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4 Answers

Askiitians_Expert Yagyadutt
askIITians Faculty 74 Points
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 point D ..then

AD.DB = CD.DE --------------------(1) ...


Now Since the length of chord is always 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 !






pranay -askiitians expert
44 Points
11 years ago

but the answer is given as c2/4 i.e. (AB)2/4

Askiitians_Expert Yagyadutt
askIITians Faculty 74 Points
11 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 ]

pranay -askiitians expert
44 Points
11 years ago

thanks much plz accept my friend request

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