An equilateral triangle of side 9cm
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An equilateral triangle of side 9 cm inscribed in a circle The radius of the circle is
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C
3
3
cm
△ABC is an equilateral triangle
AB=BC=CA=9cm
O is the circumcentre of △ABC
∴OD id the perpendicular of the side BC
In △OBD and △ODC
OB=OC (Radius of the circle)
BD=DC (D is the mid point of BC)
OD=OD (common)
∴△OBD=△ODC
⇒∠BOD=∠COD
∠BOC=2∠BAC=2×60
∘
=120
∘
(The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle)
∴∠BOD=∠COD=
2
∠BOC
=
2
120
∘
=60
∘
BD=BC=
2
BC
=
2
9
cm
In △BOD
⇒sin∠BOD=sin60
∘
=
OB
BD
∴
2
3
=
OB
2
9
⇒OB=
2
9
×
3
2
=3
3
cm
Answered by
1
Answer:
9×9×9×9=6561
yehi answer hai
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