+0  
 
0
664
1
avatar

If you fold up the sides the container holds exactly 1L. Whats the length of a?

 Oct 27, 2016
 #1
avatar+26388 
0

If you fold up the sides the container holds exactly 1L. {nl} Whats the length of a?

 

 

 

\(\begin{array}{|rclrcl|} \hline B_{\triangle} &=& \frac{a}{2}\cdot h_{\triangle} & \quad h_{\triangle}^2 + (\frac{a}{2})^2 &=& a^2\\ & & & \quad h_{\triangle}^2 &=& a^2 - (\frac{a}{2})^2 \\ & & & \quad h_{\triangle}^2 &=& a^2 - \frac{a^2}{4} \\ & & & \quad h_{\triangle}^2 &=& \frac{3}{4}a^2 \\ B_{\triangle} &=& \frac{a}{2}\cdot \frac{a}{2}\sqrt{3} & \quad \mathbf{ h_{\triangle} } & \mathbf{=} & \mathbf{\frac{a}{2}\sqrt{3}} \\ \mathbf{B_{\triangle}} &\mathbf{=}& \mathbf{\frac{a^2}{4}\cdot \sqrt{3}} \\ \hline \end{array}\)

 

\(\begin{array}{|rclrcl|} \hline \sin(60^{\circ}) = \frac{a}{2r} &=& \frac{\sqrt{3}} {2} \\ \frac{a}{2r} &=& \frac{\sqrt{3}} {2} \\ \frac{a}{r} &=& \sqrt{3} \\ \frac{r}{a} &=& \frac{1} { \sqrt{3} } \\ \mathbf{r} &\mathbf{=}& \mathbf{\frac{a} { \sqrt{3} }} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline r^2+h^2 &=& a^2 \\ h^2 &=& a^2-r^2 \\ h^2 &=& a^2 -\left(\frac{a} { \sqrt{3} }\right)^2 \\ h^2 &=& a^2 -\frac{a^2} { 3 } \\ h^2 &=& \frac{2}{3}a^2 \\ \mathbf{h} &\mathbf{=}& \mathbf{\frac{\sqrt{2}} {\sqrt{3}}a} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline V &=& \frac{1}{3} \cdot B_{\triangle} \cdot h \qquad &| \qquad \mathbf{ B_{\triangle} = \frac{a^2}{4} \sqrt{3}} \qquad \mathbf{ h=\frac{\sqrt{2}} {\sqrt{3}}a } \\ V &=& \frac{1}{3} \cdot \left( \frac{a^2}{4} \sqrt{3} \right) \cdot \left( \frac{\sqrt{2}} {\sqrt{3}}\cdot a \right) \\ V &=& \frac{1}{3} \cdot \frac{a^2}{4} \cdot \sqrt{2}\cdot a \\ V &=& \frac{\sqrt{2}}{12} \cdot a^3 \\\\ \frac{\sqrt{2}}{12} \cdot a^3 &=& V \\ a^3 &=& \frac{12} {\sqrt{2}} \cdot V \\ a^3 &=& \frac{12} {\sqrt{2}} \cdot V \cdot \frac{\sqrt{2}}{\sqrt{2}} \\ a^3 &=& \frac{12\cdot \sqrt{2} } {2} \cdot V \\ a^3 &=& 6\cdot \sqrt{2} \cdot V \\ \mathbf{a} &\mathbf{=}& \mathbf{\sqrt[3]{6\cdot \sqrt{2} \cdot V}} \qquad &| \qquad V = 1l = 1\ dm^3\\ a & =& \sqrt[3]{6\cdot \sqrt{2} \cdot 1\ dm^3} \\ a & =& \sqrt[3]{8.48528137424\ dm^3} \\ a & =& 2.03964890266\ dm \\ \hline \end{array}\)

 

The length of a is 2.04 dm

 

laugh

 Oct 27, 2016
edited by heureka  Oct 27, 2016

1 Online Users

avatar