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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