The side of the rhombus is 10√3 and the acute angle of the rhombus is 60 °. Find the radius

The side of the rhombus is 10√3 and the acute angle of the rhombus is 60 °. Find the radius of the inscribed circle of the rhombus.

The diagonals of the rhombus intersect at an angle of 90 and divide the angles at the vertices in half, which means that the triangle AOB is rectangular and the angle BAO = BAD / 2 = 60/2 = 300.

The leg BO of a right triangle lies opposite an angle of 30, which means it is equal to half the length of the hypotenuse AB. BO = AB / 2 = 10 * √3 / 2 = 5 * √3 cm, then the diagonal BD = 2 * BD = 5 * √3 * 2 = 10 * √3 cm.

Let us determine the length of the leg AO by the Pythagorean theorem.

AO ^ 2 = AB ^ 2 – BO ^ 2 = (10 * √3) ^ 2 – (5 * √3) ^ 2 = 300 – 75 = 225.

AO = √225 = 15 cm.

Then the diagonal of the rhombus AC = AO * 2 = 15 * 2 = 30 cm.

Determine the radius of the circle inscribed in the rhombus.

R = D1 * D2 / (4 * a), where D1 and D2 are the diagonals of the rhombus, and is the side of the rhombus.

R = 30 * 10 * √3 / (4 * 10 * √3) = 30/4 = 7.5 cm.

Answer: The radius of the circle is 7.5 cm.



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