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Question Number 63324 by pradyot_pathak last updated on 02/Jul/19

Answered by Hope last updated on 02/Jul/19

$$areaofpentagon=\frac{1}{4}\times 5\times {12}^2\times cot\left(\frac{{180}^o}{5}\right)$$$$=180\times cot{36}^o$$$$externalangle=\frac{{360}^o}{5}={72}^o$$$$internalangle={180}^o-{72}^o={108}^o$$$$\theta +\theta +108={180}^o\to \theta ={36}^o$$$$areaofwhitetriangle=\frac{1}{2}\times 12\times h$$$$tan{36}^o=\frac{h}{6}\to h=6tan{36}^o$$$$areaoffive\left(05\right)whitetriangle=5\times \frac{1}{2}\times 12\times 6tan{36}^o$$$$areaofshadedblue=180cot{36}^o-180tan{36}^o$$$$=180(cot36-tan36)$$$$=180\left(\frac{{cos}^{2}36-{sin}^{2}36}{sin36cos36}\right)$$$$=360\times \left(\frac{cos{72}^o}{sin{72}^o}\right)$$$$=360\times cot{72}^o$$

Commented by Hope last updated on 02/Jul/19

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