A perpendicular dropped from the top of an obtuse angle to the larger base of an isosceles trapezoid

| January 31, 2021 | education

A perpendicular dropped from the top of an obtuse angle to the larger base of an isosceles trapezoid divides it into parts having lengths 94 and 51, find the midline of this trapezoid.

Since the trapezoid is isosceles, its height divides the larger base into two segments, the length of the larger of which is equal to half the sum of the lengths of its bases.

DН = (BC + AD) / 2, which is the formula for the midline.

Then KM = DН = 94 cm.

Second way.

Let’s build the second height CP. Rectangular triangles ABН and СDР are equal in hypotenuse and acute angle, then DР = AH = 51 cm.

The quadrangle of the ВСРН is a rectangle, then НР = ВС = DН – DР = 94 – 51 = 43 cm.

Base AD = AН + DН = 51 + 94 = 145 cm.

Then KM = (BC + AD) / 2 = (43 + 145) / 2 = 188/2 = 94 cm.

Answer: The middle line of the trapezoid is 94 cm.



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