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Question
Trapezium given below; find its area.
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Solution
In the given figure, we can observe that the non-parallel sides are equal and hence it is an isosceles trapezium.
Therefore, let us draw DE and CF perpendiculars to AB.
Thus, the area of the parallelogram is given by
AB = AE + EF + FB and CD = EF = 18 cm, we have
30 = AE + 18 + FB
⇒ 30 = AE + 18 + AE
⇒ 2AE + 18 = 30
⇒ 2AE = 30 - 18
⇒ 2AE = 12
⇒ AE = 6 cm
Now, consider the right angled triangle ADE.
AD2 = AE2 + DE2
⇒ 122 = 62 + DE2
⇒ 144 = 36 + DE2
⇒ DE2 = 144 - 36
⇒ DE2 = 108
⇒ DE = `sqrt( 36 xx 3 )`
⇒ DE = `6sqrt3`
Area( `square"ABCD" ) = "Area"( Δ"ADE" ) + "Area"( square "DEFC") + "Area"( Δ"CFB" )`
⇒ Area( `square"ABCD" ) = 1/2 xx 6 xx 6sqrt3 + 18 xx 6sqrt3 + 1/2 xx 6 xx 6sqrt3`
⇒ Area( `square"ABCD" ) = 6 xx 6sqrt3 + 18 xx 6sqrt3`
⇒ Area( `square"ABCD") = 144sqrt3`= 249.41 cm2
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