From given figure , we can see that
Let AB = 16 m , OM = 6 m , AM = BM = 8 m and OA = r
In triangle AMO ,
By pythagoras theorem ,
OA2 = AM2 + OM2
OA2 = 82 + 62 = 64 + 36
OA2 = 100 ? OA = ? 100 = 10
? OA = r = 10 m
The radius of the circular lawn = 10 m
Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle.
Let CD is another chord passes through point M.
We have to prove that AB < CD.
Now join OM and draw OL perpendicular to CD.
In right angle triangle OLM,
OM is the hypotenuse.
So OM > OL
? chord CD is nearer to O in comparison to AB.
? CD > AB
? AB < CD
So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point.
Hence required answer will be option D .
Then | AX | is equal to : |
AB |
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