Question

The length of a rectangle is twice its breadth if the length is increased by 10% while breadth is decreased by 10% determine the percentage change if any in the perimeter

Answer 1

✬ Increase % = 10/3 % ✬

Step-by-step explanation:

Given:

• Length of rectangle is twice it's breadth.
• Length is increased by 10% .
• Breadth is decreased by 10%.

To Find:

• Percentage change in its perimeter.

Solution: Let the length be x and Breadth be y

∴ Length = 2 times of breadth = x = 2y......(1)

★Perimeter of rectangle = 2(Length + Breadth)★

$\small\implies{\sf }$ Perimeter = 2(x + y)

$\small\implies{\sf }$ 2 (2y + y) ......[ From equation 1 ]

$\small\implies{\sf }$ 4y + 2y = 6y Perimeter

• Length is increased by 10% •

• New Length = x + 10% of x

$\small\implies{\sf }$ x + 10x/100

$\small\implies{\sf }$ x + x/10 [ Take LCM ]

$\small\implies{\sf }$ 10x + x/10

$\small\implies{\sf }$ 11x/10 or,

$\small\implies{\sf }$ 11 (2y)/10 [ Since, x = 2y ]

$\small\implies{\sf }$ 22y/10

• Breadth is decreased by 10% •

• New Breadth = y – 10% of y

$\small\implies{\sf }$ y – 10y/100

$\small\implies{\sf }$ y – y/10 [ Take LCM ]

$\small\implies{\sf }$ 10y – y/10

$\small\implies{\sf }$ 9y/10

∴ New Perimeter = 2 { New Length + New Breadth }

$\small\implies{\sf }$ New Perimeter = 2 ( 22y/10 + 9y/10 )

$\small\implies{\sf }$ 2 x 31y/10

$\small\implies{\sf }$ 31y/5

→ Total increase in perimeter = (New perimeter – Old Perimeter)

→ Increase = 31y/5 – 6y

→ 31y – 30y/5

→ y/5

• Increase % = Increase/Original x 100 •

$\small\implies{\sf }$ Increase % = (y/5)/6y x 100 %

$\small\implies{\sf }$ (y/5 x 1/6y x 100 )%

$\small\implies{\sf }$ 100y/30y %

$\small\implies{\sf }$ 10/3 %

Hence, There is increase of 10/3 % in perimeter of rectangle.