Required circumscribed spherical surface area

First, find the center and radius of the ball.

The center and radius of a sphere should be on a known element.

First of all, my feeling should be on the hypotenuse BD of Rt△DAB.

Solution: According to the relationship between Rt△DAB edges.

BD=√ 14

It's in △BCD again.

BC^2+CD^2=BD^2

∴△BCD is a right triangle

We know that the length of the midline on the hypotenuse of a right triangle is equal to half the length of the hypotenuse.

∴: The circumscribed spherical center of the triangular pyramid is located at the midpoint of BD.

Length =√ 14/2

∴ The area of the ball =4π* 14/4= 14π.