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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