Find the distance of the center of gravity from point A on the line AB. The figure is a square of side $a$ from which two isosceles right triangles are removed from the top and bottom. The length AB is $a$, and the distance from A to the vertex of the removed triangle is $a/2$.

1. Problem Description

Find the distance of the center of gravity from point A on the line AB. The figure is a square of side $a$ from which two isosceles right triangles are removed from the top and bottom. The length AB is $a$, and the distance from A to the vertex of the removed triangle is $a/2$.

2. Solution Steps

Let $R$ be the distance of the center of gravity from point A. The area of the square is $a^2$. The area of each triangle that is removed is $(1/2) * (a/2) * (a/2) = a^2 / 8$. Since there are two such triangles, the total removed area is $2 * (a^2 / 8) = a^2 / 4$. The area of the remaining shape is $a^2 - a^2 / 4 = (3/4)a^2$.

The center of gravity of the original square is at $a/2$. The center of gravity of each removed triangle is located at $(2/3)(a/2) = a/3$ from A. The center of mass of the two removed triangles will also be at $a/3$ from A.

Let $A_1$ be the area of the square, $d_1$ be the distance of its center of gravity from A.

Let $A_2$ be the area of the two triangles, $d_2$ be the distance of their center of gravity from A.

Let $A_3$ be the area of the remaining shape, $R$ be the distance of the center of gravity of remaining shape from A.

The formula for the center of gravity of the remaining shape is

$R = \frac{A_1 d_1 - A_2 d_2}{A_1 - A_2} = \frac{A_1 d_1 - A_2 d_2}{A_3}$

$A_1 = a^2$, $d_1 = a/2$

$A_2 = a^2/4$, $d_2 = a/3$

$A_3 = (3/4) a^2$

$R = \frac{a^2 (a/2) - (a^2/4) (a/3)}{(3/4) a^2} = \frac{(a^3/2) - (a^3/12)}{(3/4) a^2} = \frac{(6a^3 - a^3)/12}{(3/4) a^2} = \frac{(5/12) a^3}{(3/4) a^2} = \frac{5}{12} * \frac{4}{3} a = \frac{5}{9} a$

3. Final Answer

The distance of the center of gravity from point A is $\frac{5}{9}a$. Therefore the answer is (3) $\frac{5}{9}R$