How to Calculate Area of Rectangle in Java

The Challenge

Calculate the area of a rectangle using the formula: Area = length × width

Solution

java

public class Main {
    public static void main(String[] args) {
        double length = 10.5;
        double width = 5.2;
        
        double area = length * width;
        
        System.out.println("Length: " + length);
        System.out.println("Width: " + width);
        System.out.println("Area: " + area);
        // Output: Area: 54.6
    }
}

Using a Method

java

public static double calculateArea(double length, double width) {
    return length * width;
}

// Usage
double area = calculateArea(10.5, 5.2);
System.out.println("Area: " + area); // 54.6

Using a Class

java

class Rectangle {
    double length;
    double width;
    
    Rectangle(double l, double w) {
        length = l;
        width = w;
    }
    
    double getArea() {
        return length * width;
    }
}

// Usage
Rectangle rect = new Rectangle(10.5, 5.2);
System.out.println("Area: " + rect.getArea()); // 54.6

Did You Know?

A square is a special type of rectangle where length equals width! You can use the same formula: Area = side × side = side²

Pro Tip

Add perimeter calculation too! Perimeter = 2 × (length + width)

Real-World Use Cases

Construction: Calculate floor area for materials
Real estate: Compute property sizes
Graphics: Calculate canvas or screen areas
Farming: Measure field dimensions

›Short versionOpen if you only have a minute.

Area is length times width, in one unit, after rejecting a negative side.
int times int is an int product. 50_000 * 50_000 wraps.
Promote one side to long before multiplying when the product must survive.
A zero side is a zero area. It is not the same bug as a negative side.

Rectangle area

Multiply length by width after validating that dimensions make sense.

Working approach

Reject a negative length or width before multiplying.
Use long area = (long) length * width when the inputs are ints that may be large.
Keep both sides in the same unit. The formula does not convert centimeters to meters.

Walk the sample

Length 8, width 5
Product 8 * 5
Area 40

The trap

The int product overflows even when you store it in a long, if both factors are still ints at the moment of multiplication.

java

static long area(int length, int width) {
    if (length < 0 || width < 0) {
        throw new IllegalArgumentException("negative dimension");
    }
    return (long) length * width;   // cast first; length * width alone wraps
}

What is 50_000 * 50_000 when both factors are ints?

The mathematical product is 2_500_000_000, past Integer.MAX_VALUE. The int multiply wraps to a negative value.

Assigning that wrapped int to a long keeps the negative value. The cast to long has to happen on one factor before the multiply.

Time O(1), space O(1).

Easy to get wrong

Multiplying two ints and only then widening the stored result.
Mixing units and still multiplying.