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A solid right prism  has a height of 16 as shown. Also, its bases are equilateral triangles with side length 12. Points X Y Z   and  are the midpoints of edges  AC, BC, DC respectively. A part of the prism above is sliced off with a straight cut through points  X Y Z. Determine the surface area of solid CXYZ ,the part that was sliced off.

the answer is not 72+9sqrt3

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 Jun 29, 2020
edited by bunnybeer  Jun 29, 2020
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The suface area is 36*sqrt(3) + 48.

 Jun 29, 2020
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The answer was 48+9sqrt3+3sqrt91

bunnybeer  Jun 30, 2020
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\(To determine the surface area of solid $CXYZ,$ we determine the area of each of the four triangular faces and sum them. Areas of $\triangle CZX$ and $\triangle CZY:$ 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 $\triangle CXY:$ This triangle is equilateral with side length $6.$ We draw the altitude from $C$ to $M$ on $XY.$ Since $\triangle CXY$ is equilateral, then $M$ is the midpoint of $XY.$ Thus, $\triangle CMX$ and $\triangle CMY$ 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 $\triangle CXY$ is$$\frac{1}{2}(6)(3\sqrt{3})=9\sqrt{3}.$$Area of $\triangle XYZ:$ We have $XY = 6$ and $XZ = YZ = 10$ and drop an altitude from $Z$ to $XY.$ Since $\triangle XYZ$ 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{align*} ZM &= \sqrt{ZX^2 - XM^2} \\ &= \sqrt{10^2-3^2} \\ &= \sqrt{91}. \end{align*}Since $XY = 6,$ the area of $\triangle XYZ$ is$$\frac{1}{2}(6)(\sqrt{91})=3\sqrt{91}.$$Finally, the total surface area of solid $CXYZ$ is$$24+24+9\sqrt{3}+3\sqrt{91}=\boxed{48+9\sqrt{3}+3\sqrt{91}}.$$\)

bunnybeer  Jun 30, 2020

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