Theory:

Let us learn how to find the perimeter and area of the shape after adding a portion to a given shape with an example.
Example:
Consider a square whose side length is \(6\) \(cm\). Let us find the area and perimeter of the square.
 
Area, \(A=s \times s\)
 
\(A=6 \times 6\)
 
\(A=36\) \(sq. cm\)
 
Perimeter, \(P=4s\)
 
\(P=4\times 6\)
 
\(P=24\) \(cm\)
 
Now, let us consider a right-angled triangle of base \(6\) \(cm\) and height \(8\) \(cm\) is attached to a rectangle. Find the perimeter and area of the given shape.
 
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Solution:
 
Given:
 
The length of the rectangle \(=9\) \(cm\).
 
The breadth of the rectangle \(=6\) \(cm\).
 
The base of the triangle \(=6\) \(cm\).
 
The height of the triangle \(=8\) \(cm\).
 
To find the perimeter and the area of the square after adding a portion to the rectangle.
 
Perimeter of the given shape \(=\) Sum of all measures of outer length
 
Perimeter \(=(6+9+9+8+10)\) \(cm\)
 
Perimeter \(=42\) \(cm\)
 
Therefore, the perimeter of the given shape is \(42\) \(cm\).
 
Area of the given shape \(=\) Area of the rectangle \(+\) Area of the right triangle
 
Area \(=(l \times b)\) \(+ \frac{1}{2}(b \times h)\)
 
Area \(=(9 \times 6) +(\frac{1}{2}(6 \times 8))\)
 
Area \(=(54)+\) \((\frac{1}{2}\times 48)\)
 
Area \(=54+24\) \(=78\) \(sq. cm\)
 
Therefore, the area of the given shape is \(78\) \(sq. \ cm\).
Hence, both the area and the perimeter are increased.