The right angle vertex P of the triangle ruler slides on the ray OM, and the two right angle sides intersect OA and ob at point CD respectively Conjecture the quantitative relationship between PC and PD

The right angle vertex P of the triangle ruler slides on the ray OM, and the two right angle sides intersect OA and ob at point CD respectively Conjecture the quantitative relationship between PC and PD

Through P, PE ⊥ ob in E, PF ⊥ OA in F, PE = PF can be easily obtained from the property of bisector, and then ≌△ dep can be proved by ASA by proving ≌△ dep by ASA
PC=PD
Through P, PE ⊥ OB to e, PF ⊥ OA to F, respectively,
∴∠CFP=∠DEP=90°,
∵ OM is the bisector of AOB,
 PE = PF, (7 points)
∵ 1 + ∠ FPD = 90 °, (right angle triangle plate)
And ∵ AOB = 90 °,
∴∠FPE=90°,
∴∠2+∠FPD=90°,
∴∠1=∠2,
In △ CFP and △ dep
 {∠CFP=∠DEP   PE=PF   ∠1=∠2,
∴△CFP≌△DEP（ASA）,
∴PC=PD．

The angle AOB = 90 degrees, OM bisects the angle AOB, and moves the vertex P of a right angle on the ray OM, and the two sides are respectively connected with the edges OA and o If both sides intersect OA and ob respectively with point C, is PC and PD equal? Please explain why

equal
Because the OM bisector angle AOB, according to the characteristics of the angular bisector (PC vertical OA, PD vertical OB)

As shown in the figure, ∠ AOB = 90 ° and OM bisect ∠ AOB. Move the vertex P of the right triangle plate on the ray OM, and the two right angles intersect with OA and ob at points c and D respectively. Is PC equal to PD? Try to explain the reason

PC is equal to PD. The reasons are as follows: pass through point P as PE ⊥ OA at point E, PF ⊥ ob at point f. ∵ om bisection ∵ AOB, point P on OM, PE ⊥ OA, PF ⊥ ob, ∵ PE = PF (the distance from the point on the bisector to both sides of the corner is equal) and ? AOB = 90 °, PEO = ∠ PFO = 90 ° and ∵ Ep

As shown in the figure, ∠ AOB = 90 ° and OM bisect ∠ AOB. Move the vertex P of the right triangle plate on the ray OM, and the two right angles intersect with OA and ob at points c and D respectively. Is PC equal to PD? Try to explain the reason

PC is equal to PD
Pass point P as PE ⊥ OA at point E, PF ⊥ ob at point F
∵ om bisection ∵ AOB, point P on OM, PE ⊥ OA, PF ⊥ ob,
﹣ PE = PF (the distance from the point on the bisector to both sides of the corner is equal)
And ? AOB = 90 °, PEO = ∠ PFO = 90 °,
The quadrilateral oepf is rectangular,
∴∠EPF=90°，
∴∠EPC+∠CPF=90°，
And ∵ cpd = 90 °,
∴∠CPF+∠FPD=90°，
∴∠EPC=∠FPD=90°-∠CPF．
In △ PCE and △ PDF,
A kind of
∠PEC＝∠PFD
PE＝PF
∠EPC＝∠FPD ，
∴△PCE≌△PDF（ASA），
∴PC=PD．

As shown in the figure, ∠ AOB = 90 ° and OM bisect ∠ AOB. Move the vertex P of the right triangle plate on the ray OM, and the two right angles intersect with OA and ob at points c and D respectively. Is PC equal to PD? Try to explain the reason

PC is equal to PD. The reasons are as follows: pass through point P as PE ⊥ OA at point E, PF ⊥ ob at point f. ∵ om bisection ∵ AOB, point P on OM, PE ⊥ OA, PF ⊥ ob, ∵ PE = PF (the distance from the point on the bisector to both sides of the corner is equal) and ? AOB = 90 °, PEO = ∠ PFO = 90 ° and ∵ Ep

As shown in the figure, ∠ AOB = 90 °, AC ‖ ob, OA = 10, AB is the arc with point o as the center, BC is the arc with point a as the center, and calculate the area of shadow part There's no way to paint

The area of arcuate AB is s' = π · 10? 2 / 4-10 × 10 △ 2 = 25 π - 50
Shadow area s = π· (10 √ 2) 2 / 8-s' = 125 π - 25 π + 50 = 100 π - 50

The radius OA and ob of circle P and sector OAB intersect at C and D respectively, and intersect with arc AB at point E. given OA = 15 and angle AOB = 60 degrees, the area of shadow part in the graph is calculated Circle P is in the sector. Find the sector minus the area of the circle. Circle P intersects with arc ab

Let the radius of circle p be r, then OP = 2R, OE = 3R = OA = 15, so r = 5, and the area of circle P is 25 π
The sector area is 225 π / 6 = 75 π / 2, so the shadow area is 25 π / 2

As shown in the figure, there are two semicircles in the sector AOB, and the area of the shaded part can be obtained by calculating the area of the sector AOB which is equal to 90 ° and OA which is equal to 4 cm

If AOB is equal to 90 °, it means that the sector is a quarter circle, OA = 4cm and diameter = 4cm
4 / 2 = 2cm2 * 2 is the square of 2, 2 * 2 * 3.14 / 2 * 2 = 12.56cm2
/2 is because there are two semicircles
The answer is 12.56

As shown in the figure, the center angle of the fan-shaped OAB is 90 ° with OA and ob as the diameters, and the semicircle is made in the fan. P and Q represent the areas of the two shadow parts respectively. Then the size relationship between P and Q is () A. P=Q B. P＞Q C. P＜Q D. Can't be sure

∵ the center angle of sector OAB is 90 ° and the sector radius is a,
The fan area is 90 × π × A2
360=πa2
4，
The semicircle area is: 1
2×π×（a
2）2=πa2
8，
∴SQ+SM =SM+SP=πa2
8，
∴SQ=SP，
That is, P = Q,
Therefore, a

The fan-shaped AOB with 90 ° center angle and COD are stacked together (as shown in the figure) to connect AC and BD. if OA = 3cm, OC = 1cm, calculate the area of shadow part synchronously Who can help me find the picture of this problem Synchronous practice 34, ask for pictures, can you send a picture?

You first draw a circle on the paper. The fixed point of the compass is point O, and then any point on the circle you draw is point A. connect OA and then make point B at an angle of 90 degrees. Then draw COD in fan-shaped AOB. The compass sheet is 1cm. In fan-shaped AOB, the fan-shaped area is the shadow part