Math please help

In triangle PQR, we have <P = 30 degrees, <Q = 60 degrees, and <R=90 degrees. Point X is on line PR such that line QX bisects <PQR. If PQ = 12, then what is the area of triangle PQX?

I have determined that i need to use the equation PR/\sqrt{3}, but i cannot find PR.

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  1. So you have the famous 30-60-90 right-angled triangle, which has sides in the ratio
    1 : √3 : 2
    (memorize those values, easy to remember that the smallest side is opposite the smallest angle and the largest side is opposite the largest angle)

    so using simple ratios
    QR : PR : 12 = 1 : √ 3 : 2

    QR/1 = 12/2 ---> QR = 6
    PR/√3 = 12/2 ---> PR = 12√3/2 = 6√3
    (did you notice that we simply multiplied each of the ratio terms by 6, keeping our new triangle similar to 1:√3:2 )

    area of PQR = (1/2)PR*QR
    = (1/2)(6√3)(6) = 18√3

    Since PXR has the same height, but only half the base ,
    its area is 9√3

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  2. Gggyy

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  3. x is not the midpoint of PR, so theres a bit more math to do. I don't believe 9 sqrt(3) is correct.

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