A.

rB = (0,0)

rC=(12,0) cm

For rA, draw a perpendicular AM from A on BC. As AB=AC, M will be the mid-point of BC. So BM=6cm

Hence x-coordinate of A will be 6.

For y-coordinate, using pythagoras thm., AM=sqrt(AB^2 - BM^2) = sqrt(10^-6^2) = sqrt(64) = 8.

So rA=(6,8) cm

Now coordinates of centre of gravity

=(200*rA+100*rB+100*rC)/(200+100+100)

=100(2rA+rB+rC)/400

={2(6,8) + (0,0) +(12,0)}/4

=(24,16)/4

=(6,4) cm

=(0.06,0.04)m

B.

Moment of inertia about an axis passing through A and perpendicular to the plane of paper is

I=200*0^2 + 100*10^2 +100*10^2 = 0 + 10000 + 10000 =20000 g*cm^2

=20000*10^-3*10^-4 kg m^2

=2*10^-3 kg m^2

C. Use the same method as in B

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