I'm using my brain now, but I could be wrong... but that triangle is only one of four possible angles inside the sphere on the north pole say. There would be four corresponding directions towards its south pole. A triangle in two dimensions has to be split into eight parts inside the sphere. No?
The red line is \(sqrt{3.14}/2=0.8860022573334675\) units in length. The pink line is 0.5 units in length. The area of that triangle is 0.2215005643333669 square units.
Interesting. Let's pretend you are correct on that. You and I already agreed the distance between the centers of the two circles was \(1.0000000000000000\), so the length of the pink line is half of that or \(Pink=0.5000000000000000\), and we can apply the Pythagorean Theorem to solve for the length of the blue line which you refuse to answer: \(Blue^2 = Red^2 + Pink^2\) \(Blue = sqrt{Red^2 + Pink^2}\) \(Blue = sqrt{0.8860022573334675^2 + 0.5000000000000000^2}\) \(Blue = sqrt{0.7850000000000000 + 0.2500000000000000}\) \(Blue = 1.0173494974687902\) So, you saying the length of the blue line is \(Blue = 1.0173494974687902\) even though the radius of the circles is only \(R=1.0000000000000000\). To bad it is impossible to fit a line of length \(Blue = 1.0173494974687902\) between the center point and the edge of the circle, if the radius of the circle is only \(R=1.0000000000000000\). Nice work, MD.Please Register or Log in to view the hidden image!
What do you think the area of that triangle is? Nice work in advance! Please Register or Log in to view the hidden image!
Sure. Start with a line that is 1.017 long and a circle of radius 1.000 and then squeeze the line in between the center point and the edge of that circle. Once you get the line to fit between those points, we can talk about area.
Which triangle? The equilateral triangle with sides of length 1 with area \(\frac{\sqrt{3}}{4}\) or the right triangle which is half of the equilateral triangle, with hypotenuse equal to 1 and base equal to \(\frac{1}{2}\) and which therefore has area \(\frac{\sqrt{3}}{8}\) ? http://lmgtfy.com/?q=area of equilateral triangle
The correct way to solve the problem is to start with \(Blue=1.0000000000000000\) because we know it is the radius of the circle. Then we can apply Pythagorus to solve for the length of the red line, instead of 'proclaiming' it to be \(sqrt{3.14}/2\) by decree, as you have done. \(Blue^2 = Red^2 + Pink^2\) \(Red^2 = Blue^2 - Pink^2\) \(Red = sqrt{Blue^2 - Pink^2}\) \(Red = sqrt{1.0000000000000000^2 - 0.5000000000000000^2}\) \(Red = sqrt{1.0000000000000000 - 0.2500000000000000}\) \(Red = sqrt{0.7500000000000000}\) \(Red = 0.8660254037844386\) And you can solve for the area of the triangle as follows: \(area_{triangle} = (Red * Pink) / 2\) \(area_{triangle} = (0.8660254037844386 * 0.5000000000000000) / 2\) \(area_{triangle} = 0.4330127018922193 / 2\) \(area_{triangle} = 0.2165063509461096\) Also note that the lengths of the sides of the square \(L_{square}\) are not what you claimed they are: \(L_{square} = 2*Red\) \(L_{square} = 2*0.8660254037844386\) \(L_{square} = 1.7320508075688772\)
If the diagram is correct, and the height of the square is \(sqrt{3.14}\) then each quarter of the square has .785 square units of area, for a total of 3.14 square units of area for the entire square. The top right square has a bottom length of \(sqrt{3.14}/2\), and the bottom line of that triangle shown is .5 units in length, which leaves \(0.3860022573334675*sqrt{3.14}/2\) for the other part of the 1/4 square along the same axis that .5 is measured.. That means that part of the area is 0.3419988713332662 square units of area. That leaves only 0.4430011286667338/2 for that triangle, in order to total the .785 square units for that 1/4 of the entire square.
Still waiting for those area answers... Should I come back tomorrow and check to see if you've figured it out yet? The crickets are starting to get rather annoying. Please Register or Log in to view the hidden image!
The blue side is of length 1, the pink side has length 1/2. It is known the red and the pink sides form a right angle. Thus the area is known without guessing what the length of the red side is. http://lmgtfy.com/?area of a right triangle with hypotenuse given \(a = \frac{1}{2} \\ c = 1 \\ A = \frac{1}{2} a \sqrt{c^2 - a^2} = \frac{1}{2} \times \frac{1}{2} \times \sqrt{1^2 - \left( \frac{1}{2} \right)^2} = \frac{1}{4} \sqrt{\frac{4}{4} - \frac{1}{4} } = \frac{1}{4} \sqrt{\frac{3}{4} } = \frac{1}{4} \times \frac{\sqrt{3}}{\sqrt{4}} = \frac{1}{4} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8} \approx 0.21651\) Not "0.2215005643333669". Thus, your assertion that the length of the red side is of length \(\frac{\sqrt{3.14}}{2}\) in [post=3222557]post #62[/post] is incompatible with the diagram as drawn. The fault was clearly explained in [post=3222271]post #8[/post], [post=3222462]post #19[/post], and [post=3222479]post #22[/post]. You can't assert the height of the square is \(\sqrt{3.14}\), the OP asked for a construction. The only lines that you drew in [post=3222223]post #8[/post] that come close to potentially explaining the height of the square are the central circle and the circle to the right of it which appear to have the relationships observed which gives the height of the square as \(\sqrt{3}\). If you meant to explain something different, it was explained to you in posts [post=3222462]#19[/post], and [post=3222479]#22[/post] that the burden was on you to explain how your diagram satisfies the OP's request for a geometric construction of \(\sqrt{3.14}\). Typically, this will require that you describe your steps and explain how you know the resulting square side is \(\sqrt{3.14}\) times the length of the radius of the circle. But as the OP indicated and as reinforced in [post=3222151]post #6[/post] and [post=3222515]post #40[/post], it's not possible to square the circle with straightedge and compass, we are just trying to construct approximations, just like 3.14 is just an approximation (and not a very good one) for π. Obviously the area of the equilateral triangle is twice this areas and \(2 A = \frac{\sqrt{3}}{4}\) just like you were told in [post=3222574]post #68[/post].
rpenner, Your best bet is to take a hint from Neddy and sign off and figure it out, before I make you look foolish. You don't get it!
A hypothetical triangle with sides of 1, 0.5, √3.14/2 = \(1, \; \frac{1}{2}, \; \frac{\sqrt{314}}{20}\) has area: \(A = \frac{1}{4} \sqrt{ 2(ab)^2 + 2(ac)^2 + 2(bc)^2 - a^4 - b^4 - c^4 } = \frac{\sqrt{ 2 \times 20^2 \times 10^2 + 2 \times 20^2 \times 314 + 2 \times 10^2 \times 314 - 20^4 - 10^4 -314^2 }}{1600} = \frac{\sqrt{125404}}{1600} = \frac{\sqrt{31351}}{800} \approx 0.2213\) This is smaller than \(\frac{\sqrt{3.14}}{8} \approx 0.2215\) which is the area of a right triangle with legs \(\frac{1}{2}\) and \(\frac{\sqrt{3.14}}{2}\), therefore as you have been told, your claim doesn't match your diagram.
Do you agree that 1/4 of the area of the total square is .785 square units of area? Please don't avoid my questions, it makes you look bad.
The (Blue) Square WXYZ has side length √3.14 times the length of AB. The line CD has length √3 times the length of AB. Thus Motor Daddy's construction (as interpreted and labeled by me) doesn't produce the desired edge length. The difference is small, but as geometric construction and proofs based on it, when done properly is as precise as algebra, we see that Motor Daddy has failed to carry his burden. The separation between line WX and line OC is only about 1/50th of the distance between A and B, so becomes readily apparent when the lines are thinner than 1/100th of the distance between A and B. Motor Daddy's lines are too thick and too crudely drawn to know if √3.14 or √3 is the side length of the red square. But there is nothing in the construction which allows the ratio 1:√3.14 to be constructed from the radius AB. Full Image: View attachment 7323 Zoomed on point C: View attachment 7324 The difference between √3.14 and √π, if shown, would require zooming in much further.
Setup: Starting with DE, we extend the line and make a circles at D and E with radius DE getting new points K and L. At point K we make a circle with radius DK getting new point J. At point L we make a circle with radius EL getting new point M. At point J we make a circle with radius JE getting new point C. At C we make a circle with radius CL getting new point B. At M we make a circle with radius KM getting new point N. At point N we make a circle with radius KN getting new point F. Finally from both points L and B draw circles which radius BL and label the intersections of he circles as points Q and R which are on the perpendicular line through C. View attachment 7325 Payoff: Draw a circle through B with radius BE getting new point A on the line DE but also getting new point H on the upper part of line QR. Draw a circle through D with radius AD getting new point G on the line DE but also getting new point I on the lower part of line QR. Then AC:HI are in the ratio 1Please Register or Log in to view the hidden image!√(3/2)+√(3/10)) and so the Blue square HIOP is less than 1/65269th larger than the red circle with radius AC. View attachment 7326
