Challenge 2: Inheritance with @ISA — Possible Solution
====================================================================

package Shape;

sub area {
    return 0;
}

package Square;
our @ISA = ('Shape');

sub new {
    my $class = shift;
    my $side  = shift;
    return bless { side => $side }, $class;
}

sub area {
    my $self = shift;
    return $self->{side} * $self->{side};
}

package main;
use strict;
use warnings;
use feature 'say';

my $sq = Square->new(4);
say $sq->area();

Output:
16

WHY THIS WORKS AS AN ANSWER
------------------------------
package Shape; with a plain sub area { return 0; } reuses the
chapter's own base-class pattern with an unimplemented/default
method — deliberately simple, since Square is expected to override it.

our @ISA = ('Shape'); inside package Square; reuses the chapter's own
inheritance declaration directly, making Square a subclass of Shape —
this is what would let Square inherit Shape's area() automatically IF
Square didn't define its own, exactly the chapter's own Dog-inherits-
new-from-Animal point.

Square's own sub new and sub area both reuse the chapter's shift-the-
class-then-shift-the-args constructor pattern and the shift-self-first
method pattern respectively — since Square defines its OWN area(),
that version wins over Shape's inherited one, per the chapter's own
explanation that "Perl searches the current package first" before
walking @ISA.

$self->{side} * $self->{side} reuses ordinary hash-value access
(Course 1, Chapter 4) through the arrow syntax (Course 1, Chapter 9)
to compute the actual square's area — Square->new(4) stores side => 4,
so area() correctly computes 4 * 4 = 16.