Question

Question #5b3a4

Answer 1

To find the median of a trapezoid we need the top and bottom lengths.
Start by constructing line BH such that BH is perpendicular to AD and H is on the line AD. We can AD using BC and a little trig. You can quickly calculate the length of AH via the right triangle ABH.

In trigonometry the cosine of an angle is the nearby side from the angle over the hypotenuse. Thus, `cos(53)=(AH)/(AB)`
`cos(53)*AB=AH`
`AB=15`
thus
`cos(53)*AB=9.027...=AH`

You then find BH using the Pythagorean theorem.
`AH^2+BH^2=225`
`BH=sqrt(225-AH^2)`
`BH=11.979`

Since the height in a trapezoid is the same everywhere, we can draw C's perpendicular to line AD and create line CF.

`CF=BH=11.979`

We have triangle CFD with 45-45-90 degrees, which means each of the short legs are 1/sqrt(2) of the hypotenuse and the short legs are equal

`FD=BH=8.47`

Now we have `AD=BC+AH+FD=29.476`

The median of a trapezoid is the average of its top and bottom sides:

`median=(29.476+8)/2=18.736`