Math Question and Answers


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 49680 by peter frank last updated on 09/Dec/18

	Answered by math1967 last updated on 09/Dec/18

	$T o t a l s u r f a c e a r e a = 2 \times 6 \times \frac{\sqrt{3}}{4} \times 20^{2} + 6 \times 20 \times 15$ $= 3878.46 {c m}^{2} (a p r o x)$ $V o l u m e = \frac{\sqrt{3}}{4} \times 6 \times 20^{2} \times 15 = 233826.85 {c m}^{3} (a p r o x)$
	Commented by peter frank last updated on 09/Dec/18

	$thank you sir . . . where this come from \frac{\sqrt{3}}{4} ?$
	Commented by math1967 last updated on 09/Dec/18

	$R e g u l a r h e x a g o s p l i t s t o 6 e q u i l a t e r a l$ $t r i a n g l e s s o a r e a o f h e x a g o n = 6 \times \frac{\sqrt{3}}{4} \times {(s i d e)}^{2}$
	Commented by peter frank last updated on 09/Dec/18

	$thank you sir once again$
	Commented by math1967 last updated on 09/Dec/18

	$w e l c o m e$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum