Question

p is the centre of the circle and seg QM is a tangent to the circle at point M if QM = 30, and PQ = 34, then find the radius of the circle

Answer 1

By property of Tangent is always = radius
Therefore, PM is perpendicular to MQ
now, angle M =90 degree
Therefore ,In triangle PMQ By Pythagoras theorem
PM^2 +MQ^2= PQ^2
PM^2+ 30^2=34^2
PM^2=1156-900
PM^2=256
PM=√256
PM=16units
HOPE IT HELPS

Answer 2

Answer:

By property of Tangent is always = radius

Therefore, PM is perpendicular to MQ

now, angle M =90 degree

Therefore ,In triangle PMQ By Pythagoras theorem

PM^2 +MQ^2= PQ^2

PM^2+ 30^2=34^2

PM^2=1156-900

PM^2=256

PM=√256

PM=16units

Step-by-step explanation: