Question

In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and ∠ A = 60°. Find: length of AB

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,


From right triangle ADP we have

cos 60° = `"AP"/"AD"`

`(1)/(2) = "AP"/(20)`

AP = 10

Similarly from the right triangle BMC we have BM = 10 cm.

Now from the rectangle PMCD we have CD = PM = 20 cm.

Therefore,
AB = AP + PM + MB = 10 + 20 + 10 = 40 cm.