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Question Number 176212 by otchereabdullai@gmail.com last updated on 15/Sep/22

$$\mathrm{A}\mathrm{solid}\mathrm{rigth}\mathrm{triangular}\mathrm{prism}\mathrm{of}$$$$\mathrm{length}12\mathrm{cm}\mathrm{and}\mathrm{a}\mathrm{cross}\mathrm{section}$$$$\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{of}$$$$6\mathrm{cm}.\mathrm{Find}\mathrm{the}\mathrm{total}\mathrm{surface}\mathrm{area}.$$

Commented by adhigenz last updated on 15/Sep/22

$$\mathrm{Area}\mathrm{of}\mathrm{base}=\frac{1}{2}\times {12}^2\times \sin 60°=36\sqrt{3}{\mathrm{cm}}^2$$$$\mathrm{Total}\mathrm{surface}\mathrm{area}=2\times \mathrm{area}\mathrm{of}\mathrm{base}+\mathrm{perimeter}\mathrm{of}\mathrm{base}\times \mathrm{length}$$$$=2\times 36\sqrt{3}+(6+6+6)\times 12$$$$=(72\sqrt{3}+216){\mathrm{cm}}^2$$

Commented by otchereabdullai@gmail.com last updated on 15/Sep/22

$$\mathrm{Am}\mathrm{much}\mathrm{grateful}\mathrm{God}\mathrm{bless}\mathrm{you}!$$

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