In trapezoid ABCD, angle A is 90 degrees. The distance between the midpoints of the larger base AD

| September 22, 2021 | education

In trapezoid ABCD, angle A is 90 degrees. The distance between the midpoints of the larger base AD and the lateral side CD is equal to the root of 18 cm, BC is equal to 6 cm.A) Find the angle CAD B) Find the distance from point D to line AC if the tangent of angle ACD is 2

Since points K and M are the middle of the sides AD and CD, then the segment KM is the middle line of the triangle ACD, then AC = 2 * KM = 2 * √18 = 6 * √2 cm.

In a right-angled triangle ABC, according to the Pythagorean theorem, we determine the length of the leg AB.

AB ^ 2 = AC ^ 2 – BC ^ 2 = 72 – 36 = 36. AB = 6 cm, therefore, triangle ABC is rectangular and isosceles, and the angles at the base of AC are 45.

Then the angle CAD = BAD = BAC = 90 – 45 = 45.

Let the length of the segment CH = X cm, then AH = AC – CH = (6 * √2 – X) cm.

In a right-angled triangle СDН tgDСН = DH / СН.

DН = СН * tgDСН = X * 2 cm.

Triangle АНD is rectangular and isosceles, then DH = АН = (6 * √2 – X).

Then (6 * √2 – X) = 2 * X.

3 * X = 6 * √2.

X = CH = 2 * √2 cm.

DН = 2 * СН = 4 * √2 cm.

Answer: Angle CAD is 45, the distance from point D to AC is 4 * √2 cm.



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