jmc

To determine the surface area of solid $CXYZ,$ we determine the area of each of the four triangular faces and sum them.

Areas of $△CZX$ and $△CZY:$

Each of these triangles is right-angled and has legs of lengths 6 and 8; therefore, the area of each is $\frac{1}{2}(6)(8)=24$.

Area of $△CXY:$

This triangle is equilateral with side length $6.$ We draw the altitude from $C$ to $M$ on $XY.$ Since $△CXY$ is equilateral, then $M$ is the midpoint of $XY.$

Thus, $△CMX$ and $△CMY$ are $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangles. Using the ratios from this special triangle, $$CM=\frac{\sqrt{3}}{2}(CX)=\frac{\sqrt{3}}{2}(6)=3\sqrt{3}.$$Since $XY=6,$ the area of $△CXY$ is $$\frac{1}{2}(6)(3\sqrt{3})=9\sqrt{3}.$$Area of $△XYZ:$

We have $XY=6$ and $XZ=YZ=10$ and drop an altitude from $Z$ to $XY.$ Since $△XYZ$ is isosceles, this altitude meets $XY$ at its midpoint, $M,$ and we have $$XM=MY=\frac{1}{2}(XY)=3.$$By the Pythagorean Theorem, $$\begin{array}{rl}ZM& =\sqrt{Z{X}^2-X{M}^2}\\ & =\sqrt{10^2-3^2}\\ & =\sqrt{91}.\end{array}$$Since $XY=6,$ the area of $△XYZ$ is $$\frac{1}{2}(6)(\sqrt{91})=3\sqrt{91}.$$Finally, the total surface area of solid $CXYZ$ is $$24+24+9\sqrt{3}+3\sqrt{91}=\boxed{48+9\sqrt{3}+3\sqrt{91}}.$$