Answer to Question #126630 in Geometry for billiy

Question

a box of chocolates has the shape of a regular pentagon each side measures 8.1 the distance to the center of the box is 9.5 what is the area of the chocolate box to the nearest square cm

Answer 1

$Area_{ABCDE}=5\cdot A_{\Delta AOE}=5\cdot{1\over 2}AE\cdot OF$

$\Delta AOE:\angle AOE={360\degree\over 5}=72\degree, AO=EO$

$AF$ is perpendicular bisector:

$OF\perp AE, AF=FE={1\over 2}AE,$

$\angle AOF=\angle EOF={1\over 2}\angle AOE=36\degree$

Right $\Delta AOF$

$\tan(\angle AOF)={AF\over OF}$

$OF={AF\over \tan(\angle AOF)}={AE\over 2\tan(\angle AOF)}$

$Area_{ABCDE}={5\over 2}AE\cdot {AE\over 2\tan(\angle AOF)}=$

$={5(AE)^2\over 4\tan(\angle AOF)}$

Given $AE=8.1 cm$

$Area_{ABCDE}={5(8.1cm)^2\over 4\tan(36\degree)}\approx113\ cm^2$

The area of the chocolate box is 113 cm².