What is the area of a regular pentagon with a side of 10 in.? Round the answer to the nearest tenth.
96.3in^2
172.0in^2
344.1in^2
192.6in^2
DONT RESPOND WITH A LINK

A = 172.0 in²

Step-by-step explanation:

Exterior angle of any side is 360 / 5 = 72°

Interior angle between any two adjacent sides is 180 - 72 = 108°

A regular pentagon can be thought of as 5 isosceles triangles with two common angles of 108 / 2 = 54°

The height of each triangle is

h = (10/2)tan54

The area of each triangle is ½(10)h = ½(10)(10/2)tan54 = 25tan54

The area of the pentagon is

A = 5(25tan54) = 125tan54 = 172.04774005...

A = 172.0 in²

where are the choices?

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