sanatan sharma
Last Activity: 13 Years ago
let the polygon circumscribed about a circle be denoted by PQRS........ &
the polygon inscribed in the circle be ABCD......... and their centres be O .
in polygon PQRS........ let the mid-point of PQ be Y so OY is the radius of the circle.
i recommend you to take pen and paper and do as directed .
let side PQ = x
angle POQ = 360/n therefore angle POY = 180/n
tan 180/n = x/2r where r is the radius of the circle
x = 2r tan 180/n
area of triangle POQ = xr/2
area of triangle POQ = (2r2)/2 tan 180/n
area of polygon PQRS............= n (area of triangle POQ) (1)-
in polygon ABCD..............
OB is the radius of the circle , D is the mid-point of AB and angle AOB= 360/n
angle BOD = 180/n
sin 180/n = y/2r where y is the length AB
y = 2r sin 180/n
similarly length OD = r cos 180/n
area of triangle AOB = y(OD)/2
area of triangle AOB = 2r2/2 (sin 180/n)( cos 180/n)
area of polygon ABCD........... = n(area of triangle AOB)
(area of polygon PQRS............)/(area of polygon ABCD...........) = 4/3
substituting the values we will get a equation as
cos 180/n = √3/2
cos 180/n = cos 30
180/n = 30
therefore n =6