Question 247799
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Insufficient evidence.  You don't say anything limiting about the shape of the triangular base.  The only requirement for a prism to be right triangular is that the base be some sort of triangle.  It could be a right triangle with legs of 6 and 1 (largest possible perimeter, roughly 13.1), or it might be an isosceles triangle with a base of 6 and sides *[tex \LARGE \sqrt{10}] (smallest possible perimeter, roughly 12.3), hence the perimeter of the base can be anything in the range *[tex \LARGE 6 + 2\sqrt{10}] to *[tex \LARGE \sqrt{37}]


The formula for the surface area of a right prism of any shape is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ SA\ =\ 2\,\cdot\,B\ +\ P\,\cdot\,h]


Where *[tex \LARGE B] is the area of the base, *[tex \LARGE P] is the perimeter of the base, and *[tex \LARGE h] is the height of the prism.


You could say that the surface area is in the range:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\ +\ 14\left(6\ +\ 2\sqrt{10}\right)\ \leq\ SA\ \leq\ 6\ +\ 14\sqrt{37}]


but that is all you can do for sure.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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