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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 60o, so angle C = 60o
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 = 60o (In a triangle, congruent sides opposite congruent angles.)
so, angle BOE = 180o - 60o - 60o = 60o
so, triangle BOE is an equilateral triangle (In a triangle, if each interior angle is 60o, then the triangle is an equilateral triangle.)
so, BE = r
Because AE is a diameter, so angle ABE = 90o (In a circle, the diameter opposite 90o angle.)
In the right triangle ABE,
AB2 + BE2 = AE2
b2 + r2 = (2r)2
so, b2 = 4r2 - r2 = 3r2
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) × b2 × (square root of 3)
= (1/4) × 3r2 × (square root of 3)
= 3(square root of 3)r2/4
so, the area of the inscribed triangle is equal to 3(square root of 3)r2/4.