Area of triangle ABC = 12.8 square units Area of triangle ACD = 18.4 square units Area of triangle AED = 14.8 square units A polygon ABCDE is shown. The vertices A and D are joined with a dashed line, and the vertices A and C are joined with a dashed line. detailed show you work :)
Question is not complete. Assuming the side lengths of a REGULAR pentagon is required.
Sum of areas of the pentagon = 12.8+14.8+18.4=55 Area of triangle formed by the centre and one of the sides = 55/5=11 sq. units. For a pentagon, the interior angle is (5-2)*180/5=108 degrees. Therefore base angle of each isosceles triangle (base = side) =108/2=54 degrees.
Let length of each pentagon side = c Area of triangle, (c/2)(c/2)tan(54 degrees)/2=11 c^2tan(54)/8 = 11 sq. units c^2=11*8/tan(54)=63.94 sq. units c=8.00 units = side length of pentagon
