24. rectangle a is 16 feet wide and is 8 feet longer than rectangle b which is 14 feet wide.
if the sum of both rectangles perimeters is 156 feet, what is the area of rectangle a?

The Area of rectangle A is  576 feet²  .

Step-by-step explanation:

Given as :

The width of rectangle A = 16 feet

The Length of rectangle A = 8 feet + The length of rectangle B

The width of rectangle B = 14 feet

Let The length of rectangle B = L feet

So, The Length of rectangle A = 8 feet + L feet

The perimeter of rectangle A + The perimeter of rectangle B = 156 feet

So, 2 × ( Length A + width A ) +  2 × ( Length B + width B ) = 156 feet

Or,  2 × ( 8 + L + 16 ) +  2 × ( L + 14 ) = 156 feet

Or,  2 ×( 24 + L ) +  2 × ( L + 14 ) = 156 feet

Or, 48 + 2 L + 2 L + 28 = 156

Or, 76 L + 4 L = 156

So, 4 L = 156 - 76

Or, 4 L = 80

∴  L = = 20 feet

So , The Length of rectangle A = 8 feet + 28 feet = 36 feet

And The width of rectangle A = 16 feet

So, Area of rectangle A = Length of rectangle A × width of rectangle A

I.e  Area of rectangle A = 36 × 16 = 576 feet²

Hence The Area of rectangle A is  576 feet²  . Answer

10 cm

step-by-step explanation:

if the polygon is a rectangle, then the perimeter is x+x+y+y. using the values that you gave, the perimeter is 3+3+2+2=10cm.

step-by-step explanation:

idk the answer i want to say 10 but then it would be equal so try 9 but then yeah so yeah we love edgen