### 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. 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.