The figure shows an isosceles triangular plat | vedclass

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(A) Let the height of the triangle be $h$. Since the apex angle is $90^o$ and the triangle is isosceles,the height $h$ is half the base length $L$. Th

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$\frac{ML^2}{8}$

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(A) Let the height of the triangle be $h$. Since the apex angle is $90^o$ and the triangle is isosceles,the height $h$ is half the base length $L$. Thus,$h = L/2$.The area density of the plate is $\sigma = \frac{M}{	ext{Area}} = \frac{M}{\frac{1}{2} 	imes L 	imes h} = \frac{M}{\frac{1}{2} 	imes L 	imes (L/2)} = \frac{4M}{L^2}$.Consider a thin strip of thickness $dy$ at a distance $y$ from the $X$-axis. The length of this strip $l(y)$ varies linearly from $0$ at the apex $(y=0)$ to $L$ at the base $(y=h)$.Using similar triangles,$l(y) = \frac{L}{h} y = \frac{L}{L/2} y = 2y$.The mass of this strip is $dm = \sigma 	imes l(y) 	imes dy = \left(\frac{4M}{L^2}ight) 	imes (2y) 	imes dy = \frac{8M}{L^2} y dy$.The moment of inertia of this strip about the $X$-axis is $dI = dm 	imes y^2 = \left(\frac{8M}{L^2} y dyight) y^2 = \frac{8M}{L^2} y^3 dy$.Integrating from $y=0$ to $y=h=L/2$:$I = \int_0^{L/2} \frac{8M}{L^2} y^3 dy = \frac{8M}{L^2} \left[ \frac{y^4}{4} ight]_0^{L/2} = \frac{8M}{L^2} 	imes \frac{1}{4} 	imes \left(\frac{L}{2}ight)^4 = \frac{2M}{L^2} 	imes \frac{L^4}{16} = \frac{ML^2}{8}$.

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 $X$-axis is

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