+1207 In PQR, we have P=30°, Q=60°, R=90°, Point X is on PR such that QX bisects PQR. If PQ equals 12, then what is the area of PQX?
+193 pr = 12*sin60 = 12 * [sqrt(3)/2] = 6 * sqrt(3)
And we can find XR thusly :
XR/ sin30 = 6/ sin60
XR = 6sin30/ sin60
XR = 6(1/2) / [sqrt(3) / 2]
XR = 3*2/ sqrt(3) = 6/sqrt(3)
So PX = PR - XR = 6*sqrt(3) - 6/sqrt(3) = [18 - 6] / sqrt(3) = 12/sqrt(3)
And the area of PQX = (1/2) *PX* RQ = (1/2) (12/sqrt(3)) * (6) = 36 / sqrt(3) square units
https://web2.0calc.com/questions/in-pqr-we-have-p-30-q-60-r-90-point-x-is-on-pr-such-that-qx-bisects-pqr-if-pq-equals-12-then-what-is-the-area-of-pqx
