The isosceles triangle of least area that can be circumscribed about a circle of radius r turns out to be not just isosceles, but also equilateral. Each side has length 2r x ( 3 )0.5 . The area is r2 x (27)0.5 . Thanks are due to litotes for pointing out that the original answer did not actually answer the question ! tpm Since the equilateral triangle is also an isosceles triangle, we can say that at least area that can be circumscribed to a circle is the area of an equilateral triangle.
If we are talking only for isosceles triangle where base has different length than two congruent sides, we can say that at least area circumscribed to a circle with radius r, is the area of an isosceles triangle whose base angles are very close to 60 degrees. Solution: Let say that the isosceles triangle ABC is circumscribed to a circle with radius r, where BA = BC. We know that the center of the circle inscribed to a triangle is the point of the intersection of the three angle bisectors of the triangle. Let draw these angle bisectors, and denote with D the point where the bisector drawn from the vertex, B, of the triangle, intersects the base AC. Since the triangle is an isosceles triangle, then BD bisects the base and it is perpendicular to the base. So that AD = DC, OD = r, and the triangles ADB and AOD are right triangles (O is the center of the circle). In the triangle ADB, we have:
tan A = BD/AD, so that AD = BD/tan A In the triangle AOD, we have:
tan A/2 = OD/AD, so that AD = r/tan A/2, and AC = 2r/tan A/2 Therefore,
BD/tan A = r/tan A/2, and
BD = (r tan A)/tan A/2 Area of triangle ABC = (1/2)(AC)(BD) = (1/2)(2r/ tan A/2)[(r tan A)/tan A/2] = (r2 tan A)/tan2 A/2 After we try different acute angles measure, we see that the smallest area would be: If the angle A= 60⁰,
then the Area of the triangle ABC = r2 tan 60⁰/tan2 30⁰ ≈ 5.1961r2 If the angle A= 59.8⁰,
then the Area of the triangle ABC = (r2 tan 59.8⁰)/tan2 29.9⁰ ≈ 5.1962r2
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