plz solve the both ques
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by applying pythagoras theorem in triangle ABC , we get AB2=AC2+BC2 =AC2+(2CD)2 (SINCE D IS THE MID POINT OF BC) =AC2+4CD2 =AC2+4(AD2_AC2) ( SINCE IN TRIANGLE ACD WE GET AD2=AC2+CD2) =AC2+4AD2_4AC2 =4AD2_3AC2 THEREFORE , AB2=4AD2_3AC2. THANK YOU .
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2. AB²=BC²+AC² BY PT
=(BD+CD)²+AC² BC=BD+CD
=BD²+CD²+2BDCD+AC²
=(AC²+CD²)+BD²+2BD.BD
=AD²+BD²+2BD² AC²+CD²=ad²
=AD²+CD²+2CD²
=AD²+3CD²
=AD²+3(AD²-AC²)
=AD²+3AD²-3AC²
=4AD²-3AC²
HENCE PROVED
=(BD+CD)²+AC² BC=BD+CD
=BD²+CD²+2BDCD+AC²
=(AC²+CD²)+BD²+2BD.BD
=AD²+BD²+2BD² AC²+CD²=ad²
=AD²+CD²+2CD²
=AD²+3CD²
=AD²+3(AD²-AC²)
=AD²+3AD²-3AC²
=4AD²-3AC²
HENCE PROVED
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