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What is the area of an equilateral triangle inscribed in a circle of radius 4cm?
Wiki User
∙ 11y ago
Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.
Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.
And so the area of the inscribed triangle is 12 cm2.
Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.
Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.
And so the area of the inscribed triangle is 12 cm2.
Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.
Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.
And so the area of the inscribed triangle is 12 cm2.
Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.
Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.
And so the area of the inscribed triangle is 12 cm2.
Wiki User
∙ 11y ago
Wiki User
∙ 11y agoTwo vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.
Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.
And so the area of the inscribed triangle is 12 cm2.
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The shortest distance from the center of the inscribed circle to the triangle's sides is the circle's?
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True
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