The smaller base of the BC of the trapezoid ABCD is 12 cm, AB = CD, the angle D = 45 degrees

| August 11, 2021 | education

The smaller base of the BC of the trapezoid ABCD is 12 cm, AB = CD, the angle D = 45 degrees, the height of the trapezoid is 8 cm, find the area of the trapezoid and the length of its midline.

Let’s build the height of the CH.

In a right-angled triangle СНD, the angle СDН = 45, then the right-angled triangle СDН is right-angled and isosceles. DH = CH = 8 cm.

Since the trapezoid ABCD is isosceles, the height CH divides the larger base into two segments, the length of the smaller of which is equal to the half-difference of the lengths of the bases.

DH = (AD – BC) / 2.

AD = 2 * DH + BC = 2 * 8 + 12 = 28 cm.

Determine the length of the midline of the trapezoid.

KM = (BC + AD) / 2 = (12 + 28) / 2 = 20 cm.

Determine the area of the trapezoid.

Savsd = KM * CH = 20 * 8 = 160 cm2.

Answer: The length of the middle line of the trapezoid is 20 cm, the area is 160 cm2.



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