PQR is an isosceles triangle inscribed in a circle. If PQ = PR = 25 cm and QR = 14 cm, calculate the radius of the circle to the nearest cm.

Given,

PQ = PR = 25 cm

In an isosceles triangle, the perpendicular from a vertex between equal sides bisects the opposite side.

Thus,

S is the mid-point of QR.

∴ QS = $\dfrac{QR}{2} = \dfrac{14}{2}$ = 7 cm.

PS is perpendicular bisector of QR and perpendicular from center bisects the chord.

Thus, centre of the circle O lies on PS. Let radius of circle OQ be r.

In right-angled triangle PSQ,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ PQ² = PS² + SQ²

⇒ 25² = PS² + 7²

⇒ 625 = PS² + 49

⇒ PS² = 625 - 49

⇒ PS² = 576

⇒ PS = $\sqrt{576}$ = 24 cm

OS = PS - OP = 24 - r

In right-angled triangle OSQ,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OQ² = OS² + SQ²

⇒ r² = (24 - r)² + 7²

⇒ r² = r² - 48r + 576 + 49

⇒ 48r = 625

⇒ r = $\dfrac{625}{48}$ = 13.02 cm ≈ 13 cm

Hence, radius of the circle = 13 cm.