AB is a diameter of a circle .M is mid point in AB such that AM=18 cm and MB=8cm .Find the length of shortest chord through M.
Answers
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174
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
Answer.
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
Answer.
Attachments:
Answered by
3
Answer:
24
Step-by-step explanation:
12×2=24
hope it helps you
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