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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.