Question 1175969: Equilateral triangle XYZ is insicribed in equilateral triangle ABC as shown. What is the ratio of the area of triangle XYZ to the area of triangle ABC.
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Found 2 solutions by MathLover1, Solver92311: Answer by MathLover1(20850) (Show Source): Answer by Solver92311(821) (Show Source):
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Angle B is a 60-degree angle by virtue of the fact that it is a vertex of an equilateral triangle and therefore is the vertex of an equiangular triangle whose measure must be one-third of 180 degrees.
Since angle BYZ is right, angle BZY must be 30 degrees (180 - 90 - 60 = 30)
Therefore triangle BYZ is a 30-60-90 right triangle, and by similar logic and the fact that triangle XYZ is equilateral as a given, triangles BYZ, AXY, and CZX are all congruent. Congruency demands that the three triangles mentioned must be equal in area.
Let the measure of segment BZ be 1 unit. Then by the properties of a 30-60-90 right triangle, segments BY, CZ, and AX must be unit, and YZ, ZX, and XY must be units.
Since segment BC is the sum of segments BZ and ZC, the measure of BC must be units, so if you drop a perpendicular to AC from vertex B at M, triangle BCM is both a 30-60-90 right triangle and is one-half of the area of triangle ABC. By the properties of a 30-60-90 right triangle, units and
Area of triangle ABC: units².
Area of triangle BYZ: . And the sum of the areas of triangles BYZ, AXY, and CZX is 3 times the area of BYZ, to wit:
The area of triangle XYZ is the difference between the area of triangle ABC and the sum of the areas of the three triangles just calculated, namely
Then the ratio of the area of XYZ to ABC is
John

My calculator said it, I believe it, that settles it
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