Three coins are placed in such a way that one coin touches the other two (in the same plane). If the radius of each coin is r, what is the area of the triangle that circumscribes this arrangement of coins.

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Solution

The correct option is B

The angle shown is 30 degrees. Thus AB=r $\sqrt{3}$

Thus side of a triangle: 2r + 2r $\sqrt{3}$ .

Area of the equilateral triangle: $\frac{\sqrt{3} (r + 2 r \sqrt{3})^{2}}{4}$

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S = n \sum r = 0 (- 1)^{r}^{n} C_{r} [\frac{2^{r}}{3^{r}} + \frac{8^{r}}{3^{2 r}} + \frac{26^{r}}{3^{3 r}} + . . . \infty]

\sum_{r = 0}^{n} {(- 1)}^{r} ({^{n} C}_{r}) [\frac{2^{r}}{3^{r}} + \frac{8^{r}}{3^{2 r}} + \frac{26^{r}}{3^{3 r}} + . . . . u p t o \infty]

1 c m

5 cm

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