Determine the area of the quadrilateral whose vertices are A(-14, -2), B(10, 10), C(-10,-10), and D(4, -24). Use Area of a quadrilateral ABCD = Area of △ABC+ Area of △ACD

To determine the area of quadrilateral \(ABCD\) using the method of dividing it into two triangles, we calculate the area of triangles \(ABC\) and \(ACD\) separately and then sum their areas.

### 1. Area of Triangle \(ABC\)
The vertices of triangle \(ABC\) are:
- \(A(-14, -2)\)
- \(B(10, 10)\)
- \(C(-10, -10)\)

The area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) can be calculated using the formula:

\[
\text{Area}_{ABC} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

Substitute the coordinates of points \(A\), \(B\), and \(C\):

\[
\text{Area}_{ABC} = \frac{1}{2} \left| (-14)(10 - (-10)) + 10(-10 - (-2)) + (-10)(-2 - 10) \right|
\]

Simplify the expression:

\[
\text{Area}_{ABC} = \frac{1}{2} \left| (-14)(20) + 10(-8) + (-10)(-12) \right|
\]

Calculate each term:

\[
(-14)(20) = -280
\]
\[
10(-8) = -80
\]
\[
(-10)(-12) = 120
\]

Add these up:

\[
\text{Area}_{ABC} = \frac{1}{2} \left| -280 - 80 + 120 \right| = \frac{1}{2} \left| -240 \right| = \frac{1}{2} \times 240 = 120
\]

So, the area of triangle \(ABC\) is \(120\) square units.

### 2. Area of Triangle \(ACD\)
The vertices of triangle \(ACD\) are:
- \(A(-14, -2)\)
- \(C(-10, -10)\)
- \(D(4, -24)\)

Using the same area formula for a triangle:

\[
\text{Area}_{ACD} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

Substitute the coordinates of points \(A\), \(C\), and \(D\):

\[
\text{Area}_{ACD} = \frac{1}{2} \left| (-14)(-10 - (-24)) + (-10)(-24 - (-2)) + 4(-2 - (-10)) \right|
\]

Simplify the expression:

\[
\text{Area}_{ACD} = \frac{1}{2} \left| (-14)(14) + (-10)(-22) + 4(8) \right|
\]

Calculate each term:

\[
(-14)(14) = -196
\]
\[
(-10)(-22) = 220
\]
\[
4(8) = 32
\]

Add these up:

\[
\text{Area}_{ACD} = \frac{1}{2} \left| -196 + 220 + 32 \right| = \frac{1}{2} \times 56 = 28
\]

So, the area of triangle \(ACD\) is \(28\) square units.

### 3. Area of Quadrilateral \(ABCD\)
Finally, sum the areas of triangles \(ABC\) and \(ACD\) to get the area of quadrilateral \(ABCD\):

\[
\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD} = 120 + 28 = 148
\]

Thus, the area of the quadrilateral \(ABCD\) is \( \boxed{148} \) square units.

Would you like further explanations or have any other questions?

Here are 5 questions related to this method:

1. How can we verify the area calculated using a different method?
2. What are the advantages of dividing a quadrilateral into triangles for area calculation?
3. How would the area change if one of the vertices was shifted?
4. How can the concept of vectors be used to find the area of the triangles?
5. What would happen to the area calculation if one of the points was collinear with two others?

**Tip:** When dividing a quadrilateral into two triangles, ensure that the diagonal used divides the shape into two non-overlapping triangles for accurate area calculation.

Learn how to calculate the area of quadrilateral ABCD with vertices at (-14, -2), (10, 10), (-10, -10), and (4, -24). Detailed step-by-step solution using the method of dividing the quadrilateral into two triangles. Suitable for students in Grades 9-12 studying coordinate geometry.

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