cyclic quadrilateral!

We will divide the cyclic Quadrilateral into 2 triangles. Since diagonals coincide at the center, Each is a right angled triangle. By the property, each angle subtended by diameter is on the arc is a right angle. Thus, we can find the other two sides of the quadrilateral. Lets assume them to be x & y. ( 4 sqrt(3))^2 = (5)^2 + x^2 (Pythagoras theorem) ( 4 sqrt(3))^2 = (4)^2 + x^2 (Pythagoras theorem)

x= sqrt(23) = 4.7 & y = 4*sqrt(2) = 5.7

Now, finding out the area of two triangles and adding them gives us the area of the quadrilateral.

Triangle 1 = 1/2 * 5 4.7 = 11.9 Triangle 2 = 1/2 * 4 5.7 = 11.4

Adding both, 23.3

Apply pythagorean theorem on $\triangle ABC$ , we have

$BC=\sqrt{(4\sqrt{3})^2-4^2}=\sqrt{32}$

Apply pythagorean theorem on $\triangle ADC$ , we have

$CD=\sqrt{(4\sqrt{3})^2-5^2}=\sqrt{23}$

The area of the quadrilateral is therefore,

$A=\dfrac{1}{2}(4)(\sqrt{32})+\dfrac{1}{2}(5)(\sqrt{23})\approx$ $\boxed{23.3}$

Note:

If a triangle is inscribed in a circle and one of its side is the diameter of the circle, the triangle is a right triangle.

The diagonal coincides of the quadrilateral are the diameter of the circle. Let, AB= 4 units and DC=5 units. The radius is 2* rootover3. So, the diameter, AC= 4* rootover3 Applying Pythagoras , we get the values of BC and AD. Thus, the area of the quadrilateral will be= area of triangle ADC + area of triangle ABC. Answer= 23.3

bretschneider's formula given sides 4, 5, 4root6 - 4, and 4root6 - 5. semiperimeter (9 + (8root6 - 9)) / 2 = S ((S-4)(S-5)(S-(4root6))(S-(4root6-5))-((4x5x(4root6 - 4)x(4root6 - 5))(1+cos180)/2))^(1/2) =23.3