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The following figure shows a square ABCD...

The following figure shows a square ABCD and an equilateral triangle DCE. Prove that :
(i) `angleADE=angleBCE=150^(@)`
(ii) `DeltaADE cong DeltaBCE`
(iii) `AE = BE`

Answer

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The adjoining figure shows a square ABCD and an equilateral triangle DEC. Prove that : (i) angleADE=angleBCE=30^(@) (ii) Delta cong DeltaBCE

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Knowledge Check

  • The give figure shows a square ABCD and an equilayeral teiangle APB. Calculate : {:((1)angleAOB,(ii)angleBPC),((iii)anglePCD,(iv)"reflex"angleAPC):}

    A
    `(i)75^(@)" "(ii)75^(@)" "(ii)15^(@)" "(iv)225^(@)`
    B
    `(i)55^(@)" "(ii)65^(@)" "(ii)35^(@)" "(iv)215^(@)`
    C
    `(i)65^(@)" "(ii)35^(@)" "(ii)45^(@)" "(iv)220^(@)`
    D
    `(i)45^(@)" "(ii)65^(@)" "(ii)35^(@)" "(iv)125^(@)`

  • In the figure below ,ABCD is a square ,MDC is an equilateral triangle .Find the value of x .

    A
    `75^(@)`
    B
    `90^(@)`
    C
    `105^(@)`
    D
    `60^(@)`

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