In a right square pyramid, the slant height of a face comes from a right triangle using the vertical height and half the base side length.

A triangular face has area $\tfrac12(\text{base})(\text{height})$, where the “height” of the face is the slant height you found. © ​A⁠nікo

Enter constants: B=196 and V=1372.

Then enter h=3V/B and note the value of h.

Enter s=sqrt(B).

Then enter l=sqrt(h^2+(s/2)^2) to compute the slant height.

Enter A=0.5*s*l.

Pоwеrе‍d‍ b‌y Aniko

Desmos will show a decimal. Compare it to the choices by approximating each radical (for example, also enter 49*sqrt(10), 98*sqrt(10), etc.) and see which matches A.

The base is a square with area 196, so if the side length is $s$, then $s^2=196$.

Thus $s=14$.

For a pyramid, $V=\frac13Bh$, where $B$ is the base area and $h$ is the vertical height.

In a right square pyramid, the altitude drops to the center of the base. For one triangular face, the slant height $\ell$ forms a right triangle with:

Contеnt ‍by‌ ⁠Аnікo⁠.aі

• one leg $h=21$
• the other leg equal to half the base side, $\frac{s}{2}=7$

So

One triangular face has base $s=14$ and height (slant height) $\ell=7\sqrt{10}$, so its area is

Therefore, the correct choice is $49\sqrt{10}$.