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As shown in the figure, in the square ABCD, ab = 1, point P is a point on the diagonal AC, and AP and PC are used as the diagonal to make the square respectively, then the sum of the perimeter of the two small squares is______ ．
Let the side length of a small square be x, then the side length of a larger square is 1-x, so the sum of the circumference of two small squares = 4x + 4 (1-x) = 4cm
As shown in the figure, in the square ABCD, ab = 1, point P is a point on the diagonal AC, with AP and PC as the diagonal respectively to make the square, and the perimeter of the two sides of the square
The square side length in AP is X. the side length in PC is y.x + y = 1
The sum of two circumferences = 4x + 4Y = 4 (x + y) = 4 * 1 = 4
As shown in the figure, points E and F are on the sides BC and CD of the square ABCD respectively. Given that the perimeter of the triangle ECF is equal to half of the perimeter of the square ABCD, the degree of the angle EAF is calculated
What's easy to prove? No,
Connect AE and AF, and rotate the triangle ADF 90 degrees counterclockwise around point a to get the triangle ABM,
AE=AE
AF=AM
EF = perimeter of triangle ECF - cf-ce
=Half the circumference of square ABCD - cf-ce
=BC+DC-CF-CE
=FD+BE=EB
So the triangle ame is equal to the triangle AFE (SSS),
So angle Mae = angle FAE,
And the angle MAF = angle mAb + angle BAF = angle fad + angle Fab = 90 degrees,
So EAF = 45 degrees
Given that point P is a point in square ABCD, connect PA, Pb and PC. turn △ PAB clockwise 90 ° around point B to the position of △ P skimming point CB
(3) If the square of PA + the square of PC = the square of 2PB, point P must be on the diagonal AC
From the rotation, PP '^ 2 = 2PB ^ 2,
∵PA^2+PC^2=2PB^2,
∴P‘A^2+PC^2=PP’^2,
∴∠PCP‘=90°,
∴∠PCB+∠PAB=90°,
∵∠PBA+∠PBC=90°,
The sum of the inner angles of the two triangles is 360 degrees,
A, P and C are in the same line,
That is, P is on AC
Point P is a point in the square ABCD, connecting PA, Pb and PC. rotate △ PAB clockwise 90 degrees around point B to the position of △ ECB. If PA = 2, Pb = 4, PC = 6, find the diagonal length
Solution tips:
Connecting PE
It is easy to know that △ PBE is an isosceles right triangle, so PE = 4 √ 2,
Because PC = 6, CE = 2, PE = 4 √ 2,
So PC ^ 2 = CE ^ 2 + PE ^ 2
So △ AEB is a right triangle, ∠ PEC = 90 degree
Therefore, BEC = 135 degree
If CF ⊥ be, then CF = EF = √ 2
therefore
CB^2＝CF^2＋BF^2
＝(√2)^2＋(√2＋4)^2
＝20＋8√2
So AC = √ 2 * BC = 2 √ (10 + 4 √ 2)
As shown in the figure, the quadrilateral ABCD is a square, and P is any point in the square. Connect PA and Pb, and rotate △ PAB clockwise around point B to △ P ′ CB. (1) guess the shape of △ PBP ′ and explain the reason; (2) if PP ′ = 22cm, find s △ PBP ′
(1) ∵ △ PAB rotates clockwise around point B to △ P ′ CB, with ∵ BP = BP ′, ∵ ABP = ∵ CBP ′, ∵ ABP + ∵ PBC = 90 degree, ∵ CBP ′ + ∵ PBC = 90 degree, ∵ PBP ′ = ∵ ABC = 90 degree, and ∵ PBP ′ = isosceles right triangle; (2) ∵ PP ′ = 22cm, the distance from point B to PP ′ = 12pp ′ =
Point P is a point in the square ABCD, connecting PA, Pb and PC. rotate △ PAB clockwise 90 ° around point B to the position of △ ECB
If PA = 2PB = 4, PC = 6, find the length of diagonal of square ABCD
Connecting PE, it is easy to know that △ PBE is isosceles right triangle
So PE = √ 2PB = 4 √ 2
Because PC = 6. CE = PA = 2. PE = 4 √ 2
So PC ＾ 2 = CE ＾ 2 + PE ＾ 2
So △ AEB is a right triangle
So ∠ BEC = ∠ PEC + ∠ PEB = 90 ° + 45 ° = 135 °
If CF ⊥ be, then CF = EF = √ 2
therefore
BC＾2=CF＾2+BF＾2
=(√2)＾2+(√2+4)＾2
=20+8√2
BC=2√(5+2√2)
So AC = √ 2 * BC = √ 2 * 2 √ (5 + 2 √ 2) = 2 √ (10 + 4 √ 2)
P is a point in the square ABCD, connecting PA, Pb and PC. rotate △ PAB clockwise 90 around point B to the position of △ ECB, PA = 2PB = 4, PC = 6, find s positive
⊿ BPE isosceles right angle, PE = 4 √ 2 PC & # - 178; = 36 = EC & # - 178; + EP & # - 178; ≠ PEC = 90 & # - 186; ∠ APB = ∠ bef = 45 & # - 186; + 90 & # - 186; = 135 & # - 186; s Square = AB & # - 178; = 2 & # - 178; + 4 & # - 178; + 2 × 2 × 4 × 1 / √ 2 = 20 + 8 √ 2 ≈ 31.314 (area unit)
P is a point inside the square ABCD, and there is a point e outside the square ABCD, which satisfies the following conditions: angle Abe = angle CBP, be = BP
If PA: Pb = 1:2, angle APB = 135 °, calculate AP: AE
Let BP = 2, AP = 1, PBE = 90, EP = root 8, BPE = 45
Because the angle APB is 135 degrees, the angle EPA is 90 degrees, so AE = 3
So AP: AE = 1:3
P is a point inside the square ABCD, P is a point inside the square ABCD, and E is a point outside the square ABCD, satisfying the following conditions: angle Abe = angle CBP, be = BP
PA ratio Pb = 1:2, ∠ APB = 135, find the value of cos ∠ PAE
EBP was isosceles RT △
If ∠ EPB = 45 °, then ∠ ape = 90 °
EP=√2,AP=2 ,AE=√6
cos∠PAE=AP/AE=√6/3
RELATED INFORMATIONS
- 1. It is known that, as shown in the figure, P is a point in the square ABCD, and there is a point e outside the square ABCD, satisfying the angle Abe = angle CBP, be = BP 2.PB⊥BE.
- 2. As shown in the figure, in rectangular ABCD, e is a point on CD, angle DEA = 30 degrees, and AE = AB = 8cm, find 1, degree of angle EBC, 2, area of triangle Abe
- 3. The ratio of the top to the bottom of a trapezoid is 3:5. The top is 1 meter shorter than the bottom, and the bottom is 25% higher. The area of the trapezoid is ()
- 4. As shown in the figure, in the diamond ABCD, ab = 2, ∠ C = 60 °, the diamond ABCD rolls to the right on the straight line l without sliding. Each rotation of 60 ° around a vertex is called an operation. After 36 such operations, the total length of the path that the diamond center o passes through is (the result retains π)______ ．
- 5. In the right angle trapezoid ABCD, AD / / BC, ∠ B = 90 °, ad = 24cm, BC = 26cm, the moving point P moves rapidly from a along the side of ad to D at the speed of 1cm per second In the right angle trapezoid ABCD, AD / / BC, ∠ B = 90 °, ad = 24cm, BC = 26cm, the moving point P starts from a and moves at the speed of 1cm / s along the ad side to D, the moving point Q starts from point C and moves at the speed of 3cm / s along the CB side to B, and P and Q start from points a and C at the same time. When one point reaches the end point, the other point also stops moving?
- 6. As shown in the figure, in the known trapezoidal ABCD, ad ‖ BC, ∠ B = 30 °, C = 60 °, ad = 4, ab = 33, then the length of the bottom BC is______ ．
- 7. As shown in the figure, a, B, C and D are the four vertices of rectangle ABCD, ab = 25cm and ad = 8cm. The moving points P and Q start from point a and C at the same time. Point P moves to point B at the speed of 3cm / s until point B, and point Q moves to point d at the speed of 2cm / S. (1) P and Q, from the start to the second, PQ ‖ ad? (2) Question: P, Q two points, from the start to a few seconds, quadrilateral pbcq area of 84 square centimeters?
- 8. The height of a trapezoid is 10 cm. If the upper bottom is increased by 6 cm and the lower bottom is increased by 4 cm, it becomes a parallelogram. Calculate the area of the original trapezoid
- 9. The top of a trapezoid is three times as big as the bottom. If the bottom is extended by 6 cm, it will become a parallelogram. How many cm are the top and bottom of the trapezoid?
- 10. The area of a parallelogram is 86 square meters, which is the same as the area of a triangle with the same base and height______ ．
- 11. In square ABCD, point E.F is on edge BC and CD respectively. If AE = 4, EF = 3 and AF = 5, what is the area of square ABCD
- 12. Let the vertex a of the square ABCD be the plane ABCD, PA = AB = a, and find the dihedral angle between the plane PAB and the plane PCD
- 13. It is known that PA is perpendicular to the plane of rectangle ABCD, and M is the midpoint of ab. if PD and ABCD form an angle of 45 degrees, it is proved that plane PCM is perpendicular to plane PCD
- 14. What do QA and QC mean?
- 15. As shown in the figure, place a rectangle in the plane rectangular coordinate system, OA = 2, OC = 3, e is the midpoint of AB, and the image of inverse scale function passes through point E and connects with BC Intersection at point F (1) Finding the analytic formula of straight line OB and the analytic formula of inverse scale function (2) Connect of and OE, point P is on the straight line ob, and s triangle AOP = 2S quadrilateral oebf, calculate the coordinates of point P
- 16. As shown in the figure, two pieces of paper of equal width are crossed and overlapped, and the quadrilateral ABCD of the overlapped part is______ Shape
- 17. In the plane rectangular coordinate system, the coordinates of the three vertices of square ABCD are a (0,0), B (- 2,0), D (0,2), then the coordinates of point C are Then the coordinates of the midpoint of the square BC are 2. If point P (a, 3-2a) is on the bisector of the second and fourth quadrants, then the value of a is
- 18. Given that the rectangle ABCD, and ab = 5, ad = 3, establish an appropriate plane rectangular coordinate system, find the coordinates of each vertex of the rectangle
- 19. If the inner angle of the diamond is 120 ° and the length of its diagonal is 12cm, the perimeter of the diamond is The answer is 48CM or 2 √ 3, remember that there are two cases
- 20. As shown in the figure, fold the rectangular ABCD paper along EF so that point d coincides with the midpoint d 'of BC side. If BC = 8, CD = 6, CF=______ ．