It is known that there are four points a, B, C and D on the sphere with radius 2. If AB = CD = 2, the maximum volume of the tetrahedral ABCD is There is one thing unclear in the analysis The analysis is as follows: make plane PCD through CD so that ab ⊥ plane PCD intersects AB and P, Set the distance from point P to CD as H, Then v = 1 / 3 × 2 × h × 1 / 2 × 2, When the diameter passes through the midpoint of AB and CD, the maximum h is 2 √ 3, so the maximum V is 4 √ 3 / 3 Why is h the largest when the diameter passes through the midpoint of AB and CD? There's no title or analysis

It is known that there are four points a, B, C and D on the sphere with radius 2. If AB = CD = 2, the maximum volume of the tetrahedral ABCD is There is one thing unclear in the analysis The analysis is as follows: make plane PCD through CD so that ab ⊥ plane PCD intersects AB and P, Set the distance from point P to CD as H, Then v = 1 / 3 × 2 × h × 1 / 2 × 2, When the diameter passes through the midpoint of AB and CD, the maximum h is 2 √ 3, so the maximum V is 4 √ 3 / 3 Why is h the largest when the diameter passes through the midpoint of AB and CD? There's no title or analysis

Let the midpoint of AB be p, the midpoint of CD be q, and the center of the ball be o. it is easy to know that P and Q must be on a sphere with the center of the ball also o but the radius is smaller than o (that is, a concentric sphere with a smaller point), and the radius is R. let the angle between Cd and plane ABQ be a, and let the angle between PQ and ab be B, then there is v_ (ABCD)=(1/3)*S_ (ABQ)*CD...

It is known that there are four points a, B, C and D on the sphere with radius 2. If AB = CD = 2, the maximum volume of the tetrahedral ABCD is ()
A. 233B. 433C. 23D. 833

If the distance from point P to CD is h, then there is v tetrahedron ABCD = 13 × 2 × 12 × 2 × H = 23h. When the diameter passes through the midpoint of AB and CD, Hmax = 222 − 12 = 23, so Vmax = 433