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It is obtained by rotating the abc around point A by 15° counterclockwise.

bab.=15°

c=30°，∠b=90°

bac=60°

b.=90°

ab.= Set ab=

is in . ab.²+

i.e. x + x = 2 times the square of the root number 2.

x>0x=2ab.=

ab=2∠c=30°

ac=2ab=4

bc²=ac²-ab²

i.e. bc = 16-4

bc = 2 times root number 3

Circumference of ABC = AB+BC+AC

2+2 times root number 3+4

6+2 times root number 3

∠bab'= 15 degrees, b'ad=60-15=45 degrees, and b'=45 degrees, ab'd is the rt isosceles triangle.

and ad=2 times the root2 ab'=2=ab (hypotenuse is 2 times the root of the right-angled side) ac=2 2=4 bc=2 times the root 3=2 times the perimeter of the root 3 abc is equal to =ab+bc+ac=4+2+2 root3 perimeter = 6+2 root3

bab'=15° so b'ad=45° ad=2 root number 2 so ab'=b'd = 2 acb = 30° so ac= 4 b'c = 2 root number 3

1.Point E is on AB, point F is on BC, let the coordinates of point E (A, 4) and the coordinates of point F (5, B).

then a=k 4 b = k 5

Coordinates of point e (k 4, 4) coordinates of point f (5, k 5).

2.The perpendicular x-axis of eh intersects with h

Area of triangle OEF = area of quadrilateral AEHO + area of trapezoidal EHCF - area of triangle AOE - area of triangle FCO.

ae*ao+1/2(ah+cf)ch-1/2ae*ao-1/2oc*cf

k+1/2(4+k/5)(5-k/4)-1/2k-1/2k

1/2(4+k/5)(5-k/4)

Area of the triangle BEF = 1 2BE*BF = 1 2(5-K 4)(4-K 5).

s = area of the triangle OEF - area of the triangle BEF.

1/2(4+k/5)(5-k/4)-1/2(5-k/4)(4-k/5)

1/2(5-k/4)(4+k/5-4+k/5)

k/5(5-k/4)

(100-20k+k²-100)/20=5-(10-k)²/20

To make s largest, then (10-k) 20 is minimum, so when k=10, s has a maximum value of 5

3 Since DG is a crease, so DG bisects EO perpendicularly, and the intersection point is Q and EC is parallel to the Y axis, so the triangle EQM is a triangle OQD, then EDOM is a parallelogram.

And because it's folded in half, od=de

then EDOM is diamond-shaped.

Since m(x,y) then the coordinates of point n are (x,0) and the coordinates of point e are (x,4).

then om=em=4-y

mn²+on²=om²

The functional relationship between y and x is y=-1 8x +2

(1) Point e: is when y=4, x=k 4. So point e(k 4,4)point f:

Yes when x=4, y=k 4. So the coordinates of point F (4,K4)(2)S = area of the rectangular OABC - area of the triangle OAE - area of the triangle OFC - 2 times the area of the triangle BEF.

4*5-1/2*4*k/4-1/2*5*k/4-2*1/2*(5-k/4)(4-k/4)

20-k/2-5k/8-(5-k/4)(4-k/4)=-k^2/16+9k/8

1 16*(k 2-18k)=-1 16[(k-9) 2-81]When k=9, the above equation has a maximum value, and the maximum value is 81 16(3)(1) is a diamond.

Solution: Root number 3 2, original formula = 2 - root number 3 + root number 3 = 2

1 a(-10,12), with the coordinate origin o as the center of the position, and the coordinates of the corresponding point a of the similarity ratio of 1:2 point a are: (-5,6),(5,-6) 2

Solution: In the rectangular ABCD, OA=ob, AOB=60°, AOB is an equilateral triangle, AB=OB, ABO=AOB=60°, AE is the bisector of BAD, AB=BE, AB=BE=BE=OBC=90°-60°=30°, BOE=1 2(180°-30°)=75°

aoe=∠aob+∠boe=135°

Hope it helps, hope. Good luck with your studies.

The rectangle can be obtained, ao=ab=bo, angle bae+angle eao+angle oad=90, angle eao=45-30=15

Since the inner misalignment angles are equal, the angle bea = 45

be=ab=bo=ao

Isosceles triangle.

Let the angle AEO be x

180= 45+2x+60+15

x = 30 angle aoe = 180-15-30

Original = (50-49) * (50 + 49) + (48-47) * (48 + 47) + ...Wei Zhi + (2 + 1) refers to Duan Ming* (2-1).

50+49+48+47+……Burning as +2+1

Hope, thank you.

Original lead Lu Xiang formula = (50 + 49) (50-49) + (48 + 47) (48-47) +1+2)(2-1)

50 (50+1) Huai Bo Xi Qi 2