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# Find the area of the triangle inside the circle

Question: An equilateral triangle inscribed in a circle. The radius of the circle is r and center of the circle is point O. Find the area of the equilateral triangle.

- Solution:
- Extended AO, which intersects BC at point D, and intersects the circle at point E.
- Because ABC is an equilateral triangle, so each interior angle is 60
^{o}, so angle C = 60^{o} - Because the point C is on the circle, so the angle C is an inscribed angle.
- Because angle C is opposite the arc AB and the angle E is also opposite the arc AB,
- so, angle E = angle C (Inscribed angles opposite the same arc are equal.)
- Connect BO, because BO = EO = r,
- so, angle OBE = angle E = 60
^{o}(In a triangle, congruent sides opposite congruent angles.) - so, angle BOE = 180
^{o}- 60^{o}- 60^{o}= 60^{o} - so, triangle BOE is an equilateral triangle (In a triangle, if each interior angle is 60
^{o}, then the triangle is an equilateral triangle.) - so, BE = r
- Because AE is a diameter, so angle ABE = 90
^{o}(In a circle, the diameter opposite 90^{o}angle.) - In the right triangle ABE,
- AB
^{2}+ BE^{2}= AE^{2} - b
^{2}+ r^{2}= (2r)^{2} - so, b
^{2}= 4r^{2}- r^{2}= 3r^{2} - so, b = (square root of 3) r
- The area (A) of the triangle ABC is,
- A = (1/2)BC × AD
- = (1/2) b × (square root of 3) b/2
- = (1/4) × b
^{2}× (square root of 3) - = (1/4) × 3r
^{2}× (square root of 3) - = 3(square root of 3)r
^{2}/4 - so, the area of the inscribed triangle is equal to 3(square root of 3)r
^{2}/4.