Find the radius of the circle inscribed in the parallelogram if its diagonals are 12cm and 3√2.

univerkov | September 26, 2021 | education

Since a circle is inscribed in a parallelogram, such a parallelogram is a rhombus.

The diagonals of the rhombus, at the intersection point O, are halved and intersect at right angles. Then AO = AC / 2 = 12/2 = 6 cm, OD = BD / 2 = 3 * √2 / 2 cm.

Determine the area of the triangle AOD.

Saod = AO * OD / 2 = 6 * 3 * √2 / 4 = 9 * √2 / 2 cm2.

By the Pythagorean theorem, we determine the length of the hypotenuse AD.

AD ^ 2 = AO ^ 2 + OD ^ 2 = 36 + 9/2 = (72 + 9) / 2 = 81/2.

AD = 9 / √2 cm.

Also Saod = AD * OH / 2.

OH = 2 * Saod / AD = 2 * (9 * √2 / 2) / (9 / √2) = √4 = 2 cm.

The height OH of the AOD triangle is the radius of the inscribed circle. R = 2 cm.

Answer: The radius of the inscribed circle is 2 cm.

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