MHB Determine the ratio of the base to the perimeter.

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The isosceles triangle has two equal sides of length 9x + 3, and its perimeter is 30x + 10. The base is calculated as b = 12x + 4 after solving the equation for the perimeter. The ratio of the base to the perimeter simplifies to (6x + 2) / (15x + 5). For the triangle to exist, x must be strictly greater than -1/3.
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8) An isosceles triangle has two sides of length $$9x+3$$. The perimeter of the triangle is $$30x+10$$

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

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

$$2(9x+3)+b=30x+10$$

So, we need to solve this for $b$, to have $b$ in terms of $x$...
 
eleventhxhour said:
8) An isosceles triangle has two sides of length $$9x+3$$. The perimeter of the triangle is $$30x+10$$

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

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 $\frac{-1}{3}$
 
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