Geometry

Pentagon inscribed in a circle => 5 isosceles triangles with central angle 72^o. 

Circle centre O with r = 2, angle subtended between A and B, B and C, C and D, D and E, E and A with centre O = 360^o / 5 = 72^o

Area of triangle ABO = (1/2)(2*2*sin 72^o) = 1.902 sq units

[ABCDE] = There are 5 triangles of equal area = (1.902 sq units)5 = 9.51 sq units.

.A,B,C,D, and E are points on a circle of radius  2 in counterclockwise order. We know AB=BC=CD=DE=2.  Find the area of pentagon ABCDE.

Hello Guest!

\(360^o-4\cdot 60^o=120^o\)

\(ABCDE=4\cdot MAB+MEA\\ =4\cdot \frac{a\sqrt{3}}{2}+MEA\\ =4\cdot \frac{2\sqrt{3}}{2}+4\cdot sin(60^o)cos(60^o)\\ =4\cdot \frac{2\sqrt{3}}{2}+4\cdot \frac{1}{2} \sqrt{3}\ \cdot \ \frac{1}{2}\\ =5\cdot\sqrt{3}\)

\(ABCDE=8.660254\)

  !

\(ABCDE\) forms an irregular pentagon, which can be split into \(5\) triangles, each with an area of \(\sqrt3\), so the answer is \(\color{brown}\boxed{5\sqrt3}\). 

Equal chords are subtended from equal angles. at the centre

We have 4 equilateral triangles.  

together that make an angles at the centre of  4*60 = 240degrees

The angle at the centre remaining for the last tirangle is 120 degrees.

The Apothem for the 4 congruent trianlges is   sqrt3

So their total area is  4sqrt3

The last triangle is isosceles with angles 120, 30,30

The length of the base is 2sqrt3 and the height is 1 so ther area is    sqrt3

So the total area is  5sqrt3 units     (which agrees with asin's answer.)

Here is the pic