The contents of the three units in this lecture are: graphic rotation, linear motion into a plane, and plane moving body.

Graphic rotation is one of the effective methods to solve geometric problems. When solving geometric problems, the rotation method satisfies the following three conditions: (1) irregular figures are transformed into regular figures by rotation, (2) the edges are equal, and (3) the angles are complementary or complementary. When you see a geometric figure that meets these conditions, you should be able to think of "rotation", which is a kind of thinking and a realm.

Trajectory: points move into lines, lines move into surfaces, and surfaces move into volumes.

60 1, what is the area of the quadrilateral in the first cell 1? Difficulty level ★★★★☆☆☆ Problem-solving ideas This problem can be worked out without rotation, as shown in figure 1, but you should learn to see equilateral and right angles.

Think about how to turn.

As shown in Figure 2 (Delta OAB rotates 90 degrees counterclockwise around O) and Figure 3 (Delta OAC rotates 90 degrees clockwise around O).

Degree). Figure 3, 12× 12 = 144.

A

Figure 2 1 Figure 3

Answer 144.

602. The figure shows the first unit 2. In △ABC, ∠ ABC = 90, AB = 3 and BC = 5. Take AC as one side, make a square outside △ABC with the center of O, and find the shadow area. Difficulty level ★★★★☆☆☆

The idea of solving problems can be solved without rotation.

12

×5×3+

14

×(3+5) = 16。

22

However, to learn how to spin well, Delta △OAB rotates 90 degrees counterclockwise around O, as shown in the figure. The right angle △OBB' is what you want (BCB' is in a straight line, which proves simple). An isosceles right triangle, knowing the length of the hypotenuse, can find the area. 5+3=8,

12

×8×4= 16。

Answer 16.

603, the first unit 3 as shown in the figure, AB =AE =4cm, BC =DC, ∞∠BAE =∠BCD = 90, AC = 10 cm, then S? ABC +S？ ACE +S？ CDE = _ _ _ cm.

2

Difficulty level ★★★★ ☆ ☆ Problem solving ideas △ ABC rotates 90 degrees counterclockwise around C, and△ ABC rotates 90 degrees clockwise around A, all of which turn to the lower part of AC, forming a square below AC, as shown in the figure. You can also turn them upside down to form a square.

Area: 10× 10 ÷ 2 = 50. Answer 50.

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604. The first unit 4, as shown in the figure, makes a right triangle ABE in a square, with the side AB of the square as the hypotenuse, ∠ AEB = 90, and the intersection of AC and BD O. It is known that the lengths of AE and BE are 3cm and 5cm respectively, and the area of △OBE is calculated. Difficulty level ★★★★★☆☆ Problem solving ideas provide two solutions. Solution 1: Suspend = Blank, Rotate

s？ OBE =

1

s？ DBE =[S？ ABD -(S？ ABE +S？ ADE ) ] 22

1

e "

122

s？ ABD =？ (3+5) = 17;

2

△ Abe is easy to find, but△ Ade is not easy to find. Turn △ADE 90 degrees clockwise around point A, and find (S? ABE +S？ ADE) becomes a right-angled trapezoid AEBE', and the area of (3+5) × 3 ÷ 2 = 12.

s？ OBE =

12

×( 17- 12) =2.5。

Solution 2: Chord diagram

See squares and right triangles, think of chord diagrams, and make three other right triangles. △△DBE base be = 5, which is higher than the side length of the central square of the chord diagram, and 5-3 = 2.

s？ OBE =

1

1 1

s？ DBE =×(×5×2) =2.5 .222

2

The answer is 2.5cm.

605. Learning plan 1 The following figure △ABC is an isosceles triangle, AB =AC, ∞∠BAC = 120, △ADE is a regular triangle, point D is on the side of BC, and BD: DC = 2: 3. When the area of △ABC is 50cm, what is the area of △ADE? Difficulty level ★★★★★☆

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2

See isosceles, see 120, think of rotation.

Rotate △ABC and △ADE around point A for two turns (120,240) to connect D, E, D', E', D "and E" to form a regular hexagon.

s？ B CC " =50×3= 150 .

B

B

B

B

s？ ADE =

1

S is a regular hexagon, but some points of the regular hexagon are suspended and the area is difficult to find. You can find the area of △ DD' d ",6.

1 1 1 1

S =S =？ (2S ) S S？ Dd "d" =, so? ADE =S？ DD“D”.？ Regular hexagon regular hexagon

6623

s？ DD " D " =( 1-

2? 377 1

? 3) ? s？ B CC " =？ s？ B CC " =？ 150=42; s？ ADE =×42= 14 .5? 525325

The bird head model is used here, 2× 3: 5× 5.

The regular hexagon in this problem needs to be proved: ∠ CAD' = ∠ bad,

∠EAD′=∠EAC+∠CAD′=∠EAC+∠BAD =∠BAC-∠DAE

= 120-60 = 60, so △EAD' is a regular triangle. The answer is 14cm.

606. Add 1 As shown in the figure, in a rectangle, it is known that the areas of two triangles are 2 and 3 respectively. What is the solution? The area of this place. Difficulty level ★★★★☆☆☆

The problem-solving ideas are displayed as auxiliary lines.

Step 1: Find 3 (butterfly wings);

Step 2: Find 4.5 (the area ratio is 2: 3);

Step 3: Half of the rectangle: 3+4.5 = 7.5,? =7.5-2=5.5。 Answer 5.5.

607. The length of the hypotenuse AB of the right triangle ABC in the second unit 3 is 10 cm, ∠ ABC = 60, BC = 5 cm. Turn △ABC 120 clockwise around point B, and point A and point C reach the positions of point E and point D.. Find the area swept by AC edge, that is, the shaded part of the graph (π takes 3). Difficulty level ★★★★☆☆☆

The thinking of solving problems moves to a plane. Rotate △EBD counterclockwise, and the shadow is the difference between two sectors.

2

120

? π? ( 10-5) =? π? 75=75。 3603

22

1

A

∠ CBA = 60 Page 3/* * Page 7

The answer is 75 square centimeters.

608. The fourth unit is shown in the picture. ABCD is a rectangle with a length of 4, a width of 3 and a diagonal length of 5. It rotates 90 clockwise around point C, and calculates the areas swept by four sides respectively.

Difficulty level ★★★★★☆☆ Problem solving ideas DC: π? 4=4π;

1

2

four

BC :π？ 3=2.25π;

1

2

four

D

2

2

Ad: (1) rotate to (2), π? (5-4) =2.25π;

1

four

2

2

AB: (3) Rotate to (4), π? (5-3) =4π。

1

four

It is found that the areas swept by opposite sides are equal.

Answer AB: 4 π, BC: 2.25 π, CD: 4 π, DA: 2.25 π. D

609, study plan 2 as shown in figure △ABC is an isosceles right triangle with a side length of 1 m. Now, with point C as the center, rotate △ABC 90 degrees clockwise, so the area swept by side AB during rotation is _ _ _ _ square meters. Difficulty level ★★★★★☆

The way to solve the problem, the swept area is not easy to imagine.

After clockwise rotation, point A turns to point A along arc AA', and point B turns along arc BB'.

Point b', point d turns to point d' along arc DD'. Because CD is the point from C to AB.

The line segment is short, so the area swept by AB is the shadow between the arcs baa' and BDD'a' in the figure. S shadow = s semicircle -s blank.

B

1 12

S squared ADCD "= CD = S? ADC +S？ ACD " =S？ ADC +S？ BCD =S？ Abc = × 1× 1 = (square meter).

22S sector DCD '= =? π? CD =？ π? =。

4428

1

2

1 1π

1ππ 13π 12

S shadow =? π? BC-South of DCD-(South? BCD +S？ CA " D " ) = - =-=0.6775 .

228282

Answer 0.6775.

6 10, lesson 3 Three round coins with a radius of 1cm are laid flat on the table side by side, so that a coin of the same size can roll along their outer contours and return to its original position. Then the point that coincides with the origin A is _ _ _ _ _ _, the coin rotates by itself, and the circumference of the center trajectory of the coin is _ _ _ _ _. Difficulty level ★★★★★☆

The solution to the problem is to calculate the length of the trajectory: three semicircles with a radius of 2.

E

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12

? (2π? 2) ? 3=6π;

E

Coin circumference: 2π? 1=2π; 6 π ÷ 2 π = 3, which is 3 weeks. Answer a, 3 weeks, 6π cm.

6 1 1, as shown in the figure, job 3 in quadrilateral ABCD,

∠A =∠C =45，∠ABC = 105，AB =CD = 15

Cm, connecting diagonal BD, ∠ Abd = 30. Seek the quartet

The area of ABCD.

Difficulty level ★★★★★☆☆☆

The idea of solving the problem is to mark the degrees of each angle first.

Found 60 and 30, 75 and 105.

Cut △BCD along BD, flip B and D, and then stick them together to form a graph like the one on the right. The area is easy to find:

12

×15×15 =112.5 (square centimeter).

The answer is 1 12.5 square centimeters.

6 12, task 5, as shown in the figure, if a right triangle rotates around BC, the volume of the cone formed is 16π, if it rotates around AC, the volume of the cone formed is 12π, if it rotates around AB, what is the volume of the geometry formed? Difficulty level ★★★★★☆☆ If BC = A and AC = B, then

? 12

πb a = 16π3

?

? 1πa 2b = 12π3

2

C

? ab =48b 4

After simplification? 2, this equation can be solved, and the two formulas are divided by: = and substituted.

a 3？ a b =36

According to Pythagorean theorem, AB = 5, and the height on AB is:

? a =3

. b =4

3? 45

=2.4。

The volume of the rotator is: π? 2.4? 5=9.6π。

1

2

three

The answer is 9.6π.

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6 13, Supplement 2: First make an equilateral triangle with a side length of 2cm, and then make an arc with three vertices as the center and a radius of 2cm to form a curved triangle (as shown in the left figure). Prepare two more such figures, fix one (the shadow on the right), and the other will roll around it, as shown on the right, starting from the state of vertex connection. Excuse me, what is the area through which this figure rolls?

Square centimeters? (π takes 3. 14)

Difficulty level ★★★★★★★

It is difficult to solve the problem, and the center of the circle has been changed six times. The rolling process and final result are shown in the figure below. B

Rolling area: 3× [Answer 25. 12.

Li Qingzhou

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1

? π? (4-2) +? π? 2]=8π=25. 12(cm ) 66

2

2

1

22

20 13. 1 1.7

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