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.
eleventhxhour
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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 $\frac{-1}{3}$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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