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.