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: The Area of rectangle A is  576 feet²  .Step-by-step explanation:Given as :The width of rectangle A = 16 feetThe Length of rectangle A = 8 feet + The length of rectangle BThe width of rectangle B = 14 feetLet The length of rectangle B = L feetSo, The Length of rectangle A = 8 feet + L feetThe perimeter of rectangle A + The perimeter of rectangle B = 156 feetSo, 2 × ( Length A + width A ) +  2 × ( Length B + width B ) = 156 feetOr,  2 × ( 8 + L + 16 ) +  2 × ( L + 14 ) = 156 feetOr,  2 ×( 24 + L ) +  2 × ( L + 14 ) = 156 feetOr, 48 + 2 L + 2 L + 28 = 156Or, 76 L + 4 L = 156So, 4 L = 156 - 76Or, 4 L = 80∴  L = [tex]\frac{80}{4}[/tex] = 20 feetSo , The Length of rectangle A = 8 feet + 28 feet = 36 feetAnd The width of rectangle A = 16 feetSo, Area of rectangle A = Length of rectangle A × width of rectangle AI.e  Area of rectangle A = 36 × 16 = 576 feet²Hence The Area of rectangle A is  576 feet²  . Answer