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Answer:
1, 2: draw x as the altitude of a right triangle
3: draw x as double the altitude of the right triangle
Explanation:
Consider the right-triangle geometry in the attachment. All of the right triangles are similar, so the ratios of corresponding sides are equal.
short side / long side = x/r = s/x
You may notice this compares to the required relation in problem 1.
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If you multiply this out, you get ...
x² = rs ⇒ x = √(rs)
This compares to the required relation in problem 2.
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So, you append the two line segments to each other so they make segment AB, find their midpoint D (using a perpendicular bisector construction), then draw the (semi-)circle centered at that midpoint. The perpendicular at X will intersect the circle at C, a distance of x from the line rs. (A perpendicular to a line at a point construction is required to create XC.)
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For problem 3, you simply need to double the length √(rs) that you found in the previous constructions. A simple way to do this is to draw a circle, not a semicircle. Then point C' on the lower half of the circle will be 2√(rs) from point C.
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In the attached, we have shown right triangle ABC for discussion purposes above. Segments AC and BC are not needed for your purpose.
Point of clarification: AX = r, BX = s, CX = x.