please answer this question

Asked by mac12303190 | 6th Mar, 2010, 12:20: AM

Expert Answer:

We have, BD = DE = EC = BC/3

In ABD,

AD2 = AB2 + BD2

AD2 = AB2 + BC2/9       ..........BD = BC/3  ................(1)

Similarly other possible right angled triangles, we can write,

AE2 = AB2 + BE2

AE2 = AB2 + (2BC/3)2

AE2 = AB2 + 4BC2/9

and

AC2 = AB2 + BC2

AC2 = AB2 + BC2     ........(2)

Now,

AE2 = AB2 + 4BC2/9

8AE2 = 8AB2 + 32BC2/9

= 5AB2 + 5xBC2/9 + 3AB2 + 27BC2/9

= 5AB2 + 5xBC2/9 + 3AB2 + 3x9BC2/9

= 5(AB2 + BC2/9) + 3(AB2 + 9BC2/9)

= 5(AB2 + BC2/9) + 3(AB2 + BC2)

= 5AD2 + 3AC....................from (1) and (2)

Regards,

Team,

TopperLearning.

 

 

 

Answered by  | 6th Mar, 2010, 02:21: PM

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