Please solve q 20?

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1 Answer
CW
May 7, 2018

#2011sqrt2# cm

Explanation:

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Let #angleBAC=x#
let #A# be the area of #DeltaABC#
#A=1/2*AB*AC*sinx#
we know that #sinx# has a maximum value of #1# ,
when #sinx=1, A# has the maximum value.
#=> x=90^@#
and given #AB=AC=2011# cm
#angleABC=ACB=45^@#
#=> BCsin45=AB#
#=> BC*1/sqrt2=2011#
#=> BC=2011sqrt2# cm

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