equilateral triangle is inscribed in a circle of radius 6 cm .find its length
Answers
Answer:
Let
ABC
ABC
be an equilateral triangle inscribed in a circle of radius 6 cm . Let
O
O
be the centre of the circle . Then ,
OA = OB = OC = 6 cm
OA=OB=OC=6cm
Let
OD
OD
be perpendicular from
O
O
on side
BC
BC
. Then ,
D
D
is the mid - point of
BC
BC
.
OB
OB
and
OC
OC
are bisectors of
\angle B
∠B
and
\angle C
∠C
respectively.
Therefore,
\angle OBD =30^o
∠OBD=30
o
In triangle
OBD
OBD
, right angled at
D
D
, we have
\angle OBD =30^o
∠OBD=30
o
and
OB=6 cm.
OB=6cm.
Therefore,
\cos (OBD)=\dfrac{BD}{OB}
cos(OBD)=
OB
BD
\implies \cos (30^o)=\dfrac{BD}{6}
⟹cos(30
o
)=
6
BD
\implies BD=6\cos 30^0
⟹BD=6cos30
0
\implies BD=6\times\dfrac{\sqrt 3}{2}=3\sqrt 3 cm
⟹BD=6×
2
3
=3
3
cm
\implies BC=2BD=2(3\sqrt 3)=6\sqrt 3 cm
⟹BC=2BD=2(3
3
)=6
3
cm
Hence, the side of the equilateral triangle is
6\sqrt 3 cm
6
3
cm
solution