The centroid of a triangle with vertices $(2a,b),(-b,a)$ and $(b,3a)$ is $(4,11)$. The area of the triangle is ___ square units.

The centroid of a triangle is the average of its three vertices. For vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, the centroid is $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.

Finding the values of $a$ and $b$:

For the $x$-coordinate of the centroid:

$\frac{2a + (-b) + b}{3} = 4$

The $-b$ and $+b$ cancel out:

$\frac{2a}{3} = 4$

$a = 6$

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For the $y$-coordinate of the centroid:

$\frac{b + a + 3a}{3} = 11$

$\frac{b + 4a}{3} = 11$

$b + 4a = 33$

Substituting $a = 6$:

$b + 24 = 33$

$b = 9$

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Finding the actual vertices:

Substituting $a = 6$ and $b = 9$ into the original vertices:

Original Vertex	Actual Coordinates
$(2a, b)$	$(12, 9)$
$(-b, a)$	$(-9, 6)$
$(b, 3a)$	$(9, 18)$

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Calculating the area:

The area formula for a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is:

$\text{Area} = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Using $(12, 9)$, $(-9, 6)$, $(9, 18)$:

$\text{Area} = \frac{1}{2}|12(6 - 18) + (-9)(18 - 9) + 9(9 - 6)|$

$= \frac{1}{2}|12(-12) + (-9)(9) + 9(3)|$

$= \frac{1}{2}|-144 - 81 + 27|$

$= \frac{1}{2}|-198|$

$= \frac{1}{2} \times 198 = 99$

The area of the triangle is 99 square units.