**Find a Ratio**

**By: Erica Fletcher**

**Part 2 of Final: Investigation 2**

Below I constructed a square
with side **s **that is inscribed in a
semicircle. The length **a **is on each
side of the square along the diameter. If you would like to construct a figure
like the one shown below use this GSP file.

**Goal:** For this assignment we want to find the ratio of **s** to **a.**

In order to begin this construction we need to first construct a square
with side lengths **s. **Then after constructing the square we
first had to construct the midpoint of any side of the square. This midpoint is
connected to the opposite vertex of the square. Now using this midpoint and
segment we can construct a circle by center + radius. Lastly, we constructed a
semicircle by constructing an arc using 3 points.

**We can see that the midpoint of the side
of the square along the diameter is also the midpoint of the diameter.
Therefore, we can say that the distance from this midpoint to one of the
opposite vertices of the square is the radius of the semicircle. **

**Now by looking at the construction
above, we notice that the radius is also half of the diameter (also by
definition). We see that d = s + 2a so it follows :**

**d = s +2a**

**r = **

**r = ****(s + 2a)**

**r = ****s + a.**

**Furthermore, we know that the triangle formed by the
radius, a side of the square, and half of the other side of the square (1/2s)
is a right triangle. Since we have a right triangle, we can use the Pythagorean
Theorem **** to solve for r.**

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**By definition,
the radius of a circle is a segment from the center of the circle to any point
of the circle. So all radii are equal. As we observe above we have both
equations equal to r, so we can set
them equal and solve for the desired**

** ratio **
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**Now divide both sides by a
**

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**Now solve for the desired ratio**

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**If we multiply by the conjugate **^{}

**We will get the golden ratio**

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**Click here if you would like to explore with the GSP file of the Inscribed square in a half circle.**

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Fletcher’s Home Page
**

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