An other way using basic knowledge is like this:
Join centres of small circle to obtain an equilateral triangle. Then simply find length of centroid using basic trigonometry or Pythagoras theorem. When that length is found add the radius of small circle to it. As the centroid of triangle is the center of bigger circle.
–
Sachin Vishwakarma
·
11 months, 2 weeks ago

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The general formula is \(\huge R = \frac{r[1 + \sin(\frac{\pi}{n})]} {\sin(\frac{\pi}{n})}\) where \(r,R\) and \(n\) are the radius of the inner circle, radius of the outer circle and number of circles inscribed within the larger circle respecticely.

Substituting we get the radius of the outer circle \(\approx 2.1547\) cm.
Try proving the formula of your own!
–
Svatejas Shivakumar
·
11 months, 2 weeks ago

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TopNewestAn other way using basic knowledge is like this: Join centres of small circle to obtain an equilateral triangle. Then simply find length of centroid using basic trigonometry or Pythagoras theorem. When that length is found add the radius of small circle to it. As the centroid of triangle is the center of bigger circle. – Sachin Vishwakarma · 11 months, 2 weeks ago

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The general formula is \(\huge R = \frac{r[1 + \sin(\frac{\pi}{n})]} {\sin(\frac{\pi}{n})}\) where \(r,R\) and \(n\) are the radius of the inner circle, radius of the outer circle and number of circles inscribed within the larger circle respecticely.

Substituting we get the radius of the outer circle \(\approx 2.1547\) cm. Try proving the formula of your own! – Svatejas Shivakumar · 11 months, 2 weeks ago

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