Physics - KE/inertia

The rigid body shown in Figure 10-66 (it is a triangle with the upper mass = to M and the bottom two corners/masses equal to 2M. The bottom side length equals .70 cm with point P in the middle of the side and the other two sides equal .55 cm)consists of three particles connected by massless rods. It is to be rotated about an axis perpendicular to its plane through point P. If M = 0.40 kg, a = 35 cm, and b = 55 cm, how much work is required to take the body from rest to an angular speed of 5.0 rad/s?
*My work so far*
I believe I need to use:
W = (1/2)(I)(ang speed)^2
I am not sure how to find I though. Would I use I = I1 + I2 + I3 and calculate I for each of the 3 masses?

For Further Reading

* Physics - KE/Inertia - bobpursley, Monday, March 26, 2007 at 4:03pm

Yes, use the addition formula to get the moment of inertia. In each case, the distance is the distance of the mass from point P.

My work:
I found I1 + I2 + I3 = Itotal
I1 = (.80)(.35)^2 = 0.098 kg*m^2
I2 = (.80)(.35)^2 = 0.098 kg*m^2
I3 = (.40)(.18)^2 = 0.01296 kg*m^2, for this r I found h of the isosceles triangle by doing h=sqrt(b^2-.25(a^2))
I added up the 3 I's and got Itotal to be 0.012946 kg*m^2.
Next I used the equation:
Work = .5*Itotal*omega^2
Work = .5(0.012946)(5.0)^2 = 2.612 Joules
This is apparently the wrong answer but I don't know why. Do you see any mistakes? Thanks!

Never mind...I forgot to take the square root when finding h. I got it now.

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