A regular triangle is inscribed in the circle, the side of which is 6 cm. Calculate the area of a square inscribed in the same circle.
Since triangle ABC is regular, its internal angles are 60.
The inscribed angle ACB rests on the arc AB, then the degree measure of the arc AB = 120.
Then the central angle AOB is also based on the arc AB, and then the angle AOB = 120.
In an isosceles triangle AOB AO = OB = R, then, by the cosine theorem:
AB ^ 2 = R ^ 2 + R ^ 2 – 2 * R * R * Cos120.
36 = 2 * R ^ 2 – 2 * R ^ 2 * (-1/2).
3 * R ^ 2 = 36.
R ^ 2 = 12.
R = 2 * √3 cm.
The segment OM = R = 2 * √3 cm and is equal to half the diagonal of the square of the CMHR. Then MР = 4 * √3 cm.
Let’s define the area of a square in terms of its diagonal.
Skv = MР ^ 2/2 = 48/2 = 24 cm2.
Answer: The area of the square is 24 cm2.
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