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(D) Let the mass of the square plate be $M$ and its area be $a^2$. The mass per unit area is $\sigma = M/a^2$.The triangle cut out has a base $a$ and

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$a / 9$

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(D) Let the mass of the square plate be $M$ and its area be $a^2$. The mass per unit area is $\sigma = M/a^2$.The triangle cut out has a base $a$ and height $a/2$. Its area is $A_t = \frac{1}{2} 	imes a 	imes \frac{a}{2} = \frac{a^2}{4}$.The mass of the triangle is $m = \sigma A_t = \frac{M}{a^2} 	imes \frac{a^2}{4} = \frac{M}{4}$.The center of mass of the triangle is at a distance $d = \frac{1}{3} 	imes 	ext{height} = \frac{1}{3} 	imes \frac{a}{2} = \frac{a}{6}$ from the center of the square.Let the center of the square be the origin $(0,0)$. The center of mass of the square is at $(0,0)$.The center of mass of the triangle is at $(-a/6, 0)$.Let $M'$ be the mass of the remaining part,$M' = M - m = M - M/4 = 3M/4$.Let $x_{cm}$ be the center of mass of the remainder. Using the principle of moments: $M(0) = M'(x_{cm}) + m(-a/6)$.$0 = (3M/4)x_{cm} - (M/4)(a/6)$.$(3/4)x_{cm} = a/24$.$x_{cm} = \frac{a}{24} 	imes \frac{4}{3} = \frac{a}{18}$.Wait,re-evaluating the geometry: The distance of the centroid of the triangle from the apex is $2/3$ of the median. The median is $a/2$. So distance is $\frac{2}{3} 	imes \frac{a}{2} = \frac{a}{3}$.Actually,the distance from the center of the square to the centroid of the triangle is $\frac{1}{3} 	imes 	ext{height} = \frac{1}{3} 	imes \frac{a}{2} = \frac{a}{6}$.Using $x_{cm} = \frac{m_1 x_1 - m_2 x_2}{m_1 - m_2}$,where $m_1 = M, x_1 = 0$ and $m_2 = M/4, x_2 = -a/6$.$x_{cm} = \frac{0 - (M/4)(-a/6)}{M - M/4} = \frac{Ma/24}{3M/4} = \frac{a}{24} 	imes \frac{4}{3} = \frac{a}{18}$.Given the options,let's re-check the triangle height. If the triangle base is $a$ and it spans from the center to the edge,the height is $a/2$. The distance of the centroid from the apex is $2/3$ of the height,which is $2/3 	imes a/2 = a/3$. No,the distance from the apex is $2/3$ of the median. The distance from the base is $1/3$ of the median. The apex is at the center. So the distance from the center is $2/3 	imes (a/2) = a/3$. Recalculating: $x_{cm} = \frac{(M/4)(a/3)}{3M/4} = \frac{a}{12} 	imes \frac{4}{3} = \frac{a}{9}$.

From a homogeneous square plate,a triangle is cut out as shown in the figure. The side of the square is $a$,and the apex of the triangle is at the center of the square. The distance from the center of the square to the center of mass of the remainder of the plate is:

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