In trapezium ABCD, the base of AD is twice the base of BC and twice the lateral side of CD

| May 22, 2021 | education

In trapezium ABCD, the base of AD is twice the base of BC and twice the lateral side of CD. The ADC angle is 60 degrees, the AB side is 6. Find the area of the trapezoid.

Let the smaller base of the trapezoid be equal to X cm, then, by condition, the larger base will be equal to 12 * X.

From the vertices B and C of the trapezoid, we lower two heights to the base of AD.

Consider a right-angled triangle СНD, which, according to the condition, has an angle СDН = 60, then the angle DСН = 180 – 90 – 60 = 30. The leg DН lies opposite the angle 30 and, accordingly, is equal to half of the CD.

DH = CD / 2 = X / 2.

Determine the length of the segment AK.

AK = AD – KN – DH = 2 * X – X – X / 2 = X / 2.

Since AK = DH, the trapezoid ABCD is isosceles, since only at the isosceles trapezoid the heights are cut off at the base of equal segments.

Then AB = BC = CD = 6 cm, and the base AD = 2 * 6 = 12 cm.

AK = AB / 2 = 6/2 = 3 cm.

Let us define the height of BK about the Pythagorean theorem.

BK ^ 2 = AB ^ 2 – AK ^ 2 = 36 – 9 = 27.

BK = √27 = 3 * √3 cm.

Determine the area of ​​the trapezoid.

S = (AD + BC) * BK / 2 = (12 + 6) * 3 * √3 / 2 = 27 * √3 cm2.

Answer: The area of ​​the trapezoid is 27 * √3 cm2.



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