Q. 34.4( 101 Votes )

In Fig. 6.58, ABC

Answer :

Using Pythagoras theorem in ΔADB, we get:

AB2 = AD2 + DB2 (i)

Applying Pythagoras theorem in ΔACD, we obtain

AC2 = AD2 + DC2

AC2 = AD2 + (DB + BC)2

AC2 = AD2 + DB2 + BC2 + 2DB x BC

AC2 = AB2 + BC2+ 2DB x BC [Using equation (i)]

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