A trapezoid is inscribed in a circle of radius R, whose lower base is twice as large as each

| September 8, 2021 | education

A trapezoid is inscribed in a circle of radius R, whose lower base is twice as large as each of the other sides. Find the area of the trapezoid.

Let AB = BC = CD = X cm, then, by condition, AD = 2 * X cm.

Since the trapezoid is isosceles, the height CH cuts off the segment DH equal to the half-difference of the bases. DН = (2 X – X) / 2 = X / 2 cm.

From the right-angled triangle СDН, according to the Pythagorean theorem, we define the leg СН.

CH ^ 2 = CD ^ 2 – DH ^ 2 = X ^ 2 – X ^ 2/4 = 3 * X ^ 2/4.

CH = X * √3 / 2 cm.

Then Savsd = (BC + AD) * CH / 2 = 3 * X * (X * √3 / 2) / 2 = X ^ 2 * 3 * √3 / 4.

Consider a triangle ACD inscribed in a triangle.

By the Pythagorean theorem, AC ^ 2 = AH ^ 2 + CH ^ 2 = (9 * X ^ 2/4) + (3 * X ^ 2/4) = 12 * X ^ 2/4.

AC = X * √3 cm.

Sasd = AD * CH / 2 = 2 * X * (X * √3 / 2) / 2 = X ^ 2 * √3 / 2 cm2.

Also Saavs = (AC * CD * AD) / 4 * R = (X * √3) * X * 2 * X / 4 * R = X ^ 3 * √3 / 2 * R.

X ^ 2 * √3 / 2 = X ^ 3 * √3 / 2 * R.

X = R.

Then Sacd = R ^ 2 * √3 / 2 cm2.

Answer: The area of ​​the trapezoid is R ^ 2 * √3 / 2 cm2.



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