Answer:
Let O be the common center of two concentric circles and let AB be a chord of larger circle touching the smaller circle at P join OP
Since OP is the radius of the smaller circle to any chrod of the circle bisects the chord.
∴ AP=BP
In right ΔAPO we have
OA
2
=AP
2
+OP
2
⇒25−9=AP
2
⇒AP
2
=16⇒AP=4
Now AB=2,AP=2×4=8[∵AP=PB]
hence the length of the chord of the larger circle which touches the smaller circle is 8cm.
Step-by-step explanation:
