Problema Solution
A star inscribed in a regular Pentagon of side 10 cm. calculate the area of the star.
Answer provided by our tutors
draw the regular pentagon in the star like on the picture here
http://mathcentral.uregina.ca/QQ/database/QQ.09.06/s/chetna2.1.gif
A,B,C,D and E are the vertices of the pentagram:
O is the center of the circle (where the vertices lie)
X is the vertex of the star that doesnt lie on the circle opposite of AB (is not one of the pentagons vertices)
A = the area of the star
A1 = the area on the polygon OAXB
A = 5*A1
A1 = Area of triangle ABO - Area of triangle ABX
notice that ABO and ABX are isosceles triangles
angle OAB = angle ABO = (180 - angle AOB)/2 = (180 - (360/5))/2 = 54 degrees
angle XAB = angle ABX = (180 - (3*180/5))/2 = 36 degrees
Area of triangle ABO = (AB * (AB/2)*(tan 54))/2 = (AB^2 (tan 54))/4 = (10^2 (tan 54))/4 = 25 (tan 54)
Area of triangle ABX = (AB * (AB/2)*(tan 36))/2 = (AB^2 (tan 36))/4 = (10^2 (tan 36))/4 = 25 (tan 36)
A1 = Area of triangle ABO - Area of triangle ABX = 25 (tan 54) - 25 (tan 36) = 25(tan 54 - tan36)
A = 5*A1
A = 5*25(tan 54 - tan36)
A = 81.23 cm^2
draw the regular pentagon and the star like on the picture here
http://mathcentral.uregina.ca/QQ/database/QQ.09.06/s/chetna2.1.gif
A,B,C,D and E are the vertices of the pentagram:
O is the center of the circle (where the vertices lie)
X is the vertex of the star that doesnt lie on the circle opposite of AB (is not one of the pentagons vertices)
A = the area of the star
A1 = the area on the polygon OAXB
A = 5*A1
A1 = Area of triangle ABO - Area of triangle ABX
notice that ABO and ABX are isosceles triangles
angle OAB = angle ABO = (180 - angle AOB)/2 = (180 - (360/5))/2 = 54 degrees
angle XAB = angle ABX = (180 - (3*180/5))/2 = 36 degrees
Area of triangle ABO = (AB * (AB/2)*(tan 54))/2 = (AB^2 (tan 54))/4 = (10^2 (tan 54))/4 = 25 (tan 54)
Area of triangle ABX = (AB * (AB/2)*(tan 36))/2 = (AB^2 (tan 36))/4 = (10^2 (tan 36))/4 = 25 (tan 36)
A1 = Area of triangle ABO - Area of triangle ABX = 25 (tan 54) - 25 (tan 36) = 25(tan 54 - tan36)
A = 5*A1
A = 5*25(tan 54 - tan36)
A = 81.23 cm^2
the are of the star is 81.23 cm^2.