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the answer is 63.7. My calculations are much bigger than the correct answer, I may have not understood the question correctly quiet possibly.

finding height of the triangle: 1/2*2*3*(sin65)=3 volume of prism: 1/2*5.5*3^2=25 volume: lwh = 3.5*5.5*5.5=106

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    $\begingroup$ You'd need to show your calculation in order for us to spot any possible errors. $\endgroup$ – Angina Seng Jun 14 '19 at 5:29
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    $\begingroup$ where has the $\sin(65^\circ)$ factor gone in "1/2*2*3*(sin65)=3"? (By the way, don't forget the degree sign) $\endgroup$ – user10354138 Jun 14 '19 at 5:44
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The base of the triangle is given by $\sqrt{2^2 + 3^2 - (2)(2)(3)\cos(65)}=2.81$

The area of the triangle is given by $\frac{1}{2}(2)(3)\sin65=2.718$.

So the volume of the triangular prism is $(2.718)(3.5)=9.51$.

Also, the volume of the rectangular prism is given by $(2.81)(5.5)(3.5)=54.1$.

Adding together these volumes gives us $63.6$.

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Hint: We get $$V_1=\frac{1}{2}3m\cdot 2m\sin(65^{\circ})\cdot 3.5m$$ (the volume of the triangular Prisma) and $$V_2=3.5m\cdot 5.5m\cdot x$$ (the volume of the rectangular Prisma) where $$x^2=(2m)^2+(5m)^2-2\cdot 2m\cdot 5m\cos(65^{\circ})$$ Can you proceed?

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Assuming the obvious missing edge is horizontal. Dividing the pentagonal face into a rectangle and the top triangle, the area of the triangle is $$ \frac{1}{2}\cdot 2\cdot 3\cdot\sin(65^\circ)=3\sin(65^\circ)\approx 2.72\,\mathrm{m}^2 $$ and the length of the missing side (the base) is $$ \sqrt{2^2+3^2-2\cdot 2\cdot 3\cdot\cos(65^\circ)}\approx 2.82\,\mathrm{m} $$ so the area of the pentagonal face is $$ 3\sin(65^\circ)+5.5\times\sqrt{2^2+3^2-2\cdot 2\cdot 3\cdot\cos(65^\circ)}\approx 18.2\,\mathrm{m}^2 $$ Hence the volume is $$ 3.5\times \left(3\sin(65^\circ)+5.5\times\sqrt{2^2+3^2-2\cdot 2\cdot 3\cdot\cos(65^\circ)}\right)\approx 63.7\,\mathrm{m}^3 $$

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