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# Regular pyramid with a square base example

Question:

P-ABCD is a regular pyramid with square base. Its height is 6. The base side is 4. If all vertices lie on a sphere, what is the surface area of the sphere? Solution:

Because P-ABCD is a regular pyramid, so, PA = PB = PC = PD. Because the base plane ABCD is a square, so, AB = BC = CD = DA. The height PO is perpendicular to the base plane ABCD. The projection of the point P on the base plane is the point O and the point O is the center of the square ABCD. Because AO lies on the base plane ABCD, so, PO is perpendicular to AO.

If all vertices of the regular pyramid lie on a sphere, then the center of the sphere will be on the height PO. Let O1 be the center of the sphere. O1A = R = O1P

O1O = PO - PO1 = 6 - R
AC = 4 (square root of 2)
AO = AC/2 = 2 (square root of 2)

Because PO is perpendicular to AO and O1 is a point on PO, so, OO1 is perpendicular to AO. So, triangle AO1O is a right triangle.

In right triangle AO1O,
O1O2 + AO2 = AO12
(6 - R)2 + (2 square root of 2)2 = R2
36 - 12R + R2 + 8 = R2
36 - 12R + 8 = 0
44 - 12 R = 0
12R = 44
R = 44/12 = 11/3

The surface area of the sphere is:

S = 4 pi R2
= 4 pi × (11/3)2
= 4 pi × 121/9
= (484/9) pi
= 168.86

Therefore, the surface area of the sphere is 168.86. Watch the video for more details.