Question

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?​

Answer 1

The Area of rectangle A is 576 feet² .

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 =
`(80)/(4)` = 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