Find the area of quadrilateral formed by joining the points (-4,2) , (1,-1) ,(4,1) and (2,5) :

Options

A. 25.4
B. 20.5
C. 24.5
D. none of these

Answer

To find the area of the quadrilateral formed by the points (-4, 2), (1, -1), (4, 1), and (2, 5), we can follow these steps:

### Step 1: Identify the Points
Let the points be:
- A = (-4, 2)
- B = (1, -1)
- C = (4, 1)
- D = (2, 5)

### Step 2: Divide the Quadrilateral into Two Triangles
We can divide the quadrilateral ABCD into two triangles by drawing a diagonal from A to C. This gives us triangles ABD and BCD.

### Step 3: Calculate the Area of Triangle ABD
We will use the formula for the area of a triangle given by the coordinates of its vertices:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For triangle ABD, we have:
- A = (-4, 2) → \(x_1 = -4, y_1 = 2\)
- B = (1, -1) → \(x_2 = 1, y_2 = -1\)
- D = (2, 5) → \(x_3 = 2, y_3 = 5\)

Substituting the values into the formula:
\[ \text{Area}_{ABD} = \frac{1}{2} \left| -4(-1 - 5) + 1(5 - 2) + 2(2 - (-1)) \right| \]
Calculating each term:
\[ = \frac{1}{2} \left| -4(-6) + 1(3) + 2(3) \right| \]
\[ = \frac{1}{2} \left| 24 + 3 + 6 \right| \]
\[ = \frac{1}{2} \left| 33 \right| = \frac{33}{2} \]

### Step 4: Calculate the Area of Triangle BCD
Now we calculate the area of triangle BCD using the same formula:
- B = (1, -1) → \(x_1 = 1, y_1 = -1\)
- C = (4, 1) → \(x_2 = 4, y_2 = 1\)
- D = (2, 5) → \(x_3 = 2, y_3 = 5\)

Substituting the values into the formula:
\[ \text{Area}_{BCD} = \frac{1}{2} \left| 1(1 - 5) + 4(5 - (-1)) + 2(-1 - 1) \right| \]
Calculating each term:
\[ = \frac{1}{2} \left| 1(-4) + 4(6) + 2(-2) \right| \]
\[ = \frac{1}{2} \left| -4 + 24 - 4 \right| \]
\[ = \frac{1}{2} \left| 16 \right| = 8 \]

### Step 5: Calculate the Total Area of Quadrilateral ABCD
Now, we add the areas of triangles ABD and BCD:
\[ \text{Area}_{ABCD} = \text{Area}_{ABD} + \text{Area}_{BCD} = \frac{33}{2} + 8 \]
Converting 8 to a fraction:
\[ 8 = \frac{16}{2} \]
Thus,
\[ \text{Area}_{ABCD} = \frac{33}{2} + \frac{16}{2} = \frac{49}{2} \]

### Final Answer
The area of the quadrilateral ABCD is:
\[ \text{Area} = \frac{49}{2} \text{ square units} = 24.5 \text{ square units} \]