Question

O is the center of circle.AB is a chordof the circle and OM perpendicular to AB ,AB=20cm ,OM=2√11cm thwn what is the radius of the circle

Answer 1

By applying pythagoras formula ....answer is 12

Answer 2

Given that  AB = 20cm  OM = 2√11 cm 

we know that perpendicular line from center is bisector of chord

then 

OM is perpendicular bisector of AB

so 

$OM^{2} + (AB/2)^{2} = OA^{2}$

by substuting values
we got that

$(2 \sqrt{11}) ^{2} + 10^{2} = OA^{2}$

44 + 100 = OA^{2} 

$\sqrt{144} = OA$

thus OA = 12cm

as OA = r 

then radius of the circle is 12cm