SOLUTION: i wasn't sure where to go with this one but here it is: Point A is equidistant from P and Q. The midpoint, M, of line PQ is 12 units form A and 7 1/2 units from the midpoint of lin

Question 38264: i wasn't sure where to go with this one but here it is: Point A is equidistant from P and Q. The midpoint, M, of line PQ is 12 units form A and 7 1/2 units from the midpoint of line AQ. Find the length of line PQ. (you have to draw your own picture, which im not very good at, i keep getting weird square roots for PQ, please help!)
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
i wasn't sure where to go with this one but here it is: Point A is equidistant from P and Q. The midpoint, M, of line PQ is 12 units form A and 7 1/2 units from the midpoint(SAY N ) of line AQ THAT IS AN =7.5. Find the length of line PQ. (you have to draw your own picture, which im not very good at, i keep getting weird square roots for PQ, please help!)
IN TRIANGLE APQ,WE HAVE
M IS MID POINT OF PQ ...GIVEN
N IS MID POINT OF AQ...GIVEN.
HENCE
MN=AP/2...OR...AP =2*MN=2*7.5=15
IN TRIANGLES APM AND APQ
PM=MQ...GIVEN
AM=AM
AP=AQ...GIVEN
HENCE THEY ARE CONGRUENT AND HENCE ANGLE AMP=ANGLE AMQ =180/2=90
HENCE IN THIS RIGHT TRIANGLE
AP^2=AM^2+MP^2
MP^2=AP^2-AM^2=15^2-12^2=225-144=81
MP=9
HENCE PQ =2*MP=2*9=18
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