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A-solid-rigth-triangular-prism-of-length-12cm-and-a-cross-section-which-is-an-equilateral-triangle-of-6cm-Find-the-total-surface-area-




Question Number 176212 by otchereabdullai@gmail.com last updated on 15/Sep/22
  A solid rigth triangular prism of     length 12cm and a cross section     which is an equilateral triangle of    6cm. Find the total surface area.
$$\:\:\mathrm{A}\:\mathrm{solid}\:\mathrm{rigth}\:\mathrm{triangular}\:\mathrm{prism}\:\mathrm{of}\: \\ $$$$\:\:\mathrm{length}\:\mathrm{12cm}\:\mathrm{and}\:\mathrm{a}\:\mathrm{cross}\:\mathrm{section}\: \\ $$$$\:\:\mathrm{which}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{of} \\ $$$$\:\:\mathrm{6cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{total}\:\mathrm{surface}\:\mathrm{area}. \\ $$
Commented by adhigenz last updated on 15/Sep/22
Area of base = (1/2)×12^2 ×sin 60° = 36(√3) cm^2   Total surface area = 2×area of base + perimeter of base×length                                           = 2×36(√3) + (6+6+6)×12                                           = (72(√3) + 216) cm^2
$$\mathrm{Area}\:\mathrm{of}\:\mathrm{base}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}^{\mathrm{2}} ×\mathrm{sin}\:\mathrm{60}°\:=\:\mathrm{36}\sqrt{\mathrm{3}}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$\mathrm{Total}\:\mathrm{surface}\:\mathrm{area}\:=\:\mathrm{2}×\mathrm{area}\:\mathrm{of}\:\mathrm{base}\:+\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{base}×\mathrm{length} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}×\mathrm{36}\sqrt{\mathrm{3}}\:+\:\left(\mathrm{6}+\mathrm{6}+\mathrm{6}\right)×\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{72}\sqrt{\mathrm{3}}\:+\:\mathrm{216}\right)\:\mathrm{cm}^{\mathrm{2}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 15/Sep/22
 Am much grateful God bless you!
$$\:\mathrm{Am}\:\mathrm{much}\:\mathrm{grateful}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}! \\ $$

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