QUADRATIC �EQUATIONS
(n – 6)
(n – 6)
– 36n
+ 36
= 0
∴ (n – 6)
(5n – 6) = 0
∴ n = 6
∴ n2 = (6)2
= 36
or 5n – 6
= 0
∴ 5n
– 6
= 0
Exterior angle of a regular polygon having n-sides is more than that of the polygon having n2 sides by 500 . Find the number of the sides of each polygon.
Q)
Sol.
Number of sides of one of the regular polygon = n
Number of sides of the other regular polygon = n2
As per the given condition,
=
+
50
Exterior angle for a regular polygon having ‘n’ number of sides =
Exterior angle for a regular polygon having ‘n2’ number of sides =
Exterior angle for a regular polygon =
Means =
After the sign 50
Multiplying throughout by n2
∴ 360n
= 360
+ 50n2
Dividing throughout by 10
∴ 36n
= 36
+ 5n2
∴ 5n2
‘n’ sided closed figure having all the sides equal
REMEMBER !!!!!
Exterior angle of any regular polygon is given by the formula 360/number of sides
Square is 4 sided regular polygon
Equilateral triangle is a 3 sided regular polygon
After leaving some space +
∴ 5n2
∴ n – 6 = 0
∴ n = 6
or 5n = 6
∴ n = 6
Number of sides cannot be a fraction.
∴ Number of sides of required polygons are 6 and 36.
Find two factors of 180 in such a way that by adding factors we get middle no. 36
Since we are adding the factors give middle term sign to both the factors.
3
6
18
–
–
36 × 5 = 180
6
+
= 36
30
– 30n
– 6n
+ 36
= 0
0
0
∴ –5n2
+ 36n
– 36
= 0
Multiplying throughout by –1
Equilateral ∆ (3 sides)
1200
600
=
360
3
1200
Square (4 sides)
=
360
4
900