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Question Number 43384 by peter frank last updated on 10/Sep/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	$c o s 30^{o} = \frac{\frac{a}{2}}{R}$ $\frac{\sqrt{3}}{2} = \frac{a}{2 R}$ $a = R \sqrt{3}$ $a = 5 \sqrt{3}$ $a r e a t r i a n g l e = \frac{\sqrt{3}}{4} \times a^{2} = \frac{\sqrt{3}}{4} \times 75$ $p e r i m e t d r t r i a n g l e = 3 a = 3 \times 5 \sqrt{3} = 15 \sqrt{3}$ $a r e a o f c i r c l e Π R^{2}$ $= Π \times 5^{2} = 25 Π$ $a r e a b e t w e d n c i r c l e a n d t r i a n g l e$ $= 25 Π - \frac{75 \sqrt{3}}{4}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	Commented by peter frank last updated on 10/Sep/18

	$c o s 30 ? w h e r e d o u g e t$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	$i n e q u i l a t e r a l t r i a n g l e e a c h a n g l e 60^{o}$ $r e f e r t h e p i c t u r e a t t a c h e d$ $p e r p e n d i c u l a r f r o m c e n t r e t o t h e s i d e o f t r i a n g l e$ $b i s e c t t h e s i d e .$ $n o w c o s 30^{o} = \frac{b a s e}{h y p o} = \frac{\frac{a}{2}}{R}$ $a = s i d e o f t r i a n g l e a n d r a d i u s R i s h y p o$
	Commented by peter frank last updated on 10/Sep/18

	$t h a n k s$
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