In an isosceles trapezoid ABCD, the angle ABC = 135 and BC is less than AD. The segments BF and CH

| March 23, 2021 | education

In an isosceles trapezoid ABCD, the angle ABC = 135 and BC is less than AD. The segments BF and CH are the heights of the trapezoid. The length of the diagonal of the square FBCH is 6√2cm. Calculate the area of a trapezoid whose bases are the midline and the smaller base of the trapezoid ABCD.

Since, by condition, FBCN is a square, then BC = CH = FH = BF, and the angle between its diagonal and the side is 450, then Sin45 = CH / FC, CH = FC * Sin45 = 6 * √2 * √2 / 2 = 6 cm.

Angle ABF = ABC – FBC = 135 – 90 = 450, then triangle ABF is isosceles and right-angled, AF = BF = 6 cm.

Since the trapezoid is isosceles, the triangles ABF and CDH are equal in hypotenuse and acute angle, which means AF = DH = 6 cm.

Base length AD = AF + FH + DH = 6 + 6 + 6 = 18 cm.

Determine the length of the middle line KM = (BC + AD) / 2 = (6 + 18) / 2 = 12 cm.

The length of the segment BE = BF / 2 = 6/2 = 3 cm, since the CM divides BF in half.

Determine the area of ​​the trapezium KВСM.

S = (BC + KM) * BE / 2 = (6 + 12) * 3/2 = 27 cm2.

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



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