Question

The diagonal of an isosceles trapezoid with angle to the base is equal to 30. Find the length of the height of the trapezoid if it is known that the midline of this trapezoid is equal to 8 cm.

Answer 1

Answer:

4.62 cm to nearest hundredth.

Step-by-step explanation:

If the parallel sides are x and y then:

x + y = 2*8 = 16

x + y = 16

If we drop a perpendicular line from one of the upper points on the trapezoid we have the height.  Let the upper point be C and the point on the base be A. Let the point on right of the base be B.

AC is the height of the trapezoid.  AB is the baseline of the triangle CAB.

In triangle CAB the angle B is 30 degrees.

As this is a 30-60-90 degree triangle

AC/AB = 1/√3 so AC = AB/ √3.

As the trapezoid is isosceles:

AB = x + 0.5(y - x)

AB = 0.5x + 0.5y

So AC = 1 /√3 (0.5x + 0.5y)

           = 1 /√3 (0.5x + 0.5(16 - x))   (Substituting for x)

            = 1 /√3 (0.5x + 8 - 0.5x)

             =8 / √3    

.              = 4.6188 cm