For a any plane figure having x and y axis on the figure plane and z axis perpendicular to the figure plane (see figure attached) it can be shown that

`I_x =I_y =(1/2)*I_z`

regardless where is located the intersection of the x,y and z axis, and regardless how x and y axis are rotated in the plane of the figure (see the figure).

For a thin square sheet having sides of length l, the momentum of inertia with respect to a perpendicular axis through its center is

`I_z =(M*l^2)/6`

Therefore, relative to the x and y axis parallel to square sides in the figure plane we have

`I_x =I_y = (M*l^2)/12` .

Since the diagonals in a square are also perpendicular to each other than we can rotate x and y axis to become x' and y' axis and the momentum above stay the same.

`I_(x') =I_(y') =(M*l^2)/12`

Now the momentum of the right triangle PQR in the figure with respect to the axis PR is just half of the above momentum (from symmetry considerations).

`I_("triangle") =(1/2)*I_(y') =(M*l^2)/24`

**The correct answer is A)** `(M*l^2)/24`

**Further Reading**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now