sides.

thus, 27=n(n-3)/2

Transpose 2 , and open the bracket.

27×2=n²-3n

54=n²-3n

0=n²-3n-54.

Do middle term splitting.......

0=n²-9n+6n-54

0=n(n-9)+6(n-9)

0=(n+6)(n-9)

thus, n+6=0, n-9=0

⇒n= -6, n=9.

Since there cant be "negative" sides, we ignore n=-6.

n=9 is ur answer.

HOPE THIS HELPS :D

Answer 2

Given :-

• A polygon has 27 diagonals.

To find:-

• Number of sides = ?

Solution :-

we know that,

$\small\pink{\bigstar}\:\:{\underline{\boxed{\bf\purple{No.of\:diagonal\:of\: polygon=\dfrac{n(n-3)}{2}}}}}$

$\rm\longrightarrow\:27=\dfrac{n(n-3)}{2}$

$\rm\longrightarrow\:27\times{2}=n(n-3)$

$\rm\longrightarrow\:54={n}^{2}-3n$

$\rm\longrightarrow\:{n}^{2}-3n-54=0$

$\rm\longrightarrow\:{n}^{2}-9n+6n-54=0$

$\rm\longrightarrow\:n(n-9)+6(n-9)=0$

$\rm\longrightarrow\:(n-9)(n+6)=0$

Now,

⤇ n - 9 = 0

⤇ n = 0 + 9

⤇ n = 9

then,

n + 6 = 0

⤇ n = 0 - 6

⤇ n = -6 [ not possible ]

Here value of n will be 9

Hence,the number of sides of polygon will be 9.