Find the area of ΔAOB3√3 un24.5√3 un29√3 un2

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] .