Find the area of the rectangle ABCD, if P is any point on AB such that DP = 18 mm, CP = 21 mm and `/_DPC = 90^@`.

Options

A. `378 mm^2`
B. `189 mm^2`
C. `126 mm^2`
D. `136 mm^2`

Answer

To find the area of rectangle ABCD given the conditions in the problem, we can follow these steps:

### Step 1: Understand the Geometry
We have rectangle ABCD with point P on side AB. We know that:
- DP = 18 mm
- CP = 21 mm
- Angle DPC = 90°

### Step 2: Identify Triangle DPC
Since angle DPC is 90°, triangle DPC is a right triangle. We can use the formula for the area of a triangle:
\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can consider DP as the base and CP as the height.

### Step 3: Calculate the Area of Triangle DPC
Substituting the values:
- Base (DP) = 18 mm
- Height (CP) = 21 mm

The area of triangle DPC is:
\[ \text{Area of triangle DPC} = \frac{1}{2} \times 18 \, \text{mm} \times 21 \, \text{mm} \]
Calculating this:
\[ = \frac{1}{2} \times 378 \, \text{mm}^2 = 189 \, \text{mm}^2 \]

### Step 4: Relate the Area of Triangle DPC to Rectangle ABCD
Since triangle DPC shares the same base (DP) and height (CP) with rectangle ABCD, we can use the relationship between the area of the triangle and the area of the rectangle. The area of the triangle is half of the area of the rectangle:
\[ \text{Area of triangle DPC} = \frac{1}{2} \times \text{Area of rectangle ABCD} \]

### Step 5: Solve for the Area of Rectangle ABCD
Let the area of rectangle ABCD be \( A \):
\[ 189 \, \text{mm}^2 = \frac{1}{2} A \]
Multiplying both sides by 2 to solve for \( A \):
\[ A = 189 \, \text{mm}^2 \times 2 = 378 \, \text{mm}^2 \]

### Conclusion
The area of rectangle ABCD is:
\[ \text{Area of rectangle ABCD} = 378 \, \text{mm}^2 \]