From 7841d158e0658ab85d80941cf3e8e0f8a34f083f Mon Sep 17 00:00:00 2001
From: 0KvinayK0 <126001522+0KvinayK0@users.noreply.github.com>
Date: Sun, 4 Jun 2023 14:59:40 +0530
Subject: [PATCH] Update Day_13.md

Fixed Day 13 question 49 solution
---
 Status/Day_13.md | 20 ++++++++++++++++++++
 1 file changed, 20 insertions(+)

diff --git a/Status/Day_13.md b/Status/Day_13.md
index 1a642c3..d0bb54a 100644
--- a/Status/Day_13.md
+++ b/Status/Day_13.md
@@ -136,6 +136,8 @@ print aSquare.area()
 
 **My Solution: Python 3**
 
+* The following solution is incorrect. The parent class's area method can be accessed if the child class does not have a method by the same name. Here, because we are instantiating the child class and setting the default length value to zero, it is printing out zero and has no relation with the area of the parent class being zero. If we need to access the parent class method, the child class should not have the same name. The interpreter looks for the method in the current class before moving up the hierarchy.
+
 ```python
 class Shape():
     def __init__(self):
@@ -157,7 +159,25 @@ print(Asqr.area())      # prints 25 as given argument
 
 print(Square().area())  # prints zero as default area
 ```
+```python
+''' Solution by 0KvinayK0
+'''
+class Shape():
+    def area(self):
+        return 0
 
+class Square(Shape):
+    def __init__(self, length):
+        super().__init__()
+        self.length = length
+
+    def area(self):
+        return self.length ** 2
+
+sq = Square(5)
+print(sq.area())  # Output: 25
+print(super(Square, sq).area())  # Output: 0
+```
 ---
 
 # Question 50