Determine the ratio of the base to the perimeter.

[eleventhxhour]

8) An isosceles triangle has two sides of length \(\displaystyle 9x+3\). The perimeter of the triangle is \(\displaystyle 30x+10\)

a) Determine the ratio of the base to the perimeter, in simplified form. State the restriction on \(\displaystyle x\)

Thanks for your help!

 

[MarkFL]

First, let's find the base. We know the two given equal sides plus the base $b$ is equal to the perimeter:

\(\displaystyle 2(9x+3)+b=30x+10\)

So, we need to solve this for $b$, to have $b$ in terms of $x$...

 

[Fermat1]

8) An isosceles triangle has two sides of length \(\displaystyle 9x+3\). The perimeter of the triangle is \(\displaystyle 30x+10\)

a) Determine the ratio of the base to the perimeter, in simplified form. State the restriction on \(\displaystyle x\)

Thanks for your help!

The base has length $b=30x+10-2(9x+3)=12x+4$. So the ratio of the base to the perimeter is $\frac{12x+4}{30x+10}=\frac{6x+2}{15x+5}$. We want the triangle to exist so the perimeter must be positive. So the restriction is that $x$ is (strictly) greater than $\frac{-1}{3}$