Answer :

[tex]The\ surface\ area:A_s=2B+(S_1+S_2+S_3)\cdot H\\B=\frac{12\cdot5}{2}=\frac{60}{2}=\boxed{30\ (yd^2)}\\\\S_1=5yd;\ S_2=12yd;\ S_3=13yd;\ h=8yd\\\\therefore\\\\A_s=2\cdot30+(5+12+13)\cdot8=60+30\cdot8=60+240=\boxed{300\ (yd^2)}[/tex][tex]Volume\ of\ the\ prism:V=B\cdot H\\\\B\ (base)\ it's\ a\ right\ triangle:B=\frac{12\cdot5}{2}=\frac{60}{2}=\boxed{30\ (yd^2)}\\\\H=8yd\\\\therefore\\\\V=30\cdot8=\boxed{240\ (yd^2)}\leftarrow\boxed{C}[/tex]
codymilner
Dan's calculations are correct, but we're not solving for the volume of the prism, we're solving for the surface area. To calculate that, we take the surface area of the two triangles on the ends, (already shown to be 30 yds2 each, so when counting both comes to 60yds), plus the surface area of the side that is 8 by 13, the side that is 8 by 12, and the side that is 8 by 5. 8x5=40, 8x12=96, 8x13=104. 40+96+104+(previously determined)60 = 300. So the surface area is 300 yds2.

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