The bisectors of obtuse angles of an isosceles trapezoid intersect at its lower base, which is 12 cm long

| January 1, 2021 | education

The bisectors of obtuse angles of an isosceles trapezoid intersect at its lower base, which is 12 cm long, find the area of the trapezoid if its acute angle is 30 degrees.

In an isosceles triangle, the bisectors of the angles cut off the isosceles triangles.
The ABK triangle is isosceles, AB = AK, the CDK triangle is isosceles, DC = DC. Since AB = CD, then AK = DK = AD / 2 = 12/2 = 6 cm.
Let us lower the BH from the top B to the height. In a right-angled triangle ABH, according to the condition, the angle BAH = 300, then the leg BH, which is the height of the trapezoid, lies opposite the angle 300, which means it is equal to half the length of AB. BH = AB / 2 = 6/2 = 3 cm.
Let’s define the leg AH of the right-angled triangle ABH. AH = AB * Cos30 = 6 * √3 / 2 = 3 * √3 cm.
Then the length of the base BC = AD – 2 * AN = 12 – 6 * √3 cm.
Then the area of ​​the trapezoid is: S = (AD + BC) * BH / 2 = (12 + 12 – 6 * √3)) * 3/2 = 36 – 9 * √3 = 9 * (4 – √3) ≈ 20 , 4 cm2.
Answer: S ≈ 20.4 cm2.



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