Last Activity: 14 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
AD.DB = CD.DE --------------------(1) ...
Now Since the length of chord is always positve...so AM > GM is applied ...
(AD + DB ) /2 > = root( AD.DB)
(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 ...
AD = r sinc
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
Last Activity: 14 Years ago
but the answer is given as c2/4 i.e. (AB)2/4
Last Activity: 14 Years ago
No Problem dude !
Your given answer is also correct...and what i have written is also correct ...
Look CD.DE = AD.DB
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 )
so Maximum case....AD.DB = (AD+DB)^2/4 => AD.DB = (AB)^2/4 [ you got it ]
Last Activity: 14 Years ago
thanks much plz accept my friend request
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