find the perimeter of a regular octagon inscribed in a circle of radius 100 cm
Answers
Answered by
9
Octagon inside a circle will have 8 equal isoslese triangle which subtend 45 at the centre (368/8)
Take any one triangle (let it be triangle AOB)and construct a perpendicular C bisecting angle 45 ( i.e AOB)-----1
Both triangels will be congruent by RHS rule
then side AB WILL AC=CB-----2
Now take one triangle (OCB )
Where angle COB=22.5
Now ...using trigonometry
Sin 22.5 = CB/100. (P/H)
0.38268=CB/100
CB = 38.268
Then AB =2CB from 2
AB = 38.268x2. = 76.536 cm
Now perimetre of a regular octagon = 8x side
8 x 76.536 cm
612.288 cm. ------your answer
Take any one triangle (let it be triangle AOB)and construct a perpendicular C bisecting angle 45 ( i.e AOB)-----1
Both triangels will be congruent by RHS rule
then side AB WILL AC=CB-----2
Now take one triangle (OCB )
Where angle COB=22.5
Now ...using trigonometry
Sin 22.5 = CB/100. (P/H)
0.38268=CB/100
CB = 38.268
Then AB =2CB from 2
AB = 38.268x2. = 76.536 cm
Now perimetre of a regular octagon = 8x side
8 x 76.536 cm
612.288 cm. ------your answer
Similar questions