Answer :

Consider a pentagon PQRST with circumcircle having centre C and radius r = 15 cm.



Join CQ and CR and draw a perpendicular CH on QR.


Since, PQRST is a regular pentagon.


We get,


Each internal angle = 108°


PQR = 108°


Also CQ acts as an internal angle bisector for each internal angle of regular pentagon.


PQC = RQC


PQR = PQC + RQC = 108°


PQR = 2( PQC) = 2( RQC) = 108°


PQC = RQC = 54°


Now in Δ QCH,



CD = sin 54° × QC


CD = sin 54° × 15 cm


(Frome table, sin 54° = 0.809)


CD = 0.809 × 15 = 12.135 cm


CD = 12.13 cm


We know that in regular pentagon all sides are equal.


Length of each side = 12.13 cm


Length of each side = 12.13 cm.


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