The figure below is made up of Square ABCD and Rectangle EFGC. DCG is a straight line. DG = 49 cm and BE = 3 cm. The perimeters of Rectangle EFGC and Square ABCD are the same. Find the area of the figure. Ans: ___

1. Express the dimensions of the rectangle in terms of the square's side length.

Let the side length of Square ABCD be $x$ cm.
Since $E$ lies on $BC$ and $BE = 3$ cm, the width of the rectangle is $EC = x - 3$ cm.
Since $DCG$ is a straight line and $DG = 49$ cm, the length of the rectangle is $CG = 49 - x$ cm.

2. Find the perimeter of Rectangle EFGC.

Perimeter of Rectangle EFGC $= 2 \times (EC + CG)$
$= 2 \times ((x - 3) + (49 - x))$
$= 2 \times (46)$
$= 92$ cm

3. Use the given condition to find the side length of the square.

The perimeter of Square ABCD is the same as the perimeter of Rectangle EFGC.
$4x = 92$
$x = 92 \div 4 = 23$ cm

4. Calculate the dimensions of the rectangle.

Width $EC = 23 - 3 = 20$ cm
Length $CG = 49 - 23 = 26$ cm

5. Calculate the total area of the figure.

Area of Square ABCD $= 23 \times 23 = 529\text{ cm}^2$
Area of Rectangle EFGC $= 20 \times 26 = 520\text{ cm}^2$
Total Area $= 529 + 520 = 1049\text{ cm}^2$