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