A-plane-is-drawn-through-the-midpoint-of-a-diagonal-of-a-cube-perpendicular-to-the-diagonal-Determine-the-area-of-the-figure-resulting-from-the-section-of-the-cube-cut-by-this-plane-if-the-edge-of-t

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Question Number 20194 by ajfour last updated on 23/Aug/17

$${A}\:{plane}\:{is}\:{drawn}\:{through}\:{the}\: \\ $$$${midpoint}\:{of}\:{a}\:{diagonal}\:{of}\:{a}\:{cube} \\ $$$${perpendicular}\:{to}\:{the}\:{diagonal}. \\ $$$${Determine}\:{the}\:{area}\:{of}\:{the}\:{figure} \\ $$$${resulting}\:{from}\:{the}\:{section}\:{of}\:{the} \\ $$$${cube}\:{cut}\:{by}\:{this}\:{plane}\:{if}\:{the}\:{edge} \\ $$$${of}\:{the}\:{cube}\:{is}\:{equal}\:{to}\:\boldsymbol{{a}}. \\ $$

Commented by ajfour last updated on 24/Aug/17

$${thank}\:{you}\:{Sir}. \\ $$

Commented by mrW1 last updated on 24/Aug/17

the section is a regular hexagon with side length b=(√2)(a/2)=a/(√2) A_S =((3(√3))/2)b^2 =((3(√3))/2)×(a^2 /2)=((3(√3))/4)a^2

$$\mathrm{the}\:\mathrm{section}\:\mathrm{is}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{hexagon}\:\mathrm{with} \\ $$$$\mathrm{side}\:\mathrm{length}\:\mathrm{b}=\sqrt{\mathrm{2}}\left(\mathrm{a}/\mathrm{2}\right)=\mathrm{a}/\sqrt{\mathrm{2}} \\ $$$$\mathrm{A}_{\mathrm{S}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{b}^{\mathrm{2}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{a}^{\mathrm{2}} \\ $$

Commented by mrW1 last updated on 24/Aug/17