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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