The number of values of b for which there is | vedclass

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(C) For a triangle to exist,the sum of any two sides must be greater than the third side,and all side lengths must be positive.Let the sides be $a = b

Vedclass · Accepted Answer

$2$

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(C) For a triangle to exist,the sum of any two sides must be greater than the third side,and all side lengths must be positive.Let the sides be $a = b+5$,$c = 3b-2$,and $d = 6-b$.For sides to be positive: $b+5 > 0 \Rightarrow b > -5$,$3b-2 > 0 \Rightarrow b > 2/3$,and $6-b > 0 \Rightarrow b Case $I$: $b+5 = 3b-2$$2b = 7 \Rightarrow b = 3.5$. Sides are $8.5, 8.5, 2.5$. Since $8.5+8.5 > 2.5$,$8.5+2.5 > 8.5$,this is a valid triangle.Case $II$: $3b-2 = 6-b$$4b = 8 \Rightarrow b = 2$.Sides are $7, 4, 4$. Since $4+4 > 7$,this is a valid triangle.Case $III$: $b+5 = 6-b$$2b = 1 \Rightarrow b = 0.5$.This contradicts the condition $b > 2/3$. Also,the side $3b-2$ would be $3(0.5)-2 = -0.5$,which is not possible.Thus,there are $2$ such values of $b$.

The number of values of $b$ for which there is an isosceles triangle with sides of lengths $b+5$,$3b-2$,and $6-b$ is

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