Construct the square root of 14 using the compass and ruler on a piece of paper by following the instructions given below and keep the construction lines:
Spoiler: Construction Instructions
I highly encourage you to actually try this using a tool like geogebra.
Part I: Constructing the square root of 2.
1: Start with a line segment AB and extend it to be a line.
2: Draw a circle with center A and radius AB. Label the point D as the other point on the circle that intersects the line.
3: Draw a circle with center D and radius BD.
4: Draw a circle with center B and radius BD.
5: Label the two intersection points from steps 4 and 5 as E and F.
6: Connect the line segment EF. Label one of the intersection point on EF with the circle from step 3 as C. Notice that steps 3-6 constructed a line EF that is perpendicular to the line AB.
7: Connect points B and C to form the segment BC. By the Pythagorean Theorem, the hypotenuse of triangle ABC has length equal to the square root of 2 (given the segment AB has length 1).
Part II: Constructing the square root of 3 from the square root of 2.
8: Extend the line segment BC to be a line.
9: Using a process similar to steps 2-6 above, create the line perpendicular to the line BC that intersects BC at the point B. Instructions 9a through 9d give a reminder of the process.
9a: Draw a circle with center B and radius BC. Label the point that intersects the line BC as G.
9b: Draw a circle with center C and radius CG.
9c: Draw a circle with center G and radius CG.
9d: Draw the line segment that intersects the two circles. This line is perpendicular to BC and intersects the point B.
10: Using a compass, measure the length AB. Then draw a circle with radius AB and center B.
11: Mark the point where the circle from step 10 intersects the line segment from step 9 as B2. The line segment should have length AB.
12: Connect points B2 and C to form triangle CBB2. By the Pythagorean Theorem, the line CB2 has length proportional to the square root of 3.
Part III: Constructing the square root of n in general:
13: Construct the square root of n - 1. Note that the point C should be shared by the hypotenuses of each triangle formed up to this point. The vertex of the right angle in the last triangle that constructs the square root of n - 1 is Bn - 3 and the remaining vertex is Bn - 2. (Note that point B should be considered point B1.)
14: Construct the line perpendicular to CBn - 2 that intersects the point Bn - 2. Refer to steps 2-6 or 9 for a reminder of the process.
15: Use the compass to measure the length AB. Draw a circle with radius AB and center Bn - 2.
16: Label the point where the circle from step 15 intersects the line from step 14 as the point Bn - 1.
17: Connect points C and Bn - 1. By the Pythagorean Theorem, this line segment has length proportional to the square root of n.
Sell it as a piece of art.
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In post 113, Ircher wrote:Construct the square root of 14 using the compass and ruler on a piece of paper by following the instructions given below and keep the construction lines:
Spoiler: Construction Instructions
I highly encourage you to actually try this using a tool like geogebra.
Part I: Constructing the square root of 2.
1: Start with a line segment AB and extend it to be a line.
2: Draw a circle with center A and radius AB. Label the point D as the other point on the circle that intersects the line.
3: Draw a circle with center D and radius BD.
4: Draw a circle with center B and radius BD.
5: Label the two intersection points from steps 4 and 5 as E and F.
6: Connect the line segment EF. Label one of the intersection point on EF with the circle from step 3 as C. Notice that steps 3-6 constructed a line EF that is perpendicular to the line AB.
7: Connect points B and C to form the segment BC. By the Pythagorean Theorem, the hypotenuse of triangle ABC has length equal to the square root of 2 (given the segment AB has length 1).
Part II: Constructing the square root of 3 from the square root of 2.
8: Extend the line segment BC to be a line.
9: Using a process similar to steps 2-6 above, create the line perpendicular to the line BC that intersects BC at the point B. Instructions 9a through 9d give a reminder of the process.
9a: Draw a circle with center B and radius BC. Label the point that intersects the line BC as G.
9b: Draw a circle with center C and radius CG.
9c: Draw a circle with center G and radius CG.
9d: Draw the line segment that intersects the two circles. This line is perpendicular to BC and intersects the point B.
10: Using a compass, measure the length AB. Then draw a circle with radius AB and center B.
11: Mark the point where the circle from step 10 intersects the line segment from step 9 as B2. The line segment should have length AB.
12: Connect points B2 and C to form triangle CBB2. By the Pythagorean Theorem, the line CB2 has length proportional to the square root of 3.
Part III: Constructing the square root of n in general:
13: Construct the square root of n - 1. Note that the point C should be shared by the hypotenuses of each triangle formed up to this point. The vertex of the right angle in the last triangle that constructs the square root of n - 1 is Bn - 3 and the remaining vertex is Bn - 2. (Note that point B should be considered point B1.)
14: Construct the line perpendicular to CBn - 2 that intersects the point Bn - 2. Refer to steps 2-6 or 9 for a reminder of the process.
15: Use the compass to measure the length AB. Draw a circle with radius AB and center Bn - 2.
16: Label the point where the circle from step 15 intersects the line from step 14 as the point Bn - 1.
17: Connect points C and Bn - 1. By the Pythagorean Theorem, this line segment has length proportional to the square root of n.
Sell it as a piece of art.
Awful. Not even a hint of colour. You barely manage to make a one.
Ircher wrote:Construct the square root of 14 using the compass and ruler on a piece of paper by following the instructions given below and keep the construction lines:
Spoiler: Construction Instructions
I highly encourage you to actually try this using a tool like geogebra.
Part I: Constructing the square root of 2.
1: Start with a line segment AB and extend it to be a line.
2: Draw a circle with center A and radius AB. Label the point D as the other point on the circle that intersects the line.
3: Draw a circle with center D and radius BD.
4: Draw a circle with center B and radius BD.
5: Label the two intersection points from steps 4 and 5 as E and F.
6: Connect the line segment EF. Label one of the intersection point on EF with the circle from step 3 as C. Notice that steps 3-6 constructed a line EF that is perpendicular to the line AB.
7: Connect points B and C to form the segment BC. By the Pythagorean Theorem, the hypotenuse of triangle ABC has length equal to the square root of 2 (given the segment AB has length 1).
Part II: Constructing the square root of 3 from the square root of 2.
8: Extend the line segment BC to be a line.
9: Using a process similar to steps 2-6 above, create the line perpendicular to the line BC that intersects BC at the point B. Instructions 9a through 9d give a reminder of the process.
9a: Draw a circle with center B and radius BC. Label the point that intersects the line BC as G.
9b: Draw a circle with center C and radius CG.
9c: Draw a circle with center G and radius CG.
9d: Draw the line segment that intersects the two circles. This line is perpendicular to BC and intersects the point B.
10: Using a compass, measure the length AB. Then draw a circle with radius AB and center B.
11: Mark the point where the circle from step 10 intersects the line segment from step 9 as B2. The line segment should have length AB.
12: Connect points B2 and C to form triangle CBB2. By the Pythagorean Theorem, the line CB2 has length proportional to the square root of 3.
Part III: Constructing the square root of n in general:
13: Construct the square root of n - 1. Note that the point C should be shared by the hypotenuses of each triangle formed up to this point. The vertex of the right angle in the last triangle that constructs the square root of n - 1 is Bn - 3 and the remaining vertex is Bn - 2. (Note that point B should be considered point B1.)
14: Construct the line perpendicular to CBn - 2 that intersects the point Bn - 2. Refer to steps 2-6 or 9 for a reminder of the process.
15: Use the compass to measure the length AB. Draw a circle with radius AB and center Bn - 2.
16: Label the point where the circle from step 15 intersects the line from step 14 as the point Bn - 1.
17: Connect points C and Bn - 1. By the Pythagorean Theorem, this line segment has length proportional to the square root of n.
Sell it as a piece of art.
Oh, you thought you could get away with not showing the picture, didn't you?
Spoiler:
And you were right, because I made it instead.
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