Interface in java

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Asked by Super Admin • Feb 03, 2026 at 05:00 • Subject: Java Programming

Question:

Write a Java program to create an interface Shape with the calculateArea() method. Create three classes Rectangle, Circle, and Triangle that implement the Shape interface. Override the calculateArea() method in each subclass to calculate and return the shape's area.

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Super Admin

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Feb 03, 2026 at 05:31

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interface Shape { double calculateArea(); } class Rectangle implements Shape { private final double length; private final double width; public Rectangle(double length, double width) { this.length = length; this.width = width; } @Override public double calculateArea() { return length * width; } } class Circle implements Shape { private final double radius; public Circle(double radius) { this.radius = radius; } @Override public double calculateArea() { return Math.PI * radius * radius; } } class Triangle implements Shape { private final double base; private final double height; public Triangle(double base, double height) { this.base = base; this.height = height; } @Override public double calculateArea() { return 0.5 * base * height; } } public class ShapeDemo { public static void main(String[] args) { Shape rectangle = new Rectangle(10, 5); Shape circle = new Circle(7); Shape triangle = new Triangle(8, 6); System.out.println("Rectangle area: " + rectangle.calculateArea()); System.out.println("Circle area: " + circle.calculateArea()); System.out.println("Triangle area: " + triangle.calculateArea()); } }

Interface in java

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