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, Qin, 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 oblique square to a large oblique square, send it to the middle oblique square, take half of the remainder after subtraction, multiply it by the square and send it to the one obtained above. The remainder after subtraction is divided by 4, and the number is "real", and 1 is "angle". After the square root, you get the area.

The so-called "real" and "angle" refer to the equation px.

2=q, 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{a^2*c^2-[(a^2+c^2-b^2)/2

]^2}

When P= 1

2=q，

△=√ 1/4{a^2*c^2-[(a^2+c^2-b^2)/2

]^2}

factoring

△

^2= 1/ 16[4a^2c^2-(a^2+c^2-b^2)^2]

= 1/ 16[(c+a)

^2-b

^2][b^

2-(c-a)^

2]

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

= 1/ 16(c+a+b)(a+b+ c-2b)(b+ c+a-2a)(b+ a+c-2c)

= 1/ 16

[2p(2p-2a)(2p-2b)(2p-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".

s=√ 1/4{a^2*c^2-[(a^2+c^2-b^2)/2

]^2}

C & gtb & gta.

According to Helen formula, it can be extended to the area operation of quadrilateral. The following questions:

It is known that quadrilateral ABCD is a quadrilateral inscribed in a circle, AB=BC=4, CD=2, and DA=6. Find the area of quadrilateral ABCD.

Here we use the generalization of Helen's formula.

S circle inscribed quadrilateral =

Under the radical symbol (p-a)(p-b)(p-c)(p-d).

(where P is the half circumference and A, B, C and D are the four sides)

Substituting into the solution results in s=8√.

three

Proof (3)

In △ABC, ∠A, ∠B and ∠C correspond to sides A, B and C..

O is the center of its inscribed circle, R is its inscribed circle radius, and P is its half circumference.

There are tana/2tanb/2+tanb/2tanc/2+tanc/2tana/2 =1.

r(tanA/2 tanB/2+tanB/2 tanC/2+tanC/2)= r

r =(p-a)tanA/2 =(p-b)tanB/2 =(p-c)tanC/2

∴

r(tanA/2 tanB/2+tanB/2 tanC/2+tanC/2)

=[(p-a)+(p-b)+(p-c)]tanA/2 tanb/2 tanc/2

=ptanA/2tanB/2tanC/2

=r

∴p^2r^2tanA/2tanB/2tanC/2=pr^3

∴s^2=p^2r^2=(pr^3)/(tana/2tanb/2tanc/2)

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

∴S=√p(p-a)(p-b)(p-c)