Rectangle A has a width of 8 cm and a length of 16 cm. Rectangle B has the same area as the first but its width is 62.5% of the
width of the first rectangle. Express the length of Rectangle B as a percent of the length of Rectangle A. What percent more or less is the length of Rectangle B than the length of Rectangle A? HELP PLEASE!!!!!
Area A = 8 * 16 = 128 cm^2 ; Area B = 128 cm^2 ; But Area B = L * [ (62.5 / 100) * 8 ] ; where L = the length of rectangle B ; 128 = L * ( 625 / 1000 ) * 8 ; 128 / 8 = L * ( 625 / 1000 ); 16 = L * ( 5 / 8 ) ; L = ( 16 * 5 ) / 8 ; L = 80 / 8 ; L = 10 cm ; p / 100 = 10 / 16 ; p / 100 = 5 / 8 ; p = 500 / 8 ; p = 62.5 ; p% = 62.5% ( the length of rectangle B is 62.5% of the length of rectangle A ) ; q / 100 = ( 16 - 10 ) / 16 ; q / 100 = 6 / 16 ; q / 100 = 3 / 8 ; q = 300 / 8 ; q = 37.5 ; q% = 37.5 % ( 37.5% more <span>the length of Rectangle B than the length of Rectangle A );</span>