Question

pqr is an isosceles triangle inscribed in a circle if PQ equal to PR equal to 25 cm and QR is equal to 14 cm calculate the radius of the circle to the nearest centimetre

Answer 1

Given : PQR is an isosceles triangle with PQ = PR = 25 cm, QR = 14 cm
Construction : Let X be the mid point of QR, Join PX
Then PX ⊥ QR  (Median of an isosceles triangle is perpendicular to the base)
⇒ Center of the circle O lies on PX  (perpendicular bisector of a chord passes through its center)
⇒ QX = RX = QR/2 = 14/2 cm = 7cm
In right Δ PXR
(PX)2 + (XR)2 = (PR)2
⇒ (PX)2 = (PR)2 - (XR)2 = (25 cm)2 - (7 cm)2 = (625 - 49) cm2 = 576 cm2
⇒ PX = 24 cm
Now, required radius = OP = OR = x cm(say)
⇒ OX = PX - OP = (24 - x) cm
In right Δ OXR
(OX)2 + (XR)2 = (OR)2
⇒ (24 - x)2 + 72 = x2
⇒ 242 + x2 - 2 × 24 × x + 49 = x2
⇒ 576 + x2 - 48x + 49 - x2 = 0
⇒ 48x = 625
⇒ x = 625/48 ~ 13.02
Hence required radius is 13.02  cm

Answer 2

Step-by-step explanation:

this is the answer I think this will help you

mark me as brainlist