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#   Askiitians_Expert Yagyadutt
11 years ago

Hii pranav ! Ok now look at the picture....I guess you remeber a theorem ...we have studied in the class IX ...If AB and CE are the two chord intersecting inside the circle...at point D ..then

Now Since the length of chord is always positve...so AM > GM is applied ...

(AD + DB ) > = 2 root( CD.DE) ---------------2(1)

So maximum value of CD.DE =  (AD + DB)^2 / 4    [ is it clear ? ]

Now our aim is to find out what will be the value of  AD + DB ...

Look in the figure i have drawn ...

DB = r sinc

The angle AOD = c ...( because of the theorem that ...angle at the centre is twice the angle at any portion on the circle )

So ...AD + DB = 2 rsinc

(AD + DB )^2 = 4.r^2sin^2(c)

(AD + DB ) ^2 / 4  =  r^2.sin^2(c)

Hence  maximum value of  CD.DE as declared above ....=  r^2.sin^2(c) .....since sin^2(c) is variable ...if c is variable...then maximum value of CD.DE =  r^2   ( because sin^2(c) = 1 )  where r is the radius of the circle..which you have not specified in the question....

Now let me know...if the answer is wrong !

Regards

Yagya Askiitians_Expert Yagyadutt
11 years ago

No Problem dude !

Your given answer is also correct...and what i have written is also correct  ...

Maximum value of CD.DE will be equal to maximum value of AD.DB ...isn't ?

(AD + DB)/2  >= root (AD.DB)  ( according to AM >GM )