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Question Number 203319 by ajfour last updated on 16/Jan/24

Commented by mr W last updated on 16/Jan/24

$4 e q u a l s p h e r e s i n s i d e a r e g u l a r$ $t e t r a h e d r o n w i t h e d g e l e n g t h a ?$
Commented by ajfour last updated on 16/Jan/24

$W e l l I f e a c h e q u a l r a d i i s p h e r e t o u c h$ $t h e n d i s t a n c e f r o m g a p c e n t r e t o s p h e r e$ $c e n t e r i s h o w m a n y t i m e s r ?$ $I t h i n k i t s s i m p l e r = \frac{a}{\sqrt{3}}$
Commented by mr W last updated on 17/Jan/24

$d i s t a n c e f r o m c e n t e r o f s p h e r e t o$ $c e n t e r o f s p h e r e = 2 r$ $d i s t a n c e f r o m c e n t r o i d t o c e n t e r o f$ $s p h e r e = a = \frac{\sqrt{6} (2 r)}{4} = \frac{\sqrt{6} r}{2}$ $\Rightarrow r = \frac{2 a}{\sqrt{6}} = \frac{\sqrt{2} a}{\sqrt{3}}$
Commented by mr W last updated on 17/Jan/24

${w h a t}^{'} s t h e e d g e l e n g t h (b) o f t h e$ $r e g u l a r c i r c u m t e t r a h e d r o n f o r$ $t h e 4 s p h e r e s w i t h r a d i u s r ?$
Commented by mr W last updated on 17/Jan/24

Commented by mr W last updated on 17/Jan/24

$i g o t b = 2 (\sqrt{6} + 1) r$
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