math/ geomery

the diagonals of a rhombus have lengths 4 and 12. what is the length of a side in simplest form.

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  1. a=sqroot(12^2+4^2)

    a=sqroot(144+16)

    a=sqroot(160)

    a=sqroot(16*10)

    a=4*sqroot(10)=12.64911
    Becouse sqroot(16)=4

    a=4*sqroot(10)=12.64911

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  2. That is the wrong answer.

    p=first diagonal=4
    q=second diagonal=12

    a=sqroot[(p/2)^2+(q/2)^2]

    a=sqroot[(4/2)^2+(12/2)^2]

    a=sqroot(2^2+6^2)

    a=sqroot(4+36)

    a=sqroot(40)

    a=sqroot(4*10)

    a=2*sqroot(10)

    OR:

    p^2+q^2=4a^2 Divide with 4

    a^2=(p^2+q^2)/4

    a=sqroot[(p^2+q^2)/4]

    a=sqroot(p^2+q^2)/sqroot(4)

    a=sqroot(p^2+q^2)/2

    a=sqroot(4^2+12^2)/2

    a=sqroot(16+144)/2

    a=sqroot(160)/2

    a=sqroot(16*10)/2

    a=4*sqroot(10)/2

    a=2*sqroot(10)

    Check this.

    In google type:
    mathworld.wolfram rhombus

    When page be open click on :
    Rhombus-from Wolfram MathWorld

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