A math problem for junior high school and two years to solve

<> analysis: according to the S trapezoidal ABGF + S ABC-S CGF, and then according to the trapezoidal and triangular area formula, the area of the shadow part can be described, by CG=BC+BG, AB=BC=CD=AD, EF=FG=GB=BE, after the same amount of substitution, the area of the shadow part can be introduced

Solution: Square ABCD and square EFGB, AB=BC=CD=AD, EF=FG=GB=BE, the side length of the square ABCD is 2, S AFC=S trapezoidal ABGF+S ABC-S CGF

1/2 ×（fg+ab）×bg + 1/2 ×ab×bc - 1/2 ×fg×cg

1/2 ×（fg+ab）×bg + 1/2 ×ab×bc - 1/2 ×fg×（bc+bg）

1/2 ×fg^2 + fg + 2 - fg - 1/2 ×fg^2

Therefore, the answer is: 2, that is, the option is A

Countless] team to answer for you, hope o( o

Connect the FB quadrilateral EFGB as a square.

fba=∠bac

FB AC ABC and AFC are triangles of the same base and equal height.

2S ABC=S positive ABCD, S positive ABCD=2 2=4S=2, so choose A

I choose A<>

There are two graphs, but there is one plot point: a, f, and c cannot form a triangle.

Subtract the area of the three triangles from the area of the rectangle.

Uh, why is there a picture again?

…Whatever I think about AFC is a straight line

2.C= and a a = b b = c c = k a c = c c = k c = ac

And a=a k c=c k a=k c let k = 2 and c = 2, then a = 8

a＞b＞c∴a₁＞b₁＞c₁

a=8, b=6, c=4 a=4 b=3 c =23, there are two triangles like this.

Because the triangle ABC is similar to the triangle A1B1C1

a/a₁=b/b₁=c/c₁=k=2 ∴a=2a₁ b=2b₁ c=2c₁

b=a1,c=b1

a=2b₁ b₁=2c₁

Substituting it into a a = b b = c c the original formula becomes 4b 2b = 2b b = 2c c = 2

The presence of the triangle abc and the triangle a1b1c1 is such that k = 2

1 Because abc is similar to the triangle a1b1c1 and the similarity ratio is k, a=ka1, b=kb1, c=kc1

And because of c a1, we get a ka1 kc

2 Triangle abc, a=6, b=10, c=12, triangle a1b1c1, a1=3, b1=5, c1=63 can't be done.

a+(bc-a) (a+b +c)(a,b,c are not equal to each other), it is found that if the value of this equation does not change when a and b are exchanged, then.

a+(bc-a²)/（a²+b²+c²）=b+(ac-b²)/（b²+a²+c²）

By dividing and removing the denominator, we get a(a + b + c ) + bc-a ) = b (b + a + c ) + ac-b )

Finishing, get (a-b)(a +b +c -a-b-c)=0

and a+b+c=1, so the above equation becomes (a-b)(a +b +c -1)=0, because a, b, c are not equal to each other, then a-b≠0, so a +b +c -1=0, so a +b +c =1

So A+(BC-A) (A+B+C)=A+BC-A

When a and b are swapped again, the value of this formula does not change; If a and c are exchanged, the value of this formula remains the same, so a+bc-a =b+ac-b =c+ba-c

So a+bc-a = (a+bc-a +b+ac-b +c+ba-c) 3

2a+2bc-2a²+2b+2ac-2b²+2c+2ba-2c²)/6

2a+2b+2c-3a²-3b²-3c²+a²+b²+c²+2bc+2ac+2ba)/6

2(a+b+c)-3(a²+b²+c²)+a+b+c)²]/6

0 so. This constant value is 0

According to the title:

m=a+(bc-a) (a +b +c )=b+(ac-b) (a +b +c )=c+(ba-c) (a +b +c), so (a-b)[1-(a+b+c) (a +b +c )]=0, (b-c)[1-(a+b+c) (a +b +c )]=0, (c-a)[1-(a+b+c) (a +b +c )]=0, and a+b+c=1, so.

a-b)[1-1 (a +b +c )]=0, (b-c)[1-1 (a +b +c )]=0, (c-a)[1-1 (a +b +c )]=0, and a, b, c are not equal to each other, so 1-1 (a +b +c )=0, a +b +c =1, so m=a+(bc-a)=b+(ac-b)=c+(ab-c), 3m=a+(bc-a)+b+(ac-b)+c+(ab-c)+ab+bc+ca-(a +b +c )=ab+bc+ca。

Again(a+b+c) =a +b +c +2(ab+bc+ca), so 2(ab+bc+ca) = (a+b+c) -a +b +c )=0, ab+bc=ca=0, so 3m=0, m=0.

Therefore, m=a+(bc-a) (a+b +c)=0.

Exchange a and b a+(bc-a2) (a2+b 2+c 2) = b+(ac-b 2) (a 2+b 2+c 2).

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

Since a, b, and c are not equal to each other, they are divided by a-b

a^2+b^2+c^2=a+b+c

then ab+bc+ca=((a+b+c) 2-a 2+b 2+c 2) 2=(1-1) 2=0

Original formula = A+(BC-A2) 1=A+BC-A 2=B+AC-B 2=C+Ab-C2 (swap A & B, swap A & C).

a+b+c+bc+ac+ab-a^2-b^2-c^2)/3=(1+0-1)/3=0

The constant value is 0

As long as another a=b=c=1 3, the requirements can be satisfied. So this invariant is 1 3

Does A+ (BC-A) do the molecule or only BC-A do the molecule?

Because m n 2, m m nm 2m, multiply m by m in the same way; n²n＝mn＋2n

The above two equations are added to give :m n =2mn+2(m+n) so m 2mn n =2(m+n).

In this case, only 2 (m+n) is required

Subtract the known two formulas to yield: m -n = n 2-m 2 = n-m So: (m + n) (m -n) = n -m

i.e. m+n=-1

So the original formula = 2 (m + n) = -2

（1）d=|4|Root number (16 9 + 1) = 12 5 (2) |2k|Root number (1+k squared - 2k+1) = |k|k = 1 + root number 3 or 1 - root number 3

3) The ortho-complement angle of tan abc = -4 3 = 2 tan aco (1-tan aco squared) tan aco = 2 or -1 2 (round).

So c(-2,0) or (2,0) so ac:y=-2x+4 so the distance d=4 root number 5 5

Let y1 mx, y2 n(x 2).Then y (m n) x 2n When x = -1, y = 2 ∴m＋n）－2n＝2……When x=2, y=5.

2（m＋n）－2n＝5……Solve the system of equations to obtain: m 5 2, n 3 2 The functional relationship between y and x is y x 3

Untie; The hypotenuse c=20cm, the circumference is 48cm, and the sum of the right-angled sides is a+b=28

and a 2+b 2=c 2=400 (a+b) 2-2ab=a 2+b 2=c 2=400

2ab=28^2-400=384 s=ab/2=96

Let the length of the two right-angled sides be a,b

a+b+20=20*2+8

a^2+b^2=20^2

Solve the equation!!

My process is not standardized, you better sort it out yourself.

The circumference is 48, and the sum of the two right-angled sides is 28, which can be divided into 12+16, and the square of 12 plus the square of 16 is equal to the square of 20, which is a right-angled triangle that meets the conditions, so the area is 12*16 2, that is, 96

Let the two right-angled sides of the right-angled triangle be long-length x, y, respectively, and get:

xy/2) -x+y) )2 = x^2+y^2

x 2 y 2 4) -xy(x+y) +x+y) 2 = x 2 + y 2 (perfect syd squared formula).

x 2 y 2 4) -xy(x+y) +x 2 + y 2 + 2xy = x 2 + y 2 (perfectly squared selling pants Zheng formula).

x 2 y 2 4) -xy(x+y) +2xy = 0 (shift pure dry).

x 2 y 2 - 4xy(x+y) +8xy = 0 (the left and right sides of the equation are multiplied by 4 at the same time).

xy - 4(x+y) +8 = 0 (both sides of the equation are divided by xy).

xy - 4(x+y) +16 = 8 (add 8 to both sides of the equation).

x-4) y-4) =8 (cross multiplication).

For x,y to be positive integers, both (x-4) and (y-4) must be positive integers, and since 8=1 8=2 4, (1) x=1+4=5, y=8+4=12;

2) x=2+4=6，y=4+4=8.

To solve this problem, the three sides of a right triangle are: a, b, and the root number (a 2 + b 2).

It can be known from the inscription of Qinfu Shed that 1 2AB=A+B+Root Number (A 2+B 2)

That is: the root number (a 2 + b 2) = 1 2ab - (a + b), square its two sides to obtain a 2 + b 2 = 1 4a 2b 2-ab (a + b) + a 2 + b 2 + 2 ab

1/4a^2b^2-ab(a+b)+2ab=0

ab(1 4ab-a-b+2)=0, because ab≠0, therefore, 1 4ab-a-b+2=0

ab-4a-4b+8=0

a(b-4)=4b-8

a=(4b-8)/(b-4)=(4b-16+8)/(b-4)=4+8/(b-4)

Since the three sides of a triangle are integers, and 8 has four divisors: 1, 2, 4, and 8, there are four ways to take the value of b. Namely: 5, 6, 8, 12. They are discussed separately below:

1, b = 5, a = 12, at this time, the hypotenuse is 13, the area is 1 2 * 5 * 12 = 30, the perimeter is: 5 + 12 + 13 = 30, in line with the requirements.

2. b = 6, a = 8, the hypotenuse is 10, and the area and perimeter are 24.

3. b = 8 and a = 6, which is essentially the same as the situation in 2 above.

4. b = 12, a = 5, which is exactly the same as in 1.

To sum up, there are two cases for triangles that meet all requirements.

The trilateral chambers are:

8. Brigade MengliangLet the two knowing sides be a and b, and according to the circumference equal to the area, a+b-1 4ab=2 is obtained, and b=4-(8 (4-a)) is introduced.