Solve the question

Dear student,

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

Regards!

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