How to find area of a pentagon?

Answer 1

There are many ways to find the area. We restrict the answer to a regular pentagon with is one with equal sides and angles.

One method uses the apothem which is a line segment from the center to the midpoint of any of the sides. We use the length of the apothem. In this case, the area is 1/2 x apothem x perimeter.

This method works for any regular polygon.

Here is a slightly more complicated method and explanationl

We have n sides, all equal length a

We have n interior angles, all equal measure beta

Alpha is the central angle subtending one side

P is the Perimeter

K is the Area

Let R be the radius of circumscribed circle and

r be the radius of inscribed circle.

The for any regular polygon we have:

beta = Pi(n-2)/n radians = 180o(n-2)/n

alpha = 2 Pi/n radians = 360o/n

alpha + beta = Pi radians = 180o

P = na = 2nR sin(alpha/2)

K = na2 cot(alpha/2)/4

= nR2 sin(alpha)/2

= nr2 tan(alpha/2)

= na sqrt(4R2-a2)/4

R = a csc(alpha/2)/2

r = a cot(alpha/2)/2

a = 2r tan(alpha/2) = 2R sin(alpha/2)

For a regular pentagon

Number of sides n = 5

Internal angles beta = 3/5 radians = 108 degrees

Central angles alpha = 2/5 radians = 72 degrees

Perimeter P = 5a = 5R sqrt(10-2 sqrt[5])/2

Area K = 5a2 sqrt(1+2/sqrt[5])/4

= 5R2 sqrt(10+2 sqrt[5])/8

= 5r2 sqrt(5-2 sqrt[5])

= 5a sqrt(4R2-a2)/4

Circumradius R = a sqrt(2+2/sqrt[5])/2

Apothem r = a sqrt(1+2/sqrt[5])/2 = R(1+sqrt[5])/4

Side a = 2r sqrt(5-2 sqrt[5]) = R sqrt(10-2 sqrt[5])/2

Answer 2

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