Answer:
The area of the ΔAOB is [tex]9\sqrt{3}\ unit^{2}[/tex] .
Step-by-step explanation:
As given the figure
OA = OB = 6 unit
(Radius of the circle.)
In ΔAOB
As the two sides of a triangle thus there opposite angles are also equal .
∠A = ∠B
Let us assume that ∠A = ∠B = x°
∠A + ∠O + ∠B = 180°
(By using the angle sum property of a triangle.)
∠O = 60°
x° + 60° + x° = 180°
2x = 180 - 60
2x = 120
[tex]x = \frac{120}{2}[/tex]
x = 60
∠A = ∠B = 60°
Thus
∠A = ∠B = ∠O = 60°
As all the angles of the ΔAOB are 60° thus ΔAOB is a equilateral triangle .
Also all the sides of the ΔAOB are also equal i.e AB = OA = OB = 6 unit .
Formula
[tex]Area\ of \ a\ equilateral\ triangle = \frac{\sqrt{3}}{4} a^{2}[/tex]
Where a is the side of the equilateral triangle .
a = 6 unit
Put in the above formula
[tex]Area\ of \ a\ equilateral\ triangle = \frac{\sqrt{3}}{4}\times 6^{2}[/tex]
6² = 36
[tex]Area\ of \ a\ equilateral\ triangle = \frac{\sqrt{3}}{4}\times 36[/tex]
[tex]Area\ of \ a\ equilateral\ triangle = 9\sqrt{3}\ unit^{2}[/tex]
Therefore the area of the ΔAOB is [tex]9\sqrt{3}\ unit^{2}[/tex] .
