A isosceles trapezoid has a base of length 12 and a top of length 4 . Find its height (length of the red line) if $ AC \perp BD $.
Call the intersection of AC and BD = M
Because AC is perpendicular to BD, then in triangle BMC, angle BMC is right
And CM = BM = 12/sqrt (2)= legs of BMC
So the area of this triangle is product of the legs / 2 = (1/2) (12/sqrt 2) ^2 = 144/ 4 = 36
But the area of this triangle also = (1/2) 12 * altitude
36 = (1/2) 12 (altitude
36 = 6 * altitude
6 = altitude
And triangle DMA is similar to to triangle BMC
So the height of triangle DMA = 4/12 * 6 = 2
So..... the height of the trapezoid = 6 + 2 = 8