An isosceles trapezoid is described around a circle of radius 3. The area of a quadrilateral whose vertices

| January 3, 2021 | education

An isosceles trapezoid is described around a circle of radius 3. The area of a quadrilateral whose vertices are the points of tangency between the circle and the trapezoid is 12. Find the area of the trapezoid.

In the MНRK quadrangle, the diagonals are perpendicular, then Smnrk = MR * NK / 2.
НK is the diameter of the inscribed circle, then НK = 2 * R = 2 * 3 = 6 cm.
Then 12 = MP * 6/2.
MR = 24/6 = 4 cm.Then ME = MR / 2 = 4/2 = 2 cm.
In the triangle MOE, ME = 2 cm, OM = R = 3 cm, then SinMOE = ME / OM = 2/3.
Angle MOE = 180 – MOE.
In the AMOK quadrangle, the angle ОМА = ОКА = 900, then the angle МАК = 360 – 90 – 90 – 180 + MOE = MOE, then SinBAK = 2/3.
ВL = NK = 6 cm, then SinBAK = 2/3 = VL / AВ.
AB = 3 * ВL / 2 = 3 * 6/2 = 9 cm.
AB = СD = 9 cm.
Since a circle is inscribed into the trapezoid, AB + СD = BC + AD = 18 cm.
Then Savsd = (ВС + AD) * НK / 2 = 18 * 6/2 = 54 cm2.
Answer: The area of ​​the trapezoid is 54 cm2.



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