Why do we need super() in an extended class? code example

Example 1: super in javascirpt

class Rectangle {
  constructor(height, width) {
    this.name = 'Rectangle';
    this.height = height;
    this.width = width;
  }
  sayName() {
    console.log('Hi, I am a ', this.name + '.');
  }
  get area() {
    return this.height * this.width;
  }
  set area(value) {
    this._area = value;
  }
}

class Square extends Rectangle {
  constructor(length) {
    this.height; // ReferenceError, super needs to be called first!

    // Here, it calls the parent class's constructor with lengths
    // provided for the Rectangle's width and height
    super(length, length);

    // Note: In derived classes, super() must be called before you
    // can use 'this'. Leaving this out will cause a reference error.
    this.name = 'Square';
  }
}

Example 2: extended class call method from super in javascript

// this is how we call method from a super class(i.e MAIN CLASS) to extended class
class Person {
 constructor(name, age) {
  this.name = name;
  this.age = age;
 }
greet() {
  console.log(`Hi, all my name is ${this.name}`);
 }
}

class Employee extends Person {
 constructor(name, age, employer) {
  super(name, age);  // NOTE : super must call before accessing (this)
  this.employer = employer;
 }

 greeting() {
  super.greet();  // this is how we call
  console.log(`I'm working at ${this.employer}`);
 }
}

let Steve = new Employee('Xman', 25 , 'Skynet');
Steve;
Steve.greeting(); // check by consoling MATE