Question

Total surface area of the following prism is

• A. $48m^2+56\sqrt{2}{m}^2$
• B. $64+48\sqrt{2}{m}^2$
• C. $256+96\sqrt{2}{m}^2$
• D. $256+48\sqrt{2}{m}^2$

Answer 1

The correct option is C $256+96\sqrt{2}{m}^2$

Total surface area of a triangular prism:

Total surface area = 2$\times$(area of base)+(perimeter)(height)

Total surface area $=2\left(\frac{1}{2}\right)\left(l\right)\left(w\right)+(l+w+y)\left(h\right)$

Total surface area $=\left(l\right)\left(w\right)+(l+w+y)\left(h\right)$

$l$ = length of base
$w$ =width of base
$y$ = hypotenuse of base
$h$ = height of prism

Length of the hypotenuse is $y$ m

$y^2=8^2+8^2\Rightarrow y^2=128\Rightarrow y=8\sqrt{2}m$

Substituting the values

Total surface area$=\left(8\right)\left(8\right)+(8+8+8\sqrt{2})\left(12\right)m^2$

Total surface area$=64+(16+8\sqrt{2})\left(12\right)m^2$
Total surface area$=64+192+96\sqrt{2}{m}^2$
Total Surface Area $=256+96\sqrt{2}{m}^2$