Question

In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and A = 60^°. Find: distance between AB and DC.

Answer 1

First, draw two perpendiculars to AB from point D and C respectively. Since AB || CD therefore PMCD will be a rectangle.

Consider the figure,

Again from the right triangle APD we have

sin 60° = `"PD"/(20)`

`sqrt(3)/(2) = "PD"/(20)`

PD = `10sqrt(3)`

PD = 10 × 1.732

PD = 17.32

Therefore, the distance between AB and CD is 17.32.