# What is the area of a regular pentagon if its apothem has a length of 4 feet and each

• Sep 5th 2009, 02:48 PM
Leo34005
What is the area of a regular pentagon if its apothem has a length of 4 feet and each
What is the area of a regular pentagon if its apothem has a length of 4 feet and each side has a length of 5.8 feet?

A) 139.2 ft2
B) 58 ft2
C) 116 ft2
D) 69.6 ft2
E) 92.8 ft2
• Sep 5th 2009, 04:08 PM
mr fantastic
Quote:

Originally Posted by Leo34005
What is the area of a regular pentagon if its apothem has a length of 4 feet and each side has a length of 5.8 feet?

A) 139.2 ft2
B) 58 ft2
C) 116 ft2
D) 69.6 ft2
E) 92.8 ft2

If the centre of the pentagon is joined to the vertices, the regular pentagon is divided into 5 isosceles triangles. The apothem of the pentagon is the altitude (height) of each isosceles triangle. The side of the pentagon is the length of the base of each isosceles triangle.

Calculate the area of one of the isosceles triangles and then multiply by 5.

Edit: The data given in the question is consistent, to one decimal place. If you've learned trigonometry you should attempt to show that, in fact, if the sidelength of the regular pentagon is 5.8' then the apothem is actually 3.9915' (correct to four decimal places).
• Sep 5th 2009, 05:16 PM
Leo34005