On this topic, we can think of it this way: if all the vertices of a cube are on a sphere, you will definitely do it. The key is to find its center.

The same topic is no exception. Let's first find out the center of the triangle ABC (that is, the three points of △ABC are equal in distance).

Passing point A makes AD perpendicular to BC, and makes the distance of AD 2, then DB=BC=2, which is its center. Then we can find the center of the triangular prism. Because the height is 2, we raise the D point vertically by 1 to the E point, and then the E point is the center of the triangular prism, that is, the center of the ball. Then the distance of EA is the radius of the ball, so AE = √ 65438. +2? =√5=R

And the surface area of the ball is 4πR? =20π