Number of sides of a polygon? Each Side of a polygin is >1
(i)Ratio of number of diagonals to number of sides is\frac{2}{3}
(ii)Number of diagonals of the polygon is 9
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(i)Ratio of number of diagonals to number of sides is\frac{2}{3}
(ii)Number of diagonals of the polygon is 9
[Reveal] Spoiler: Answer
D
[Reveal] Spoiler: Solution
The relation between number of diagonals and number of sides of a polygon is
Number of diagonals =\frac{n(n-3)}{2}
Solving with (i) assume there are 2x sides and 3x diagonals. substitute these values in the formula. Number of diagonals =\frac{n(n-3)}{2}
3x = \frac{(2x)(2x-3)}{2}
6x = 4x^2 -6x
4x^2 -12x= 0
x can be 0 or 3. Side can not be zero. Hence consider x=3 and this gives us sides ( 2x) = 2(3) = 6 and diagonals (3x) =3(2)= 9
(i) is sufficient
Using (ii)
No.of diagonals given as 9. So 9 = \frac{n(n-3)}{2}
n^2-3n-18 =0 gives n=-3 or 6 Sides can not be negative. Hence number of sides (n)= 6
(ii) is also sufficient.
Therefore answer is D, each statement alone is sufficient.
Number of diagonals =\frac{n(n-3)}{2}
Solving with (i) assume there are 2x sides and 3x diagonals. substitute these values in the formula. Number of diagonals =\frac{n(n-3)}{2}
3x = \frac{(2x)(2x-3)}{2}
6x = 4x^2 -6x
4x^2 -12x= 0
x can be 0 or 3. Side can not be zero. Hence consider x=3 and this gives us sides ( 2x) = 2(3) = 6 and diagonals (3x) =3(2)= 9
(i) is sufficient
Using (ii)
No.of diagonals given as 9. So 9 = \frac{n(n-3)}{2}
n^2-3n-18 =0 gives n=-3 or 6 Sides can not be negative. Hence number of sides (n)= 6
(ii) is also sufficient.
Therefore answer is D, each statement alone is sufficient.