circle of radius 2cm is inscribed in an equilateral triangle find the area of the triangla

We have the following situation-

In triangle OMC,

Therefore,

So area of the triangle ABC is-

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 root3/4*2*2

root3 is the area of triangle

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Would be easier with a diagram but here it goes

Let the triangle be ABC
Let the circle be DEF (the 3 points of contact) (D lies on AB, E on BC, F on AC)
O will be the centre

Join OD , OE, OF and OB , OA , OC
 In Triangle OBE and Triangle OBD

OD is perpendicular to AB and OE is perpendicular to BC, So 900
OE = OD (Radii)
OB = OB (Common) 
Therefore Triangle OBE is congruent to Triangle OBD
Hence, Angle OBE = Angle OBD

Similarly, in the other triangles , we get
Angle OCE = Angle OCA
Angle OAF = Angle OAD

Now Angle CAB, ABC, BCA = 60 degrees (eq. triangle)

Which means Angle OBD = OBE = 30 degrees (half)
Do Tan (Angle OBE)
Tan30 = 1/√3

1/√3 = radius (2 cm) / BE
Therefore BE = 2√3
Similarly, CE = 2√3

So BC = 4√3

Ar(eq. Triangle) = [(Side) 2 √3] / 4
= 48√3 / 4
= 12√3 cm2

Answer is lengthy coz of that Angle, just keep in mind that when you join OA, OB and OC, Angle A, B and C are bisected, the congruence shown above is its proof.

:)
 

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