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rajan

· started a discussion

· 1 Months ago

d (Question:133466.0)

in deno.there will be root2

Question:
2a’ and 2b’ are the length of two chords which intersect at right angle. If the distance between the center of the circle and the intersecting point of the chords is ‘C’ then the radius of the circle is:
Options:
A) \(\cfrac{\sqrt{a^2 b^2 c^2 }}{2}\)
B) \(\sqrt{a^2+b^2+c^2}\)
C) \({\sqrt\frac{{a^2 +b^2 +c^2 }}{2}}\)
D) None of these
Solution:

Let AB = 2a and PQ = 2b & ‘O’ is the centre

OM ⊥ PQ and ON ⊥ AB

∴ PM = MQ = \(\cfrac{2b}{2}\) = b

and AN = NB = \(\cfrac{2a}{2}\) = a

 

in quadrilateral DMON,

∠D = ∠M = ∠N = 90°

∴ ∠NOM = 90°

it mean DMON is a rectangle/square

∴ OM = DN = x (let)

and ON = DM = y (let)

Let radius = OQ =r cm

in ∆OMQ, r2 = x2 + b2 … (i)                           

in ∆ONA, r2 = y2 + a2 … (ii)

(i) + (ii) 2r2 = a2 + b2 + (x2 + y2)

= a2 + b2 + c2

⇒ r =  \(\cfrac{\sqrt{a^2 +b^2 +c^2 }}{2}\)

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