### Find volume of regular tetrahedron given its height

Question Sample Titled 'Find volume of regular tetrahedron given its height'

If the height of a regular tetrahedron is ${7}$ $\text{cm}$ , then the volume of the tetrahedron is

A
$\dfrac{{343}}{{8}}\sqrt{{3}}$ $\text{cm}^{{3}}$ .
B
${7}$ $\text{cm}^{{3}}$ .
C
$\dfrac{{343}}{{8}}\sqrt{{6}}$ $\text{cm}^{{3}}$ .
D
$\dfrac{{1029}}{{8}}\sqrt{{3}}$ $\text{cm}^{{3}}$ .

${a}$${A}$${B}$${C}$${D}$${M}$${H}$

 Let the length of each side of the tetrahedron be ${a}$, i.e. ${A}{B}={B}{C}={a}$ Let ${M}$be a point on ${B}{C}$ such that ${A}{M}\bot{B}{C}$ . ∴  ${B}{M}$ $={M}{C}=\dfrac{{a}}{{2}}$ prop. of isos. △ ${A}{M}$ $=\sqrt{{{a}^{{2}}-{\left(\dfrac{{a}}{{2}}\right)}^{{2}}}}=\dfrac{\sqrt{{{3}}}}{{2}}{a}$ Pyth. theorem Similarly, ${D}{M}=\dfrac{\sqrt{{{3}}}}{{2}}{a}$ Let ${H}$ be the foot of perpendicular from ${A}$ to plane ${B}{C}{D}$ . ${A}{H}$ $={7}$ given ${H}$ is the in-centre of $\triangle{B}{C}{D}$ . Consider $\triangle{B}{M}{H}$ , $\angle{H}{B}{M}$ $=\dfrac{{60}^{\circ}}{{2}}={30}^{\circ}$ ${{\tan{{30}}}^{\circ}}$ $=\dfrac{{{M}{H}}}{{\dfrac{{a}}{{2}}}}$ ${M}{H}$ $=\dfrac{{1}}{\sqrt{{{3}}}}{\left(\dfrac{{a}}{{2}}\right)}=\dfrac{{a}}{{{2}\sqrt{{3}}}}$ ${{\tan{{30}}}^{\circ}=}\dfrac{{1}}{\sqrt{{{3}}}}$ Consider $\triangle{A}{M}{H}$ , ${7}^{{2}}$ $={\left({A}{M}^{{2}}\right)}-{\left({M}{H}\right)}^{{2}}$ Pyth. theorem ${7}^{{2}}$ $={\left(\dfrac{\sqrt{{{3}}}}{{2}}{a}\right)}^{{2}}-{\left(\dfrac{{a}}{{{2}\sqrt{{3}}}}\right)}^{{2}}$ ${49}$ $=\dfrac{{2}}{{3}}\cdot{a}^{{2}}$ ${a}$ $=\dfrac{{{7}\sqrt{{{6}}}}}{{2}}$  The volume of the tetrahedron $=\dfrac{{1}}{{3}}\times{7}\times{\left[\dfrac{{1}}{{2}}{\left({a}\right)}{\left({a}\right)}{\left({\sin{{60}}}^{\circ}\right)}\right]}$ Volume of tetrahedron$=\dfrac{{1}}{{3}}\times{h}\times\text{base area}$; For base area, Heron's Formula can also be used $=\dfrac{{1}}{{3}}\times{7}\times{\left[\dfrac{{1}}{{2}}{\left(\dfrac{{{7}\sqrt{{{6}}}}}{{2}}\right)}{\left(\dfrac{{{7}\sqrt{{{6}}}}}{{2}}\right)}{\left(\dfrac{\sqrt{{3}}}{{2}}\right)}\right]}$ $=\dfrac{{343}}{{8}}\sqrt{{3}}$

 For any regular tetrahedron with side length ${a}$ and height ${h}$, the relationship is listed below. ${a}$ $={h}\cdot\sqrt{{\dfrac{{3}}{{2}}}}$ ${V}$ $=\dfrac{{a}^{{3}}}{{{6}\sqrt{{2}}}}=\dfrac{\sqrt{{{3}}}}{{8}}{h}^{{3}}$

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