Question

AB is a diameter in circle M m(

Answer 1

Solution:-

Given : AM = 18 cm and MB = 8 cm

So,  

Diameter AB = AM + MB

= 18 + 8 = 26 cm

Now, we have to draw a perpendicular on point M. As that meet on both sides of the circle at C and D. So, smallest chord that passes through point M is CD.

Then, radius AO = OB = 13 cm

So, in OMC

we know that  

OC = 13 cm = radius

OM = AM - AO

= 18 - 13

= 5 cm

By using Pythagoras theorem, we get.

OC² = OM² + CM²

13² = 5² + CM²

CM² = 169 - 25

CM = √144

CM = 12 cm

SO, CD = 12*2  

CD = 24 cm

So, length of the smallest chord that passes through M is 24 cm