In the rectangular coordinate system, the vertices of a triangle have

Approximate by supposing that the vertex (3,2) is on the x-axis forming an isosceles triangle:
(3,2) -> (something>3;0)

Then, the area of this approximate isosceles triangle would be 6*11/2 = 33
The area of the actual triangle must be a little bit less than that of the approximate triangle. Thus 30, answer D.

Here is an alternate and a bit easier approach provide you know how to solve a 3x3 matrix => USE THE DETERMINANT METHOD.
Area of a triangle => 1/2|Determinant Value|


Lets get the determinant of the matrix =>

1 -3 0
1 3 2
1 0 11

Determinant => 1(33) - (-3)(11-2) + 0(0-3) => 33+27=60

Hence the area of the triangle => 1/2 * 60 => 30



SMASH THAT D.

area of triangle = ½ |y₁ (x₂ - x₃) + y₂ (x₃ - x₁) + y₃ (x₁ - x₂)|
=1/2 { 27 + 33}
=30

Using Matrix Method
|-3 0 1|
| 3 2 1|
| 0 11 1|

-3(2-11)+33=60
taking half of matrix=30

1/2 ( -3(2-11)+3(11-0)+0(0-2)) = 30

To find the area of a triangle given its coordinates, you can use the following formula:

Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

In this formula, (x1, y1), (x2, y2), and (x3, y3) are the coordinates of the three vertices of the triangle.

For your triangle with coordinates (-3,0), (3,2), and (0,11), you can plug in the values into the formula:

x1 = -3, y1 = 0
x2 = 3, y2 = 2
x3 = 0, y3 = 11

Area = 1/2 * |-3(2 - 11) + 3(11 - 0) + 0(0 - 2)|

Now, calculate each part of the expression:

Area = 1/2 * |-3(-9) + 3(11) + 0|

Area = 1/2 * (27 + 33)

Area = 1/2 * 60

Area = 30 square units

So, the area of the triangle with coordinates (-3,0), (3,2), and (0,11) is 30 square units.

Hence D