**Problem**

Find the area of the regular five-pointed star inscribed in a circle of radius 20 cm.

**Solution**

From the figure shown:

$2\theta = \frac{1}{5}(360^\circ) = 72^\circ$

$\theta = 36^\circ$

$\frac{1}{2}\theta = 18^\circ$

**From triangle ABO:**

$\theta + \frac{1}{2}\theta + \beta = 180^\circ$

$36^\circ + 18^\circ + \beta = 180^\circ$

$\beta = 126^\circ$

**By Sine Law**

$\dfrac{a}{\sin \frac{1}{2}\theta} = \dfrac{r}{\sin \beta}$

$\dfrac{a}{\sin 18^\circ} = \dfrac{20}{\sin 126^\circ}$

$a = 7.64 \, \text{ cm }$

**Area of triangle ABO:**

$A_{ABO} = \frac{1}{2}ar \, \sin \theta$

$A_{ABO} = \frac{1}{2}(7.64)(20) \, \sin 36^\circ$

$A_{ABO} = 44.903 \, \text{ cm}^2$

**Area of Pentagram (the five-pointed star)**

$A = 10\,A_{ABO}$

$A = 10(44.903)$

$A = 449.03 \, \text{ cm}^2$ *answer*

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