The radii of the circles with Centers A and B are 3 cm and 4 cm respectively. Find d (A, B)
Answers
Answer:
may be my answer helps you
Step-by-step explanation:
Given
OandO
′
aretwocirclesitersectingatP&Qrespectively.
OP&O
′
ParetangentstothecircleswithcentersO
′
&O
respectively.OP=OQ=3cmandO
′
P=O
′
Q=4cm.
Tofindout−
ThelengthofthecommonchordPQ.
Solution−
OP&O
′
Paretangentstothecirclesatthepointofcontact
ofthetangetsPandalsotheyarelinesjoiningthecentres
tothepointofcontact.
∴∠OPO
′
=90
o
.
i.eΔOPO
′
isarightonewithOO
′
ashypotenuse.
(SimilarlyΔOQO
′
isarightonewithOO
′
ashypotenuse.)
⟹O
′
O
2
=OP
2
+O
′
P
2
⟹O
′
O=
3
2
+4
2
cm=5cm..........(i)
NowbetweenΔOQO
′
&ΔOPO
′
hypOO
′
iscommon,sideO
′
P=O
′
Q
ΔOQO
′
&ΔOPO
′
arecongruent.
⟹∠POM=∠QOM.Buttheyarelinearpair.
∴∠POM=∠QOM=90
o
⟹ΔPOM&ΔQOMarerighttriangles
withOP&OQashypotenuses.
SobetweenΔPOM=ΔQOMwehave
hypotenusesOP=OQand
sideOP=OQ(radiiofthesamecircle)
∴ΔPOMiscongruenttoΔQOM
⟹∠OMP=∠OMQ.Buttheyareadjacentangles.
∴∠OMP=OMQ=90
o
.......(ii)andPM=QM⟹2PM=PQ....(iii)
NowbetweenΔOO
′
P&ΔOPMwehave
∠OMP=∠OO
′
P=90
o
(fromii),∠POMcommon.
SoΔOO
′
P&ΔOPMaresimilar.
∴
3
PM
=
OO
′
O
′
P
⟹PM=3×
5
4
cm=2.4cm.(fromi&iii)
i.ePQ=2PM=2×2.4cm=4.8cm
Ans−4.8cm.
Solution :
given AC=3cm
and BC=4cm
∴ABC=5cm
Let CP=ycm
and AP=xcm
CP^2+AP^2=AC^2
=> y^2+x^2=9
=> y^2=9−x^2
y^2+(5−x)^2=16
y^2=16−(5−x)^2
y^2=16−25−x^2+10x
9−x^2=−9−x^2+10x
18=10x
y^2=9−(1.8)^2
=24cm
CD=4.8 cm