Question Number 148868 by DELETED last updated on 01/Aug/21
Answered by DELETED last updated on 01/Aug/21
$$\mathrm{given}\:\mathrm{that},\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{AB}=\mathrm{AC}=\mathrm{BC}=\mathrm{6}\:\mathrm{cm}, \\ $$$$\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{10}\:\mathrm{cm}\:\mathrm{hight},\:\mathrm{calculate}\:\mathrm{it}'\mathrm{s}\:\mathrm{volume}. \\ $$
Commented by DELETED last updated on 01/Aug/21
$$\mathrm{Answer}:\: \\ $$$$\mathrm{Volume}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}×\mathrm{6}×\mathrm{sin}\:\mathrm{60}×\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{6}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:×\mathrm{10}=\mathrm{30}\sqrt{\mathrm{3}}\:\mathrm{cm}^{\mathrm{3}} \\ $$
Answered by Olaf_Thorendsen last updated on 01/Aug/21
$$\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{blue}\:\mathrm{triangle}\:: \\ $$$$\mathrm{A}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} \:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{6}^{\mathrm{2}} \:=\:\mathrm{9}\sqrt{\mathrm{3}} \\ $$$$\mathrm{V}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{A}{h}\:=\:\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{9}\sqrt{\mathrm{3}}×\mathrm{10}\:=\:\mathrm{30}\sqrt{\mathrm{3}} \\ $$