SAT Math 2 subject test question?

Refer to the figure below which refers to the pyramid.

In ΔABC:
AB = 8 in
BC = 4/2 in = 2 in

AC² = AB² + BC²  (Pythagorean theorem)
AC = √(8² + 2²) in
Length of an altitude of one of the Δ faces, AC = 8.25 in

The answer: D) 8.25

Use pythagoras theorema     a²+b²=c²
from the center to a side is 2 inch
2²+8²=8.25²
side face altitude is 8.25 inch

What you described would not be a regular pyramid, as only one of its five faces is a regular polygon. However, it could be a right pyramid. In that case, all lateral faces would be isosceles triangles. The distance from the pyramid vertex to the midpoint of one of the base edges would then be one altitude of a lateral face.

Let the pyramid vertex be V.
Let the center of the base be O.
Let M be the midpoint of one base edge.

Length VM is an altitude of a lateral face.

OV = 8 in
OM = 2 in
∠VOM = 90°

VM² = OV² + OM²
VM² = (8 in)² + (2 in)²
VM² = 68 in²
VM = √(68) in ≈ 8.25 in

A diagram should convince you that there is a right triangle whose hypotenuse is the desired altitude.  The legs of this triangle are 2 and 8 inches.  So, the Pythagorean Theorem gives √(2^2 + 8^2) = √68 = about 8.25 (D).