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Byju's Answer
Standard VII
Mathematics
Area of a Circle
Prove that th...
Question
Prove that the least perimeter of an isosceles triangle in which a circle of radius
r
can be inscribed, is
6
√
3
r
.
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Solution
REF.Image
Let ABC be isosceles is AB=AC
r is radius of circle
A
F
2
+
B
F
2
=
A
B
2
⇒
(
3
r
)
2
+
(
y
2
)
=
x
2
________ (1)
From
Δ
A
O
D
(
2
r
)
2
=
r
2
+
(
A
D
)
2
=
3
r
2
=
A
D
2
=
A
D
=
√
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)
∴
(
3
r
)
2
+
(
x
−
√
3
r
)
=
x
2
9
r
2
+
x
2
+
3
r
2
−
2
√
3
r
x
=
x
2
=
9
r
2
+
3
r
2
−
2
√
3
r
x
=
0
=
12
r
2
=
2
√
3
r
x
=
6
r
=
√
3
x
∴
x
=
6
√
3
r
From (2)
y
/
2
=
(
6
/
√
3
)
r
−
√
3
r
=
y
/
2
=
(
3
√
3
/
3
)
r
⇒
y
=
2
√
3
r
perimeter =
2
x
+
y
=
2
(
6
/
√
3
)
r
+
2
√
3
r
=
(
12
r
+
6
r
)
/
√
3
=
(
18
/
√
3
)
r
=
6
√
3
r
Hence proved.
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Similar questions
Q.
Prove that the least perimeter of an isosceles triangle in which a circle of radius
r
can be inscribed is
6
√
3
r
.
Q.
Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is
6
3
r.