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the ratio of the area of the regular polygon of n sides circumscribed about a circle to the area of the regular polygon of equal no. of sides inscribed in the circle is 4:3.find the value of n

sanatan sharma
26 Points
9 years ago

if you want i can send ou to your e-mail

vivek kumar
57 Points
9 years ago

yes,please send it and explain the steps

Sathya
35 Points
9 years ago

Hi Vivek kumar,

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

Pls approve my answer if it helped!!!