Question

HELP ME FAST BECAUSE I AM TIMED

If NTM is congruent to OUM, SP is congruent to SQ, S is the center of the circle, and OM = 18, find NT. Round the answer to the hundredths place.

Answer 1

we know that , in a circle radius perpendicular to chord will bisect the chord.
OM=18, so OQ=QM=18/2=9.
Given QU=8
from figure OQU is a right angled triangle , so OU^2=OQ^2 + QU^2
OU^2 = 9*9 + 8*8 = 81+72=153;
OU=sqrt(153) = 12.37 =13(approx);
From given statements of congruent NT and OU will also be congruent or identical. So, NT=OU=13