+1612 Help with tricky geometry please.
Triangle PQR is equilateral. A semicircle with center O is drawn with its diameter on PR so that one end is at P and the curved edge touches QR at X. The radius fo the semicircle is sqrt(3).
What is the length of QX?
+27 This problem is very similar:
https://www.quora.com/A-point-P-is-given-on-the-circumference-of-a-circle-of-radius-R-The-chord-QR-is-parallel-to-the-tangent-at-P-What-is-the-maximum-possible-area-of-a-triangle-PQR
See if that helps you
+128732 Draw OX = sqrt (3) = OP
Since QR is a tangent meeting the circle at X , then angle OXR = 90
And since the triangle is equilateral, then angle PRQ = 60
So triangle ORX is a 30 -60 -90 right triangle
So XR = OX /sqrt 3 = sqrt (3) / (sqrt (3) = 1
OR = 2
PR = OR + OP = 2 + sqrt (3) = QR
QR - XR = QX
2 + sqrt 3 - 1 = QX = 1 + sqrt 3
