Particles of weight 4 N, 8 N, P N and Q N are placed at the points (4, 5), (6, -4), (10, 13) and (-6, 5) respectively - Leaving Cert Applied Maths - Question 6 - 2018

Now to find q, we use the y-coordinate of the centre of gravity:

The equation for the y-coordinate is given by:
$y_{cg} = \frac{\sum m_iy_i}{\sum m_i}$
Thus, substituting the known values:

$p + 4(5) + 8(-4) + 13(q) = 12 + p + q$

From prior calculations, we get:
$4(5) + 8(-4) + 13p + 5q = 12 + p + q$
Substituting p = 1 gives:

$13 + q = 1\cdot 5 + 12 + p + q$
Solving gives:
$q = 3$.

We begin by calculating the area of the square ABCD:

$\text{Area}_{ABCD} = 12 \times 12 = 144$

Next, we determine the area of triangle EBC:

$\text{Area}_{EBC} = \frac{1}{2}(6)(12) = 36$

Then, we calculate the area of the circular portion with radius 3:

$\text{Area}_{circle} = \pi r^2 = \pi(3)^2 = 36\pi$

Total area of the remaining lamina is:

$\text{Area}_{lamina} = 144 - 36 - 36\pi$

The centroid of the square is at (6, 6). For triangle EBC it is at (x, y):

The coordinates of centroid of triangle EBC:
$x = \frac{(0 + 12 + 12)(1/2)}{36} = \frac{12}{36}$

And $y$ calculated as follows:\

$y = \frac{(0 + 0 + 12)(1/2)}{36} = 0$
Substituting into overall area gives us:
$y = 5.84$
Thus, the coordinates are (x, y) = (3, 5.84).