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Question Number 148868 by DELETED last updated on 01/Aug/21

Answered by DELETED last updated on 01/Aug/21

$$\mathrm{given}\mathrm{that},\mathrm{lenght}\mathrm{of}\mathrm{AB}=\mathrm{AC}=\mathrm{BC}=6\mathrm{cm},$$$$\mathrm{and}{\mathrm{it}}^{\prime }\mathrm{s}10\mathrm{cm}\mathrm{hight},\mathrm{calculate}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{volume}.$$

Commented by DELETED last updated on 01/Aug/21

$$\mathrm{Answer}:$$$$\mathrm{Volume}=\frac{1}{3}\times \frac{1}{2}\times 6\times 6\times \sin 60\times 10$$$$=6\times \frac{1}{2}\sqrt{3}\times 10=30\sqrt{3}{\mathrm{cm}}^3$$

Answered by Olaf_Thorendsen last updated on 01/Aug/21

$$\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{blue}\mathrm{triangle}:$$$$\mathrm{A}=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times 6^2=9\sqrt{3}$$$$\mathrm{V}=\frac{1}{3}\mathrm{A}h=\frac{1}{3}\times 9\sqrt{3}\times 10=30\sqrt{3}$$

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