pls answer this question. this is the second time i am asking this question
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AQ = AR (tangents from external points to a circle are equal)
BQ = BP (,,)
PC = CR (,,)
Now perimeter of ABC = AB + BC + AC
= AB + BP + PC + AC
= AB + BQ + CR + AC
= AQ + AR
= 2AQ ( Because AQ = AR )
Hence AQ = 1/2 (Perimeter ABC)
Hope this helps!
BQ = BP (,,)
PC = CR (,,)
Now perimeter of ABC = AB + BC + AC
= AB + BP + PC + AC
= AB + BQ + CR + AC
= AQ + AR
= 2AQ ( Because AQ = AR )
Hence AQ = 1/2 (Perimeter ABC)
Hope this helps!
Answered by
2
Hey... I think this can be ur answer!!
Given :AQ & AR are tangents
To prove : AQ=1/2(PERIMETER OF ABC)
Proof :
BP =BQ and CP =CR (tangents)
Now,
Perimeter of ABC = AB +AC +BC
AQ=AB+BQ
AR =AC +CR
Adding both the equations,
AQ+AR =AB +BQ+AC +CR
2AQ=AB +AC +BP +CP
2AQ=AB +AC +BC
2AQ =PERIMETER OF ABC
AQ =1/2(PERIMETER OF ABC
HENCE PROVED...
Hope it helps you dear ☺️☺️
Given :AQ & AR are tangents
To prove : AQ=1/2(PERIMETER OF ABC)
Proof :
BP =BQ and CP =CR (tangents)
Now,
Perimeter of ABC = AB +AC +BC
AQ=AB+BQ
AR =AC +CR
Adding both the equations,
AQ+AR =AB +BQ+AC +CR
2AQ=AB +AC +BP +CP
2AQ=AB +AC +BC
2AQ =PERIMETER OF ABC
AQ =1/2(PERIMETER OF ABC
HENCE PROVED...
Hope it helps you dear ☺️☺️
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