The greater base of an isosceles trapezoid is 2.9 dm, the acute angle is 45 degrees. the lateral sides of this trapezoid are extended to their mutual intersection at point d. calculate the length of the perpendicular dropped from point d to the given base of the trapezoid
Let us draw the required height DH. Since DH is perpendicular to AE, the triangle DH is NOT rectangular.
According to the condition, the angle at a larger base is 45, then the angle NDE = 180 – 90 – 45 = 45, and therefore the triangle HDE is isosceles, DH = EH.
Since the angles at the base of an isosceles trapezoid are equal, then the angle BAE = CEA = 45, and then the triangle ADE is isosceles, then the height DH is also the median of the triangle, and then EH = AH = AE / 2 = 2.9 / 2 = 1.45 dm. Then DH = EH = 1.45 dm.
Answer: The height is 1.45 dm.
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