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Dinesh giri

· started a discussion

· 1 Months ago

Dinesh (Question:109034.0)

PM=.5 caise ho raha hai

Question:
The two chords AB and CD of a circle cut each other perpendicularly at point O.   OA = 2 cm, OC = 3 cm and OB = 6 cm, then what will be the diameter of the circle?
Options:
A) \(\cfrac{\sqrt[]{145}}{2}\)
B) \(\sqrt[]{130}\)
C) \(\cfrac{\sqrt[]{130}}{2}\)
D) \(\sqrt[]{65}\)
Solution:

OA = 2 cm 

OB = 6 cm 

OC = 3 cm 

OA × OB = OC × OD

\(\Rightarrow OD = \cfrac{2\times6}{3}=4\) cm

M  and N are the midpoints of chord AB and CD respectively.

P is the centre of the circle.

PM = ON = 0.5 cm

MB = 4 cm

PB =\(\sqrt[]{PM^2+MB^2}\)    = \(\sqrt[]{(0.5)^2+(4)^2}=\sqrt[]{16.25}\)

Diameter = \(2\times PB = \sqrt[]{4\times 16.25}=\sqrt[]{65}\) cm

Knowledge Expert

· commented

· 1 Months ago

Dear student
We are given with OD = 4 cm
DC = OC + OD
DC = 3 + 4
DC = 7 cm
N is the midpoint so
DC/2 = 7/2 = 3.5 cm
3.5 = DN = NC
NC - OC = ON = (ie. 3.5 - 3 = 0.5 )
ON = PM = 0.5

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