Question

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

Answer 1

An isosceles triangle has two of its sides equal.

The sides are $b+5,3b-2,6-b$

When $b+5=3b-2,b=3.5$, the third side is $6-b=6-3.5=2.5$.

Such a triangle exists.

When $b+5=6-b,b=0.5$, the third side is $3b-2=1.5-2=-0.5$.

Such a triangle does not exist since the length of the third side is negative.

When $3b-2=6-b,b=4$, the third side is $b+5=4+5=9$.

The triangle with sides $2,2,9$ does not exist, since the length of the third side is greater than the length of the other 2 sides.

The the only value of b is $b=3.5$. There is only one value of b.