The chords AB and CD of a circle are parallel to each other and lie on the
opposite sides of the centre of the circle. If AB = 8 cm, CD = 6 cm and the
radius of the circle is 5 cm. Find the distance between the chords AB and
CD
Answers
ANSWER
Given−
AB=8cmandCD=6cmaretwoparallelchorddsof
acirclewithcetreO.
ThedistancebetweenAB&CDis1cm.
Tofindout−
Theradiusofthegivencircle=?
Solution−
WedropperpendicularOMonABfromO.OMmeets
ABatM.
OMisextendedtomeetCDatN.
OA&OCarejoined.
∴OA&OCareradiiofthegivencircle.
OM⊥AB⟹∠AMO=90
o
=∠CNO
(correspondinganglesoftwoparallellines)
SoON⊥CD.
SoMNisthedistancebetweenAB&CD
i.eMN=1cm.
LetOM=xcm,thenON=(x+1)cm.
NowOM⊥AB⟹AM=
2
1
AB=
2
1
×8cm=4cm
(sincetheperpendicular,fromthecentreofacircle
toanyofitschords,bisectsthelatter).
AgainOM⊥AB.
∴ΔOAMisarightonewithOAashypotenuse.
So,byPythagorastheorem,weget
OA=
OM
2
+AM
2
=
x
2
+4
2
......(i).
Similarly,ON⊥CD⟹CN=
2
1
CD=
2
1
×6cm=3cm
(sincetheperpendicular,fromthecentreofacircle
toanyofitschords,bisectsthelatter).
AgainON⊥CD.
∴ΔOCNisarightonewithOC=OAashypotenuse.
So,byPythagorastheorem,weget
OC=OA=
ON
2
+CN
2
=
(x+1)
2
+3
2
......(ii).
Comparing(i)&(ii)weget
x
2
+4
2
=
(x+1)
2
+3
2
⟹2x=6
⟹x=3cm.
So,consideringΔOMA
∠OMA=90
o
.
∴ΔOMAisarightonewithOAashypotenuse.
So,byPythagorastheorem,weget
OA=
OM
2
+AM
2
=
x
2
+4
2
=
3
2
+4
2
cm=5cm.
So the radius of the given circle is 5cm.
Ans− 8 cm