Split into two equal volumes
[vc_row][vc_column][vc_column_text]A cube is sliced vertically along the line shown in the diagram and the smaller part is thrown away.
The remaining prism is going to be sliced vertically downwards again by a line going through corner $D$.
Where would the slice have to be to split it into two equal volumes?[latexpage][/vc_column_text][/vc_column][/vc_row][vc_row][vc_column][lvca_accordion][lvca_panel panel_title="Reveal Solution"]
This is a challenging task.
There are a number of possible approaches. This is just one of them.
The trick is to realise that it is only necessary to divide the area of the top surface into two equal sections.
Area of trapezium $=(30 \times 30) - (\sfrac{1}{2} \times 30 \times 15)= 900-225= 675cm^2$.
Area of triangle $ABX=675/2=337.5$.
Area of triangle $=\sfrac{1}{2}, ab \sin C$.
The right angled triangle that was initially removed can be used to find $\sin \theta$.
Hypotenuse $=\sqrt{15^2+30^2}=15\sqrt{5}$.
So $\sin \theta =\frac{30}{15\sqrt{5}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$.
Using the formula $=\sfrac{1}{2}, ab \sin C$, the area of triangle $ABX$ could also be written as $\sfrac{1}{2} \times 30 \times x \times \sin \theta$.
Which simplifies to $15x \times \frac{2\sqrt{5}}{5}=6\sqrt{5}x$.
So $6\sqrt{5}x=337.5$ giving $x=\frac{45\sqrt{5}}{4}$ cm or $x=25.2$ cm$^2$ to 3s.f.
So the cut should go from corner $B$ to a point that is $25.2$ cm along the line $AM$.[/lvca_panel][/lvca_accordion][vc_column_text]
