X^2+4y^2=36 the lenght of the major axis is?
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Check the picture below.
so, to get the area of the triangles, we can simply run a perpendicular line from the top to the base, and end up with a right-triangle with a base of 22 and a hypotenuse of 34, let's find the altitude.
so then the surface area of the triangular prism is,
![\bf \stackrel{\textit{left and right}}{2(34\cdot 76)}~~+~~\stackrel{\textit{bottom}}{(44\cdot 76)}~~+~~\stackrel{\textit{front and back}}{2\left[\cfrac{1}{2}(44)(\sqrt{672}) \right]} \\\\\\ 8512~~+~~(44)(\sqrt{672})\qquad \approx\qquad 9652.61036291978](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bleft%20and%20right%7D%7D%7B2%2834%5Ccdot%2076%29%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Bbottom%7D%7D%7B%2844%5Ccdot%2076%29%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Bfront%20and%20back%7D%7D%7B2%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%2844%29%28%5Csqrt%7B672%7D%29%20%20%5Cright%5D%7D%0A%5C%5C%5C%5C%5C%5C%0A8512~~%2B~~%2844%29%28%5Csqrt%7B672%7D%29%5Cqquad%20%5Capprox%5Cqquad%209652.61036291978)
so, to get the area of the triangles, we can simply run a perpendicular line from the top to the base, and end up with a right-triangle with a base of 22 and a hypotenuse of 34, let's find the altitude.
so then the surface area of the triangular prism is,