The figure shows an isosceles triangular plat | vedclass

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(C) Let the height of the triangle be $h$. Since the apex angle is $90^o$ and the triangle is isosceles,the height $h$ bisects the base $L$. Thus,$h =

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$\frac{ML^2}{6}$

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(C) Let the height of the triangle be $h$. Since the apex angle is $90^o$ and the triangle is isosceles,the height $h$ bisects the base $L$. Thus,$h = L/2$.The area density $\sigma = M / (0.5 	imes L 	imes h) = M / (0.5 	imes L 	imes L/2) = 4M/L^2$.Consider a strip of thickness $dy$ at a distance $y$ from the apex. The length of the strip $l(y)$ is $2y$ (since the angle is $90^o$,the sides are $y=x$ and $y=-x$).The mass of the strip $dm = \sigma 	imes l(y) 	imes dy = (4M/L^2) 	imes (2y) 	imes dy = (8M/L^2) y dy$.The moment of inertia of this strip about the $z$-axis is $dI_z = dI_{cm} + dm 	imes y^2 = (dm 	imes l(y)^2 / 12) + dm 	imes y^2 = (dm 	imes (2y)^2 / 12) + dm 	imes y^2 = (dm 	imes y^2 / 3) + dm 	imes y^2 = (4/3) dm 	imes y^2$.Substituting $dm$: $dI_z = (4/3) 	imes (8M/L^2) y^2 	imes y dy = (32M / 3L^2) y^3 dy$.Integrating from $y=0$ to $y=h=L/2$: $I_z = \int_0^{L/2} (32M / 3L^2) y^3 dy = (32M / 3L^2) [y^4 / 4]_0^{L/2} = (8M / 3L^2) (L/2)^4 = (8M / 3L^2) (L^4 / 16) = ML^2 / 6$.

The figure shows an isosceles triangular plate of mass $M$ and base $L$. The angle at the apex is $90^o$. The apex lies at the origin and the base is parallel to the $X$-axis. The moment of inertia of the plate about the $z$-axis is

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