As shown in the figure, in the triangular pyramid p-abc, plane ABC is perpendicular to plane APC, ab = BC = AP = PA = radical 2, angle ABC = angle APC = 90 degrees Q: how to find the area of triangle PBC?

Because angle ABC = angle APC = 90 degrees
So AC = 2, PC = radical 2 (Pythagorean theorem)
Crossing point P and point B to make AC perpendicular to point O (congruent proof)
Because plane ABC is perpendicular to plane APC
So Pb = root 2
PC = BC = BP = radical 2
S triangle PBC = root 3 / 2

In the triangular pyramid p-abc, PA ⊥ plane ABC, ∠ BAC = 90 °, ab = AC = AP = 2, D is the midpoint of AB, e is the midpoint of BC, then the distance from point d to line PE is equal to______ ．

Because PA ⊥ plane ABC, AC ⊂ plane ABC, so PA ⊥ AC, and ab ⊥ AC, and PA ∩ AB = a, so AC ⊥ plane PAB, because D and E are the midpoint of AB and BC respectively, so de ∥ AC, so de ⊥ plane PAB, because PD ⊂ plane pad, so PD ⊥ De, that is, △ PDE is a right triangle, because AB = AC = AP = 2, so de = 1, PD = 5, PE = 6, let DF be the distance from point d to line PE, Then in the right angle △ PDE, from the equal area, we can get: DF = PD · depe = 306, so the distance from point d to straight line PE is equal to 306, so the answer is: 306

As shown in the figure, in the triangular pyramid p-abc, PA ⊥ plane ABC, ∠ BAC = 60 °, PA = AB = AC = 2, e is the midpoint of PC, and the cosine value of the angle formed by the straight lines AE and Pb on the different planes is calculated

Take the midpoint F of BC and connect EF, ∵ E and F are the midpoint of PC and BC respectively, ∵ EF ∥ Pb ∧ AEF is the angle formed by the out of plane straight line AE and Pb. ∵, ∧ BAC = 60 °, ab = AC = 2, ∧ ABC is positive △ AF = 3; ∵ PA ⊥ plane ABC, ∧ PA ⊥ AC, PA ⊥ abpb = 22, EF = 2; in RT △ PAC, PA = AC = 2

As shown in the figure, in the triangular pyramid p-abc, PA ⊥ plane ABC, ∠ BAC = 60 °, PA = AB = AC = 2, e is the midpoint of PC
(1) The cosine value of the angle between AE and Pb
(2) Find the volume of a-ebc
Take the midpoint F of BC and connect EF and AF, then EF ‖ Pb, so ∠ AEF or its complementary angle is the angle formed by the out of plane straight line AE and Pb. ∵ BAC = 60 °, PA = AB = AC = 2, PA ⊥ plane ABC,
∴AF=√3 AE=√2 EF=√2
cos∠AEF=2+2−3/（2×√2 ×√2）=1/4

This is the application of cosine theorem in the solution of triangle

As shown in the figure, in the triangular pyramid p-abc, plane ABC is perpendicular to plane APC, ab = BC = AP = PA = radical 2, angle ABC = angle APC = 90 degrees Q: how to find the area of triangle PBC?