For lovers of 3D geometry - 2 (Area Calculations)

Let $\triangle ABC$ be the cross-section of the pyramid, and label the points and segments as follows:

By the Pythagorean Theorem on $\triangle AMC$ , $l = \sqrt{h^2 + (\frac{s}{2})^2} = \frac{1}{2}\sqrt{4h^2 + s^2}$ .

By the angle bisector theorem, $\frac{AC}{AB} = \frac{CP}{BP}$ , or $\frac{l}{s} = \frac{l - k}{k}$ , which solves to $k = \frac{sl}{s + l}$ .

Since $\triangle BNP \sim \triangle BMC$ by AA similarity, $\frac{NP}{BP} = \frac{CM}{BC}$ , or $\frac{h_2}{k} = \frac{h}{l}$ , which solves to $h_2 = \frac{hk}{l}$ . Also, $\frac{NP}{BN} = \frac{CM}{BM}$ , or $\frac{h_2}{p} = \frac{h}{\frac{s}{2}}$ , which solves to $p = \frac{sh_2}{2h}$ .

By the Pythagorean Theorem on $\triangle ANP$ , $j = \sqrt{h_2^2 + (s - p)^2}$ .

Now we'll transfer the variables to the three-dimensional picture of the the lower pyramid:

The total surface area of the lower pyramid is the sum of the area of the square base, the area of the two congruent side triangles, and the areas of the large and small trapezoids, or $A_{\text{pyramid}} = A_{\text{square}} + 2A_{\text{triangle}} + A_{\text{large trapezoid}} + A_{\text{small trapezoid}} = s^2 + 2 \cdot \frac{1}{2}s\sqrt{p^2 + h_2^2} + \frac{1}{2}j(s + s - 2p) + \frac{1}{2}k(s + s - 2p)$ , or:

$A_{\text{pyramid}} = s^2 + s\sqrt{p^2 + h_2^2} + j(s - p) + k(s - p)$ .

In this question, $s = 12$ and $h = 10$ , so:

$l = \frac{1}{2}\sqrt{4h^2 + s^2} = \frac{1}{2}\sqrt{4 \cdot 10^2 + 12^2} = 2\sqrt{34}$

$k = \frac{ls}{l + s} = \frac{12 \cdot 2\sqrt{34}}{12 + 2\sqrt{34}} = -204 + 36\sqrt{34}$

$h_2 = \frac{hk}{l} = \frac{10(-204 + 36\sqrt{34})}{2\sqrt{34}} = 180 - 30\sqrt{34}$

$p = \frac{12 \cdot (180 - 30\sqrt{34})}{2 \cdot 10} = 108 - 18\sqrt{34}$

$j = \sqrt{h_2^2 + (s - p)^2} = \sqrt{(180 - 30\sqrt{34})^2 + (12 - (108 - 18\sqrt{34}))^2} = 12 \sqrt{578 - 99 \sqrt{34}}$

$A_{\text{pyramid}}$

$= s^2 + s\sqrt{p^2 + h_2^2} + j(s - p) + k(s - p)$

$= 12^2 + 12\sqrt{(108 - 18\sqrt{34})^2 + (180 - 30\sqrt{34})^2} + (12 \sqrt{578 - 99 \sqrt{34}})(12 - (108 - 18\sqrt{34})) + (-204 + 36\sqrt{34})(12 - (108 - 18\sqrt{34}))$

$\approx \boxed{360.1}$ .

First, we'll label the vertices of the solid.

All we have to find are the coordinates of point F, and for that, we have to construct the equations of the three planes on which it lies. Assume point $A$ is the origin of the $xy$ plane, then point $B = (12, 0, 0)$ . The angle between the triangular faces of the original pyramid and its base is $\theta_1 = \tan^{-1} \frac{10}{6} = \tan^{-1} \frac{5}{3}$ , while the angle between the cutting plane and the base is $\theta_2 = \frac{1}{2} \theta_1$

Therefore, the equations of the three planes meeting at point $F$ are,

Plane $FAB$ : $(0, - \sin \theta_1, \cos \theta_1) \cdot (x, y, z) = 0$

Plane $FBC$ : $(\sin \theta_1, 0, \cos \theta_1) \cdot (x - 12, y, z ) = 0$

Plane $FAD$ : $(-\sin \theta_2 , 0, \cos \theta_2 ) \cdot (x, y, z) = 0$

Solving the above system of $3$ equations in the $3$ unknowns results in,

$F = (x, y, z) = (8.957134107, 3.042865893, 5.071443155)$

From symmetry, it follows that point $E$ coordinates are given by

$E = (8.957134107, 12 - 3.042865893, 5.071443155) = (8.957134107, 8.957134107, 5.071443155)$

We also have,

$A = (0,0,0) , B = (12, 0, 0) , C = (12, 12, 0) , D = (0, 12, 0)$

The total area = $[ABCD] + 2 [FAB] + [AFED] + [BCEF]$

$[ABCD] = 12^2 = 144$

$[FAB] = \frac{1}{2} | B \times F |$

$[AFED] = \frac{1}{2} ( | F \times E | + | E \times D | )$

$[BCEF] = \frac{1}{2} ( | (C - B) \times (E - B) | + | (E - B) \times (F - B) | )$

Performing the above cross products and taking the magnitudes of the resulting vectors and adding it all up, we get

Total Area = $360.14361 \approx 360.1$