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]#