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Question Number 62456 by ajfour last updated on 21/Jun/19

Commented by ajfour last updated on 21/Jun/19

$I f t h e r e g u l a r t e t r a h e d r o n w i t h$ $e d g e l e n g t h u n i t y i n s c r i b e s a$ $h e m i s p h e r e s u c h t h a t t h r e e f a c e s$ $a r e t a n g e n t t o t h e s p h e r i c a l$ $s u r f a c e w h i l e f o u r t h s e r v e s t o$ $b e t h e d i a m e t e r p l a n e, t h e n f i n d$ $r a d i u s o f h e m i s p h e r e .$
	Answered by mr W last updated on 21/Jun/19

	Commented by mr W last updated on 21/Jun/19

	$e d g e l e n g t h o f t e t r a h e d r o n = a$ $v o l u m e o f t e t r a h e d r o n V = \frac{\sqrt{2} a^{3}}{12}$ $a r e a o f e a c h f a c e A_{S} = \frac{1}{2} \times a \times \frac{\sqrt{3} a}{2} = \frac{\sqrt{3} a^{2}}{4}$ $v o l u m e o f s m a l l p y r a m i d O - A B C = V_{1}$ $V_{1} = \frac{A_{S} R}{3} = \frac{\sqrt{3} a^{2} R}{12}$ $3 V_{1} = V$ $3 \times \frac{\sqrt{3} a^{2} R}{12} = \frac{\sqrt{2} a^{3}}{12}$ $\Rightarrow R = \frac{\sqrt{6} a}{9}$
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