Question

Which of the combined rectangles has an area of 40 square​

Answer 1

The figure A has an area of 40 squared feet.

Why?

In order to know which of the combined rectangles has an area of 40 feet squared, we need to calculate the areas each one of the figures.

So, calculating we have:

Figure A:

Two medium and equals rectangles and one big rectangle.

$SmallRectangles=(2ft*6ft)*2=12ft^{2}*2=24ft^{2}\\\\BigRectangle=(8ft-2ft-2ft)*4ft=16ft^{2}\\\\CombinedArea=24ft^{2}+16ft^{2}=40ft^{2}$

The figure A is the correct optiopn.

Figure B:

Two medium and equals rectangles and one small rectangle.

$MediumRectangles=(2ft*7ft)*2=14ft^{2}*2=28ft^{2}\\\\SmallRectangle=2ft*3ft=6ft^{2}\\\\CombinedArea=28ft^{2}+14ft^{2}=34ft^{2}$

The figure B is discarded, it's area is less than 40 squared feet.

Figure C:

One medium rectangle and one big rectangle.

$MediumRectangle=3ft*9ft=27ft^{2}\\\\BigRectangle=3ft*6ft=18ft^{2}\\\\CombinedRectangle=45ft^{2}$

The figure C is discarded, it's area is larger than 40 squared feet.

Figure D:

One medium rectangle and one big rectangle:

$SmallRectangle=3ft*8ft=24ft^{2} \\\\BigRectangle=5ft*5ft=25ft^{2}\\\\CombinedRectangle=24ft^{2}+25ft^=49ft^{2}$

The figure D is discarded, it's area is larger than 40 squared feet.

Hence, we have that the Figure A is the correct option.

Have a nice day!