Question

In the figure, O is the center. If radius 3 cm and <AOB = 60º, then find the perimeter of ∆AOB.

Answer 1

As radius=3cm,
AO=BO=radius=3cm

Two sides are equal and one angle is 60 °
Therefore it is an equilateral triangle.

Perimeter of AOB=3a
Perimeter = 3(3)
Hence, Perimeter=9 Cm

Hope it helps you

Answer 2

Answer:

OA=OB (radii of the circle)

∠OBA=∠OAB (Opposite angles of equal sides are equal)

In ∆AOB

=  ∠OAB + ∠AOB + ∠OBA= 180 (Angle sum property of triangle)

= 2∠OAB+60=180 (from eq 1)

=  2∠OAB =180-60

= ∠OAB = 120/2

= ∠OAB = 60 =  ∠OBA

Since ∠OAB = ∠AOB = ∠OBA = 60

Therefore AOB is an equilatreral triangle

AB=BO=BA= 3 cm

Perimeter of ∆AOB = AB+BO+BA

                                = 3+3+3

                                = 9 cm

Therefore the perimeter of ∆AOB is 9 cm

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