Question Number 13010 by tawa tawa last updated on 10/May/17 | ||
Answered by sandy_suhendra last updated on 10/May/17 | ||
$$\mathrm{BC}\:=\:\sqrt{\mathrm{5}^{\mathrm{2}} −\left(\mathrm{7}−\mathrm{4}\right)^{\mathrm{2}} }\:=\:\mathrm{4}\:\mathrm{cm} \\ $$$$\mathrm{let}\:\mathrm{A}_{\mathrm{ABCD}} \:=\:\mathrm{area}\:\mathrm{of}\:\mathrm{ABCD} \\ $$$$\mathrm{and}\:\mathrm{P}_{\mathrm{ABCD}} \:=\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{ABCD} \\ $$$$\mathrm{A}_{\mathrm{prism}} \:=\:\mathrm{2A}_{\mathrm{ABCD}} \:+\:\mathrm{P}_{\mathrm{ABCD}} ×\mathrm{h} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}×\frac{\mathrm{4}+\mathrm{7}}{\mathrm{2}}×\mathrm{4}\:+\:\left(\mathrm{7}+\mathrm{4}+\mathrm{4}+\mathrm{5}\right)×\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{44}\:+\:\mathrm{240}\:=\:\mathrm{284}\:\mathrm{cm}^{\mathrm{2}} \\ $$ | ||
Commented by tawa tawa last updated on 10/May/17 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$ | ||