Helen formula proves

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Reply: legend fan- Beginner in Jianghu Level 2 2-20 15:32

Sabc= 1/2*b*c*sinA

= 1/2*b*c*( 1-cosa^2)^0.5

cosA=(b^2+c^2-a^2)/2ac

Sabc=((p-a)(p-b)(p-c)p)^0.5

Defendant: lm 2222222222- Trainee Magician Level 3 2-20 15:36.

It's detailed here. See for yourself.

References:

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1 The building was delivered correctly.

okay

Now I really miss the days when I used to study math ~

Respondent: Qiao Yu 05- Assistant Level 2 2-20 15:38

Qin, a mathematician in the Song Dynasty in China, also put forward the "Tridiagonal Quadrature". It is basically the same as Helen's formula. In fact, there is already a formula for finding a triangle in "Nine Chapters of Arithmetic", which is "base times half height". When actually measuring the land area, it is not easy to find because the land area is not triangular. So they thought of three sides of a triangle. In this case, it is much more convenient to find the area of the triangle. But how do you find the area of a triangle according to the length of three sides? Until the Southern Song Dynasty, Jiu Shao, a famous mathematician in China, put forward "Tridiagonal Quadrature".

Qin called the three sides of a triangle small, medium and large. "Art" is the method. Tridiagonal quadrature is to add a small diagonal to a large diagonal, send it to the diagonal, take half of the remainder after subtraction, multiply it by a large diagonal and send it to the one obtained above. After subtraction, the number obtained by dividing the remainder by von 4 is regarded as "real", 1 is regarded as "angle", and the square root is used to obtain the area.

The so-called "real" and "angle" mean that in the equation px 2=qk, p is "angle" and q is "real". δ, A, B and C are used to represent triangle area, large dip angle, middle dip angle and small dip angle, so

q = 1/4[c2a 2-(c % | 2+a2-B2/2)2]

When p = 1, △ 2 = q,

s△= √{ 1/4[c2a 2-(C2+a2-B2/2)2]}

factoring

1/ 16[(c+a)2-B2][b62-(c-a)2]

= 1/ 16(c+a+b)(c+a-b)(b+ c-a)(B- c+a)

= 1/8S(c+a+b-2b)(b+ c+a-2a)(b+ a+c-2c)

=p(p-a)(p-b)(p-c)

From this, we can get:

S△=√[p(p-a)(p-b)(p-c)]

Where p= 1/2(a+b+c)

This is completely consistent with Helen's formula, so this formula is also called "Helen-Qin formula".

Trigonometric function proof

Different from Helen's original proof in Metrica, here we use trigonometric formula and formula deformation to prove it. Let the diagonals of three sides A, B and C of a triangle be A, B and C respectively, then the cosine theorem is

cosC = (a^2+b^2-c^2)/2ab

S= 1/2*ab*sinC

= 1/2*ab*√( 1-cos^2 C)

= 1/2*ab*√[ 1-(a^2+b^2-c^2)^2/4a^2*b^2]

= 1/4*√[4a^2*b^2-(a^2+b^2-c^2)^2]

= 1/4*√[(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)]

= 1/4*√[(a+b)^2-c^2][c^2-(a-b)^2]

= 1/4 *√[(a+b+c)(a+b-c)(a-b+c)(-a+b+c)]

Let p=(a+b+c)/2.

Then p = (a+b+c)/2, p-a = (-a+b+c)/2, p-b = (a-b+c)/2, p-c = (a+b-c)/2,

The above formula = √ [(a+b+c) (a+b-c) (a-b+c)/16]

=√[p(p-a)(p-b)(p-c)]

Therefore, the area of the triangle ABC is S=√[p(p-a)(p-b)(p-c)].

Interviewee: A new generation of old people-Dusi Band 6 2-20 15:49

Helen's formula has also been translated into Helen's formula. Legend has it that this formula was discovered by Xilun II, the ancient king of ancient Sula, and the triangle area was calculated by using three sides of the triangle. However, according to the book published by Maurice Klein 1908, this formula was actually discovered by Archimedes and published under the name of Tosilon II.

Suppose there is a triangle with sides a, b and c, and the area s of the triangle can be obtained by the following formula:

S=\sqrt{s(s-a)(s-b)(s-c)}

And s in the formula:

s=\frac{a+b+c}{2}

Since any polygon with n sides can be divided into n-2 triangles, Helen's formula can be used as a formula to find the polygon area. For example, when measuring the land area, you don't need to measure the height of the triangle, just measure the distance between two points, and you can easily deduce the answer.

[edit] proof

Different from Helen's original proof in Metrology, here we use trigonometric formula and formula deformation to prove it. Let the diagonals of three sides A, B and C of a triangle be A, B and C respectively, then the cosine theorem is

\ cos(c)= \frac{a^2+b^2-c^2}{2ab}

So there is

sin(c)= \sqrt{ 1-\cos^2(c)} = \ frac { \sqrt{-a^4 -b^4 -c^4 +2a^2b^2 +2b^2c^2 +2c^2a^2} } { 2ab }

So the area s of the triangle is

s = { 1 } { 2 } ab \ sin(C)

= \frac{ 1}{4}\sqrt{-a^4 -b^4 -c^4 +2a^2b^2 +2b^2c^2 +2c^2a^2}

= \sqrt{s(s-a)(s-b)(s-c)}

The last equal sign part can be obtained by factorization.

Given that the three sides of a triangle are A, B and C, the area of the triangle is:

△ = s(s-a)(s-b)(s-c) under the radical sign, where s= 1/2(a+b+c).

This formula is called [Heron formula].

Helen's formula appears in her book Geodesy. This formula has always been attributed to Helen. But Vander Waals Deng supported Bell's view that this formula was actually discovered by Archimedes [287- 2 12]. However, there is a proof in Helen's theodolite and measurement.

Qin (1022- 12 1), a great mathematician in China, used the same formula as Helen's formula in the second topic of volume 5 of Shu Jiu Zhang (1247), and its significance is as follows.

δ =1/4 {a2b2-{(a2+b2-c2)/2] 2} under the radical sign.

This formula is equivalent to Helen formula.

Responder: magic Ian 4869- Trainee Magician Level 2-20 16:5 1

Of course you can,

Respondent: Shangshang Card-Assistant Level 2-20 19: 15.

Hello, Li Liuyingxue!

I hope my answer can help you!

Suppose there is a triangle with sides a, b and c, and the area s of the triangle can be obtained by the following formula:

S=\sqrt{s(s-a)(s-b)(s-c)}

And s in the formula:

s=\frac{a+b+c}{2}

[edit] proof

\ cos(c)= \frac{a^2+b^2-c^2}{2ab}

So there is

sin(c)= \sqrt{ 1-\cos^2(c)} = \ frac { \sqrt{-a^4 -b^4 -c^4 +2a^2b^2 +2b^2c^2 +2c^2a^2} } { 2ab }

So the area s of the triangle is

s = { 1 } { 2 } ab \ sin(C)

= \frac{ 1}{4}\sqrt{-a^4 -b^4 -c^4 +2a^2b^2 +2b^2c^2 +2c^2a^2}

= \sqrt{s(s-a)(s-b)(s-c)}

The last equal sign part can be obtained by factorization.

Given that the three sides of a triangle are A, B and C, the area of the triangle is:

△ = s(s-a)(s-b)(s-c) under the radical sign, where s= 1/2(a+b+c).

This formula is called [Heron formula].

Qin (1022- 12 1), a great mathematician in China, used the same formula as Helen's formula in the second topic of volume 5 of Shu Jiu Zhang (1247), and its significance is as follows.

δ =1/4 {a2b2-{(a2+b2-c2)/2] 2} under the radical sign.

This formula is equivalent to Helen formula.

Interviewee: I know everything about astronomy and geography-manager level 5 2-2 1 15: 17.

/Article_Show.asp？ ArticleID=206 1

/view/ 1279.htm

Suppose there is a triangle with sides a, b and c, and the area s of the triangle can be obtained by the following formula:

S=\sqrt{s(s-a)(s-b)(s-c)}

And s in the formula:

s=\frac{a+b+c}{2}

[edit] proof

\ cos(c)= \frac{a^2+b^2-c^2}{2ab}

So there is

sin(c)= \sqrt{ 1-\cos^2(c)} = \ frac { \sqrt{-a^4 -b^4 -c^4 +2a^2b^2 +2b^2c^2 +2c^2a^2} } { 2ab }

So the area s of the triangle is

s = { 1 } { 2 } ab \ sin(C)

= \frac{ 1}{4}\sqrt{-a^4 -b^4 -c^4 +2a^2b^2 +2b^2c^2 +2c^2a^2}

= \sqrt{s(s-a)(s-b)(s-c)}

The last equal sign part can be obtained by factorization.

Given that the three sides of a triangle are A, B and C, the area of the triangle is:

△ = s(s-a)(s-b)(s-c) under the radical sign, where s= 1/2(a+b+c).

This formula is called [Heron formula].

Qin (1022- 12 1), a great mathematician in China, used the same formula as Helen's formula in the second topic of volume 5 of Shu Jiu Zhang (1247), and its significance is as follows.

δ =1/4 {a2b2-{(a2+b2-c2)/2] 2} under the radical sign.

This formula is equivalent to Helen formula.

Interviewee: According to the law-4 2-2 1 15:45.

The area formula of any triangle (Helen formula) = √ p (p-a) (:sp-b) (p-c), p = a+b+c/2, a.b.c, which are three sides of the triangle.

Prove:

Prove Pythagorean Theorem

Analysis: Starting with the most basic calculation formula of triangle S△ABC = aha, Helen's formula is deduced by Pythagorean theorem.

Proof: As shown in figure ha⊥BC, according to Pythagorean theorem, we get:

x = y =

Ha = = =

∴ S△ABC = aha= a× =

At this time, S△ABC is deformation ④, so it is proved.

Certificate 2: Smith Theorem

Analysis: On the basis of the first proof, ha can be obtained directly by using Smith theorem.

Smith theorem: take any point d on the BC side of △ABC,

If BD=u, DC = v and AD = t, then

t 2 =

Proof: from the first proof, u = v =

∴ ha 2 = t 2 = -

∴ S△ABC = aha = a ×

This is the deformation ⑤ of S△ABC, so it is proved.

Proof 3: Cosine Theorem

Analysis: According to the deformation 2S =, it is proved by cosine theorem c2 = a2+b2 -2abcosC.

Prove: Prove S =

It is necessary to prove that S =

= ab×sinC

At this time, S = ab×sinC is a triangular calculation formula, so it is proved.

Certificate 4: Identity

Analysis: consider using S△ABC =r p, because there is the radius of the inscribed circle of a triangle, consider applying the identity of trigonometric function.

Identity: If ∠ A+∠ B+∠ C = 180χ, then

tg tg + tg tg + tg tg = 1

Proof: As shown in the figure, tg = ①.

tg = ②

tg = ③

According to this identity, we get:

+ + =

① ② ③ Substitute, and you get:

∴r2(x+y+z) = xyz ④

As shown in the figure, A+B-C = (X+Z)+(X+Y)-(Z+Y) = 2x.

∴x = in the same way: y = z =

Substituting into ④, we get: r 2 =

Multiply the two sides to get:

r 2 =

On both sides of the square, we get: r =

On the left is r = r p = s △ ABC, and on the right is the deformation of Helen formula, so it is proved.

Prove 5: Half Angle Theorem

Half-angle theorem: tg =

tg =

Proof: According to TG = ∴ R =× Y ①

Similarly, r = × z2r = × x③.

①× ②× ③, so r3 = ×xyz.

Heron's formula

Helen formula has also been translated into Helen formula, Hailong formula, Hero formula and Helen-Qin Jiushao formula. The legend was discovered by Heron II, the ancient king of Syracuse, and the triangle area was calculated by using three sides of the triangle. However, according to the textual research of Maurice Klein published in 1908, this formula was actually discovered by Archimedes and published under the name of Tosilon II (not verified). Qin, a mathematician in the Song Dynasty in China, also put forward the "triclinic quadrature method", which is basically the same as Helen's formula.

Suppose there is a triangle with sides a, b and c, and the area s of the triangle can be obtained by the following formula:

S=√[p(p-a)(p-b)(p-c)]

P in the formula is half circumference:

p=(a+b+c)/2

——————————————————————————————————————————————

Note: "Metrica" ("Metrology") manuscript uses S as the semi-circumference, so

S=√[p(p-a)(p-b)(p-c)] and S=√[s(s-a)(s-b)(s-c)] are all possible, but usually P is used as the semi-circumference.

——————————————————————————————————————————————

Proof (1):

cosC = (a^2+b^2-c^2)/2ab

S= 1/2*ab*sinC

= 1/2*ab*√( 1-cos^2 C)