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[parent] compass and straightedge construction of square (Algorithm)

One can construct a square with sides of a given length $ s$ using compass and straightedge as follows:

  1. Draw a line segment of length s. Label its endpoints $ P$ and $ Q$.

    \begin{pspicture}(-3,-1)(4,1) \rput[a](-2,0.04){.} \psline[linecolor=blue](-2,0)... ...dots(-2,0)(2,0) \rput[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \end{pspicture}
  2. Extend the line segment past $ Q$.

    \begin{pspicture}(-3,-1)(4,1) \rput[a](-2,0.04){.} \rput[r](3.5,0){.} \psline(-2... ...dots(-2,0)(2,0) \rput[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \end{pspicture}
  3. Erect the perpendicular to $ \overrightarrow{PQ}$ at $ Q$.

    \begin{pspicture}(-3,-2)(4,5) \rput[b](2,-2){.} \rput[a](2,4.9){.} \rput[r](3.5,... ...dots(-2,0)(2,0) \rput[a](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \end{pspicture}
  4. Using the line drawn in the previous step, mark off a line segment of length $ s$ such that one of its endpoints is $ Q$. Label the other endpoint as $ R$.

    \begin{pspicture}(-3,-2)(4,5) \rput[b](2,-2){.} \rput[a](2,4.9){.} \rput[r](3.5,... ...](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \rput[b](2.2,4.2){$R$} \end{pspicture}
  5. Draw an arc of the circle with center $ P$ and radius $ \overline{PQ}$.

    \begin{pspicture}(-3,-2)(4,5) \rput[l](-2.6946,3.94){.} \rput[b](2,-2){.} \rput[... ...](-2.2,-0.2){$P$} \rput[a](2.2,-0.2){$Q$} \rput[b](2.2,4.2){$R$} \end{pspicture}
  6. Draw an arc of the circle with center $ R$ and radius $ \overline{QR}$ to find the point $ S$ where it intersects the arc from the previous step such that $ S \neq Q$.

    \begin{pspicture}(-3,-2)(4,5) \rput[l](-2.6946,3.94){.} \rput[b](2,-2){.} \rput[... ...a](2.2,-0.2){$Q$} \rput[b](2.2,4.2){$R$} \rput[b](-2.2,4.2){$S$} \end{pspicture}
  7. Draw the square $ PQRS$.

    \begin{pspicture}(-3,-2)(4,5) \rput[l](-2.6946,3.94){.} \rput[b](2,-2){.} \rput[... ...a](2.2,-0.2){$Q$} \rput[b](2.2,4.2){$R$} \rput[b](-2.2,4.2){$S$} \end{pspicture}

This construction is justified because $ PS=PQ=QR=QS$, yielding that $ PQRS$ is a rhombus. Since $ \angle PQR$ is a right angle, it follows that $ PQRS$ is a square.

If you are interested in seeing the rules for compass and straightedge constructions, click on the link provided.



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Cross-references: compass and straightedge constructions, right angle, rhombus, intersects, point, radius, center, circle, arc, line, erect the perpendicular, endpoints, line segment, straightedge, compass, length, sides, square
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This is version 2 of compass and straightedge construction of square, born on 2007-06-24, modified 2007-06-25.
Object id is 9671, canonical name is CompassAndStraightedgeConstructionOfSquare.
Accessed 2109 times total.

Classification:
AMS MSC51-00 (Geometry :: General reference works )
 51M15 (Geometry :: Real and complex geometry :: Geometric constructions)

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