In trapezoid ABCD, the base of AD is twice the base of BC and twice the size of the side of CD.

| April 8, 2021 | education

In trapezoid ABCD, the base of AD is twice the base of BC and twice the size of the side of CD. Angle ADC is 60, side AB is 4. Find the area of the trapezoid.

Let us lower the heights of the trapezoid from vertices B and C to the base of AD.

Consider a right-angled triangle CC1D, whose angle C1 = 90, angle D = 60, then the angle C1CD = 180 – 90 – 0 = 30. Then the leg C1D, which lies opposite the angle 30, is equal to half of the hypotenuse CD.

Let BC = X, then by condition, AD = 2 * X, CD = X, and according to calculations, C1D = X / 2.

B1C1 = BC = X.

Then AD = 2 * X = AB1 + X + X / 2.

AB1 = 2 * X – X – X / 2 = X / 2.

AB1 = C1D = CD / 2 = 4/2 = 2 cm, then AB = CD = 4 cm, and the trapezium ABCD is isosceles.

AD = 2 + 4 + 2 = 8 cm.

Let us determine the length of the height CC1 by the Pythagorean theorem.

CC1 ^ 2 = CD ^ 2 – C1D ^ 2 = 16 – 4 = 12.

CC1 = √12 = 2 * √3.

Then the area of ​​the trapezoid is:

S = CC1 * (BC + AD) / 2 = 2 * √3 * (4 + 8) / 2 = 12 * √3 cm2.

Answer: The area of ​​the trapezoid is 12 * √3 cm2.



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