Find the area of Quardi lateral formed by four lines `3x+4y=24," "8x+15y=120` x axis y axis

Answer

To find the area of the quadrilateral formed by the lines \(3x + 4y = 24\), \(8x + 15y = 120\), the x-axis, and the y-axis, we can follow these steps:

### Step 1: Find the intersection points of the lines with the axes.

1. **For the line \(3x + 4y = 24\)**:
- When \(x = 0\):
\[ 3(0) + 4y = 24 \implies 4y = 24 \implies y = 6 \]
So, the point is \((0, 6)\).
- When \(y = 0\):
\[ 3x + 4(0) = 24 \implies 3x = 24 \implies x = 8 \]
So, the point is \((8, 0)\).

2. **For the line \(8x + 15y = 120\)**:
- When \(x = 0\):
\[ 8(0) + 15y = 120 \implies 15y = 120 \implies y = 8 \]
So, the point is \((0, 8)\).
- When \(y = 0\):
\[ 8x + 15(0) = 120 \implies 8x = 120 \implies x = 15 \]
So, the point is \((15, 0)\).

### Step 2: Identify the vertices of the quadrilateral.

The vertices of the quadrilateral formed by these lines and the axes are:
- \(A(0, 6)\) from the first line with the y-axis.
- \(B(8, 0)\) from the first line with the x-axis.
- \(C(15, 0)\) from the second line with the x-axis.
- \(D(0, 8)\) from the second line with the y-axis.

### Step 3: Calculate the area of the quadrilateral.

To find the area of the quadrilateral \(ABCD\), we can divide it into two triangles: \(AOB\) and \(COD\).

1. **Area of triangle \(AOB\)**:
- Base \(AB = 8\) (from \(A(0, 6)\) to \(B(8, 0)\)).
- Height from point \(A\) to the x-axis is \(6\).
\[ \text{Area}_{AOB} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 6 = 24 \]

2. **Area of triangle \(COD\)**:
- Base \(CD = 15\) (from \(C(15, 0)\) to \(D(0, 8)\)).
- Height from point \(D\) to the x-axis is \(8\).
\[ \text{Area}_{COD} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \times 8 = 60 \]

### Step 4: Calculate the total area of the quadrilateral.

The total area of the quadrilateral \(ABCD\) is the area of triangle \(COD\) minus the area of triangle \(AOB\):
\[ \text{Area}_{ABCD} = \text{Area}_{COD} - \text{Area}_{AOB} = 60 - 24 = 36 \]

### Final Answer:
The area of the quadrilateral formed by the lines and the axes is \(36 \, \text{cm}^2\).

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