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# Pyramid question

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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. off-topic
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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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 haveXM = 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 is24+24+9\sqrt{3}+3\sqrt{91}=\boxed{48+9\sqrt{3}+3\sqrt{91}}.