If in $DeltaABC$ , $(a+b+c)(a+b-c)-kab=0$ , in what range does $k$ lie?

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Answer 1 · 2016-12-14T13:14:55

$k$ belongs to interval $[0,4]$

ABC is a triangle in which

$(a+b+c)(a+b-c)-kab=0$

$=>(a+b)^2-c^2-kab=0$

$=>a^2+b^2+2ab-c^2-kab=0$

$=>a^2+b^2-c^2=kab-2ab$

$=>(a^2+b^2-c^2)/(2ab)=k/2-1$

Now for $DeltaABC$

$cosC=(a^2+b^2-c^2)/(2ab)$

So $k/2-1=cosC$

As $-1<=cosC<=1$

$=>-1<=k/2-1<=1$

$=>-1+1<=k/2-1+1<=1+1$

$=>0<=k/2<=2$

$=>0<=k<=4$

i.e. $0<=k<=4$

Hence $k$ belongs to interval $[0,4]$

If in DeltaABCDeltaABC, (a+b+c)(a+b-c)-kab=0(a+b+c)(a+b-c)-kab=0, in what range does kk lie?