prove that the least perimeter of an isosclus triangle in which a circle of radius root 3 can be inscribed in 18r
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REF.Image
Let ABC be isosceles is AB=AC
r is radius of circle
AF
2
+BF
2
=AB
2
⇒(3r)
2
+(y
2
)=x
2
________ (1)
From ΔAOD
(2r)
2
=r
2
+(AD)
2
=3r
2
=AD
2
=AD=
3
r
Now BD=BF & EC=FC (as tangents are equal from an external point)
AD+DB=x
=
3
r+y
2
=x
⇒y
2
=x=
3
r -(2)
From (1) & (2)
∴(3r)
2
+(x−
3
r)=x
2
9r
2
+x
2
+3r
2
−2
3
rx=x
2
=9r
2
+3r
2
−2
3
rx=0
=12r
2
=2
3
rx
=6r=
3
x
∴x=
3
6
r
From (2)
y/2=(6/
3
)r−
3
r
=y/2=(3
3
/3)r⇒y=2
3
r
perimeter = 2x+y=2(6/
3
)r+2
3
r
=(12r+6r)/
3
=(18/
3
)r
=6
3
r
Hence proved.
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