A pentagon’s radius (measure from center to vertex) measures 4cm, what is the pentagon’s area ?

Question

I have to use a method which is

EXAMPLE: Find the area of a nonagon with a side of 8 inches.

Find the measure of one base angle.

   central angle = 360/9 = 40

   2 * side angle + central angle = 180

   2 * side angle + 40 = 180

   2 * side angle = 140

   side angle = 70°

Use a trig ratio to find the height of one triangle.

   tan(70°) = h/(8/2)

   h = 4*tan(70°)

   h = 10.99 in

Use A = 1/2 bh to find the area of one triangle.

   A = 1/2 (8)(10.99)

   A = 43.96 in2

Find the area of the entire polygon.

   A = 43.96*9

   A = 395.64 in²

Answer 1

You are not given a side; you are given the measurement from the center to a vertex.

draw a diagram

draw the height and name it "h"

name a side "s"

360/5=72°

72/2=36°

you have five congruent isosceles triangles whose base angles are each 54° (180-72=108, 108/2=54°)

drawing the height divides each triangle into two congruent right triangles

take one right triangle...

one angle is 90°, one angle is 36° and one angle is 54°

using the 54° angle and the sine function...

sine=opposite/hypotenuse

sin(54°)=h/4

h=4*sin(54°)

h=4*(0.809017)

h=3.236 cm

using the Pythagorean Theorem...

(s/2)^2+(3.236)^2=4^2

(s/2)^2=16-10.4717)

(s/2)^2=5.5283

s/2=√5.5283

s/2=2.35123

s=2*2.35123

s=4.70246

A=(1/2)*b*h

A=(1/2)(4.70246)(3.236)

A=(1/2)*(15.21716)

A=7.60858 sq cm

there are 5 triangles

multiply the answer by 5

total area of the pentagon is 5*7.60858

A=38.0429 sq cm