Question

the length of rectangle exceeds its width by 5m .If width is incresed by 1 m and length 8s decresed by by 2m, area of new rectangle is 4sq.m less than area of orignal rectangle .

Answer 1

Let the width of rectangle = b

its length = b+5

area = b(b+5)

Now, width is incresed by 1 m and length is decresed by 2m,
new width = b+1

new length = (b+5) - 2 = b+3

new area = (b+3)(b+1)

According to question, area of new rectangle is 4sq.m less than area of orignal rectangle

$b(b+5) - (b+3)(b+1) = 4\\ \\ \Rightarrow {b}^{2} + 5b - ( {b}^{2} + 4b + 3) = 4 \\ \\ \Rightarrow {b}^{2} + 5b -{b}^{2} - 4b - 3 = 4 \\ \\ \Rightarrow b - 3 = 4 \\ \\ \Rightarrow b = 3 + 4 = 7$

$Thus, \ width =7m\ \\ \\ and\ length\ = 7+5= 12m$

Answer 2

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Let the width of rectangle = b

its length = b+5

area = b(b+5)

Now, width is incresed by 1 m and length is decresed by 2m,

new width = b+1

new length = (b+5) - 2 = b+3

new area = (b+3)(b+1)

According to question, area of new rectangle is 4sq.m less than area of orignal rectangle

$b(b+5) - (b+3)(b+1) = 4\\ \\ \Rightarrow {b}^{2} + 5b - ( {b}^{2} + 4b + 3) = 4 \\ \\ \Rightarrow {b}^{2} + 5b -{b}^{2} - 4b - 3 = 4 \\ \\ \Rightarrow b - 3 = 4 \\ \\ \Rightarrow b = 3 + 4 = 7$

$Thus, \ width =7m\ \\ \\ and\ length\ = 7+5= 12m$

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