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Question Number 212827 by Spillover last updated on 25/Oct/24

	Answered by A5T last updated on 25/Oct/24

	Commented by A5T last updated on 25/Oct/24

	$\frac{y}{\frac{2 r}{\sqrt{3}}} = \frac{r}{\frac{4 r}{\sqrt{3}}} \Rightarrow y = \frac{2 r \sqrt{3}}{3} \times \frac{r \sqrt{3}}{4 r} = \frac{r}{2}$ $\Rightarrow {(2 r)}^{2} = \frac{r^{2}}{4} + {(8 - 2 r \sqrt{3} + \sqrt{r^{2} - \frac{r^{2}}{4}})}^{2}$ $\Rightarrow r = 4 \sqrt{3} - \frac{4 \sqrt{15}}{3}$ $\Rightarrow [g r e e n] = \frac{8 \times 8 s i n 60}{2} - \frac{3 π r^{2}}{2} - \frac{3 \times 2 r \times \frac{2 r}{\sqrt{3}}}{2}$ $= 64 \sqrt{15} - \frac{400 \sqrt{3}}{3} + (48 \sqrt{5} - 112) π$
	Commented by Spillover last updated on 25/Oct/24

	$g r e a t . t h a n k s$
	Commented by ajfour last updated on 25/Oct/24

	$\frac{r - y}{2 r} = \frac{r - \frac{r}{2}}{2 r} = \sin θ = \frac{1}{4}$ $8 = \frac{2 r}{\sqrt{3}} + 2 r \cos θ + r \cos \frac{π}{6} + \frac{r}{\sqrt{3}}$ $\Rightarrow r (\frac{3 \sqrt{3}}{2} + \frac{\sqrt{15}}{2}) = 8$ $r = \frac{4 (3 - \sqrt{5})}{\sqrt{3}}$ $G = 16 \sqrt{3} - \frac{3}{2} \times \frac{{(2 r)}^{2}}{\sqrt{3}} - \frac{3 π r^{2}}{2}$ $= 16 \sqrt{3} - (2 \sqrt{3} + \frac{3 π}{2}) \times \frac{32}{3} (7 - 3 \sqrt{5})$
	Commented by ajfour last updated on 25/Oct/24
	right, its okay now, thank you.
	Commented by A5T last updated on 25/Oct/24

	$G \approx 2.2636$
	Answered by mr W last updated on 25/Oct/24

	Commented by mr W last updated on 25/Oct/24

	${(2 r)}^{2} = {(r + \frac{2 x}{\sqrt{3}})}^{2} + {(r + \frac{x}{\sqrt{3}})}^{2} - (r + \frac{2 x}{\sqrt{3}}) (r + \frac{x}{\sqrt{3}})$ $x^{2} + \sqrt{3} r x - 3 r^{2} = 0$ $\Rightarrow x = \frac{\sqrt{3} (\sqrt{5} - 1) r}{2}$ $\frac{4 r}{\sqrt{3}} + \frac{\sqrt{3} (\sqrt{5} - 1) r}{2} + \frac{2 r}{\sqrt{3}} = 8$ $\Rightarrow r = \frac{4 \sqrt{3} (3 - \sqrt{5})}{3} \approx 1.764$ $g r e e n a r e a$ $= \frac{\sqrt{3} \times 8^{2}}{4} - \frac{3 \times 4 r^{2}}{2 \times \sqrt{3}} - \frac{3 π r^{2}}{2}$ $= \frac{\sqrt{3} \times 8^{2}}{4} - \frac{(4 \sqrt{3} + 3 π) r^{2}}{2}$ $= 16 \sqrt{3} - \frac{16 (4 \sqrt{3} + 3 π) (7 - 3 \sqrt{5})}{3}$
	Commented by A5T last updated on 25/Oct/24

	Commented by A5T last updated on 25/Oct/24

	$T h i s i s w h a t w e g e t c o n s t r u c t i n g w i t h t h e v a l u e$ $o f t h e r y o u g o t .$
	Commented by mr W last updated on 25/Oct/24

	$t h a n k s f o r c h e c k i n g! i^{'} v e f i x e d a n$ $e r r o r i n c a l c u l a t i o n .$
	Commented by A5T last updated on 25/Oct/24

	$Y o u a l s o f o r g o t t o s u b t r a c t t h e a r e a o f t h e t h r e e$ $r i g h t t r i a n g l e s .$
	Commented by mr W last updated on 25/Oct/24

	$y o u a r e r i g h t a g a i n .$
	Commented by Spillover last updated on 25/Oct/24

	$g r e a t . t h a n k s$
	Answered by Spillover last updated on 25/Oct/24

	Answered by Spillover last updated on 25/Oct/24

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