Repeat the experiment described in task 2, but this time dampen the paper with hot water.

| January 11, 2021 | education

Since the trapezoid is inscribed in a circle, this trapezoid is isosceles.
The ВН height divides the base of the AD into two segments. The length of the smaller of them is equal to the half-difference, and the larger half to the half-sum of the lengths of the bases.
AH = (BP – BC) / 2 = (20 – 4) / 2 = 8 cm, DН = (BP + BC) / 2 = (20 + 4) / 2 = 12 cm.
According to the Pythagorean theorem, in the triangle ABН, AB^2 = BH^2 + AH^2 = 144 + 64 = 208.
BH = 4 * √13 cm.
By the Pythagorean theorem, in the triangle ВDН, ВD^2 = ВH^2 + DН^2 = 144 + 144 = 288.
ВD = 12 * √2 cm.
Determine the area of ​​the AED triangle. Savd = BP * ВН / 2 = 20 * 12/2 = 120 cm.
A circle is described near the AED triangle, then R = AB * ВD * AD / 4 * Savd = 4 * √13 * 12 * √2 * 20/4 * 120 = 2 * √26 cm.
Answer: The radius of the circumscribed circle is 2 * √26 cm.



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