Please solve q 95 ?

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Answer 1 · 2018-05-12T04:54:40

The length of the longest side is $21$ .

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In a $DeltaABC$ ,

$rarrcosA=(b^2+c^2-a^2)/(2bc)$

$rarrArea=(1/2)a*bsinC$

Now, $Area$ of $DeltaABD=(1/2)*9*8*sinx=36sinx$

$Area$ of $DeltaADC=(1/2)*8*18*sinx=72sinx$

$Area$ of $DeltaABC=(1/2)*9*18*sin2x=81sin2x$

$rarrDeltaABC=DeltaABD+DeltaADC$

$rarr81sin2x=36*sinx+72*sinx=108*sinx$

$rarr81*2cancel(sinx)*cosx=108*cancel(sinx)$

$rarrcosx=(108)/162=2/3$

Applying cosine law in $DeltaABC$ , we get,

$rarrcos2x=(9^2+18^2-a^2)/(2*9*18)$

$rarr2cos^2x-1=(405-a^2)/324$

$rarr2*(2/3)^2-1=(405-a^2)/324$

$rarr2*(4/9)-1=(405-a^2)/324$

$rarr-36=405-a^2$

$rarra^2=405+36=441$

$rarra=21$

Also, note that

$rarrsin2x=2sinxcosx$

$rarrcos2x=2cos^2x-1$