Question

if the length of the rectangle is increased by 33.33% and width is decreased by 16.66% then what will be the percentage change to it's area ?​

Answer 1

Step-by-step explanation:

Reduction in length by 33% or 1/3rd ,will require 1/(3–1) or 50% increase in width to kep area constant.

Example :

Original length = 120

Original width = 100

Original area = 120 x 100 = 12000

Let length reduce by 33% or 1/3 rd

New length = 120 - 40 = 80

New width = 12000/80 = 150

Hence, width increases by (150 - 100)/100 * 100 = 50% or 1/2 part.

Answer 2

Percentage of change in area = 10.39%

Given: Length of the rectangle is increased by 33.33%

Width of the rectangle decreased by 16.66%

To find: percentage of change in Area  

Solution: Let $l$ and $b$ be the length and breadth of the rectangle  

Then area of the rectangle = $lb$

Given that Length $l$ is increased 33.33% of $l$ = $\frac{33.33}{100}l$ = 0.333$l$  

Length of rectangle after increasing = $l + 0.3333l$ = 1.3333$l$

Increased length of rectangle = 1.33$l$    [take upto 2 decimals ]

Given that width $b$ is decreased 16.66% of $b$ = $\frac{16.66}{100} b$ = 0.1666$b$

width of decreasing = $b - 0.1666b$ = 0.8334$b$

Decreased width of the rectangle = 0.83$b$

Therefore, area of rectangle =  (1.33$l$)(0.83$b$ ) = 1.1039$lb$

Change in area = 1.1039$lb$ - $lb$  = 0.1039$lb$  

percentage of change = $\frac{0.1039lb}{lb} (100) =$ 10.39%

Percentage of change in area = 10.39%

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