In the adjoining figure o is the centre of the circle AB = 16
, CD =14 seg om perpendicular srg AB and seg on perpendicular seg CD if om =6 then find seg on
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Given- AB=16 cm and CD=14 cm are the chordsof a circle with centre at O.
OM(=6 cm)⊥AB at M and ON⊥CD at N.
To find out -
If the length of ON=
m
cm, then m=?
Solution-
We join OCand OA.
ΔOAM and ΔOCN are right ones, since OM⊥AB at M and ON⊥CD at N.
Now AM=
2
1
AB=
2
1
×16 cm =8 cm and CN=
2
1
CD=
2
1
×14 cm =7 cm since the perpendicular from the centre of a circle to a chord bisects the latter.
So, in ΔOAM, by Pythagoras theorem, we have
OA=
OM
2
+AM
2
=
6
2
+8
2
cm =10 cm and it\ is the radius of the circle.
∴OC=OA=10 cm.
Again in ΔOCN, by Pythagoras theorem, we have
ON=
OC
2
−CN
2
=
10
2
−7
2
cm =
51
cm
But given that ON=
m
.
∴ m=51cm
2
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