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I was reviewing when I came upon a problem that requires me to construct a line with length $\sqrt7R$, where $R$ is the radius of a given circle.

I can do a line with a $\sqrt7$ unit length just fine but this one just confuses me.

Hope you guys can point me in the right direction.

Reese
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  • Take for example a right triangle $\triangle ABC$ of sides $(1,\sqrt6,\sqrt7)$ and multiply the leg $1=\overline{BC}$ by $R$ so you get a new vertex $C'$. Tracing parallel to the side $\overline{AC}$ passing by $C'$ you get the vertex $A'$. As a result you have $\overline{A'C'}=R\sqrt7$ – Piquito Sep 23 '16 at 17:59
  • "I can do a line with a 7–√ unit length just fine" That's the exact same thing. How was your "unit" length determined? Supposedly from an arbitrary line segment declared to equal "1". Line segment~ circle with radius... same thing . Set your compass at the center and arc to the circle. That's a distance of R or "1". Construct a segment of length $\sqrt 7$ times it. – fleablood Sep 23 '16 at 18:57

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Hint: think Pythagoras and remember:

$$1^2 + 1^2 = (\sqrt 2)^2$$ $$(\sqrt 2)^2 + 1^2 = (\sqrt 3)^2$$ $$(\sqrt 3)^2 + 2^2 = (\sqrt 7)^2$$

Using a straightedge and a compass you can bisect (and double) line segments and construct right angles.

If you're required to start from a circle (without that centre being marked), then stay calm, draw any two non parallel chords. The perpendicular bisectors of these chords will meet at the centre, and once you've found the centre you can immediately draw the diameter line (and you know both radius and diameter).

Deepak
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