In a parallelogram ABCD with an angle A = 45 and side AD = 10 roots of 2 (dm), inscribed in a circle

| January 12, 2021 | education

In a parallelogram ABCD with an angle A = 45 and side AD = 10 roots of 2 (dm), inscribed in a circle: a) find the radius of the circle b) find the sum of the distances from the vertex D to the points of tangency of the circle with lines AD and DC.

Since a circle is inscribed in a parallelogram, this parallelogram is a rhombus.
Then AB = BC = CD = AD = 10 * √2 cm.
Determine the area of ​​the parallelogram.
Savsd = AВ * AD * Sin45 = 10 * √2 * 10 * √2 * √2 / 2 = 100 * √2 cm2.
Also Savsd = AD * MK.
MK = Savsd / AD = 100 * √2 / 10 * √2 = 10 cm.
The radius of a circle inscribed in a rhombus is half its height.
R = OK = MK / 2 = 10/2 = 5 cm.
In a parallelogram, the sum of adjacent angles is 1800, then the angle ADС = 180 – 45 = 1350.
OD is the bisector of the ADС angle, then the ADO angle = 135/2 = 67.50.
In a right-angled triangle KOD, tg67,5 = OK / DK.
DC = OK / tg67.5 = 5 / 2.41 = 2.07 cm.
Then DK + DN = 4.14 cm.
Answer: The radius of the circle is 5 cm, the sum of the segments is 4.14 cm.



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