Two chords PQ and MN
of length 11 cm and 5 cm respectively of a circle are parallel to each other and are on the opposite side of its Centre is the distance between chord MN and chord PQ is 6 cm find the radius of the circle
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Step-by-step explanation:
Given−
Oisthecentreofacirclewithradius=r.
PQ&MNaretwoparallelchordswhen
PQ=11cmandMN=5cm.
ThedistanceABbetweenPQ&MN,throughthecentre
is6cm.
Tofindout−
r=?
Solution−
ThelineABistheperpendicuardistancebtweenPQ&MN.
AtthesametimeOA&OBarethedistancesofPQ&MN
fromthecenterO.
∴∠OAQ=∠OBN=90
o
sincetheperpediculardistance
ofachordfromthecentrebisectsthechordatrightangle.
i.eΔOAQ&ΔOBNarerighttriangleswithhypotenusesas
OQ=randON=rrespectively.
AlsoOA=
2
11
cmandOB=
2
5
cm.
LetustakeOA=x.ThenOB=AB−OA=6−x.
Then,applyingPythagorastheorem,wehave
OA
2
+AQ
2
=r
2
andOB
2
+BN
2
=r
2
.
∴OA
2
+AQ
2
=OB
2
+BN
2
⟹x
2
+(
2
11
)
2
=(6−x)
2
+(
2
5
)
2
⟹x=1.
∴r=
OA
2
+AQ
2
=
x
2
+(
2
11
)
2
=
1
2
+(
2
11
)
2
cm=
2
5
5
cm
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