1 thought on “What will be the change in area of a rectangle,<br />if its length is decreased by 10% and breadth is<br />increased by 20%?”
let the length be l units and breadth be b units
old area = l × b
length is decreased by 10%, new length = (l × 90) / 100
breadth is increased by 20%, new breadth = (b × 120) / 100
[tex]new \: area = \frac{l \times 90}{100} \times \frac{b \times 120}{100} \\ new \: area = l \times b \times ( \frac{90 \times 120}{100 \times 100} ) \\ new \: area = old \: area \times \frac{9 \times 12}{100} \\ new \: area = old \: area \times \frac{108}{100} \\ new \: area = old \: area \times ( 1 + \frac{8}{100} ) \\ new \: area = old \: area + old \: area \times \frac{8}{100} \\ area = old \: area + 8 \: percent \: of \: old \: area[/tex]
let the length be l units and breadth be b units
old area = l × b
length is decreased by 10%, new length = (l × 90) / 100
breadth is increased by 20%, new breadth = (b × 120) / 100
[tex]new \: area = \frac{l \times 90}{100} \times \frac{b \times 120}{100} \\ new \: area = l \times b \times ( \frac{90 \times 120}{100 \times 100} ) \\ new \: area = old \: area \times \frac{9 \times 12}{100} \\ new \: area = old \: area \times \frac{108}{100} \\ new \: area = old \: area \times ( 1 + \frac{8}{100} ) \\ new \: area = old \: area + old \: area \times \frac{8}{100} \\ area = old \: area + 8 \: percent \: of \: old \: area[/tex]
New area increases by 8%