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