Thank you so much sir ❤

Why is it 2 that the external point connects to?

nice one ADTW Study

How do I make a regular pentagon of side 2.5 cm.??

Ok

Can you please explain me the reason that how that first side you get as one fifth of the circumference? I cannot get it don't know which property is used here .. please tell...

@ADTWstudy

7 ай бұрын

Dividing the diameter of a circle into five equal parts, as illustrated, results in a distance between point A and 2-dash that is approximately equivalent to the length of one side of a pentagon. This constructional approach is employed to approximate the general shape of a pentagon. For precise length calculations, the formula 2Rsin36(degree) can be utilized, where R represents the diameter of the outer circle.

can you draw a 7 sided regular polygon with this method?

@ADTWstudy

Жыл бұрын

Yes, divide the horizontal line into 7 parts and draw a vertical line passing through point 2 on the horizontal line...

@lanceraltria

Жыл бұрын

@@ADTWstudy pffffffftttttttt

@radhabai2666

11 ай бұрын

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@wiilli4471

11 ай бұрын

@@lanceraltria No you cannot. It will be an approximation but it will be slightly off. Try it yourself.

@wiilli4471

11 ай бұрын

@@ADTWstudy No you cannot. it will be a close approximation but not correct

NOT a general method. While this does correctly inscribe a triangle, square, or hexagon in a circle, it does not work for other regular polygons. It does produce remarkably close results for a pentagon (.065% angle error), heptagon (.17% error), and octagon (.42% error), but the error continues to grow as the sides increase. By the time it reaches a 25-gon, the error is large enough to not even fit 24 sides in the circle. It's fine if you want to draw an n-gon approximation with a small number of sides, but don't use this on a math test.

@laslaminas

6 ай бұрын

we use it all over spain, not as an exact methow but as aproximate method

@Goofayball

Ай бұрын

So what’s the best method?

@mjeffery

Ай бұрын

@@Goofayball To inscribe an exact regular n-gon in a circle with straight-edge and compass? It's not possible because most regular n-gons cannot be constructed with straight-edge and compass. What you can do is inscribe a regular p-gon into a circle where p is a Fermat prime (a prime of the form 1 + 2^{2^n}), take an inscribed m-gon and n-gon and inscribe an LCM(m,n)-gon, and double the sides of an inscribed n-gon to get an inscribed 2n-gon. Combining the above allows the construction of a k-gon when k = 2ⁿ·p₁·p₂·p₃..., where the pᵢ values are 0 or more distinct Fermat primes. No other n-gon can be constructed. Note that there are only 5 known Fermat primes (3, 5, 17, 257, and 65537). So it's possible to construct an 8-gon (octagon) and a 60-gon, since 8 = 2³ and 60 = 2²·3·5, but not a 7-gon (heptagon) or 9-gon (nonagon) because 7 is a non-Fermat odd prime and 9 = 3² is the result of multiplying two non-distinct Fermat primes (3 with itself). Doubling a regular polygon just involves bisecting a side or angle and intersecting it with the circumscribed circle. Inscribing an LCM(m,n)-gon involves inscribing an m- and n-gon in a circle with a common vertex, then finding a pair of the nearest non-common vertexes between the m- and n-gon and copying that around the circle. Inscribing a Fermat p-gon in a circle can be done by memorizing the known cases-it's unlikely you'll need anything beyond the triangle and pentagon in practice. I'm sure there is a marvelous method to construct any Fermat p-gon which this comment is too narrow to contain.

I HATE YOU MAKING ME DOING SOMETHING WRONG WITH MY WORK