In an equilateral triangle,the inradius (r) a | vedclass

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(B) For any triangle,the relation between inradius $r$ and circumradius $R$ is given by $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$

Vedclass · Accepted Answer

$r = R/2$

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(B) For any triangle,the relation between inradius $r$ and circumradius $R$ is given by $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.In an equilateral triangle,all angles are equal to $60^\circ$,so $A = B = C = 60^\circ$.Substituting these values,we get $r = 4R \sin(30^\circ) \sin(30^\circ) \sin(30^\circ)$.Since $\sin(30^\circ) = \frac{1}{2}$,we have $r = 4R 	imes (\frac{1}{2}) 	imes (\frac{1}{2}) 	imes (\frac{1}{2})$.$r = 4R 	imes \frac{1}{8} = \frac{R}{2}$.

In an equilateral triangle,the inradius $(r)$ and the circumradius $(R)$ are connected by which of the following relations?

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