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The-area-of-a-rectangle-is-255-m-2-If-its-length-is-decreased-by-1-m-it-becomes-a-square-The-perimeter-of-the-square-is-m-

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Question Number 13490 by aamaguestt last updated on 20/May/17
The area of a rectangle is 255 m^2 . If its length is decreased by 1 m, it becomes a square. The perimeter of the square is ____ m.
$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{is}\:\mathrm{255}\:\mathrm{m}^{\mathrm{2}} .\:\mathrm{If} \\ $$$$\mathrm{its}\:\mathrm{length}\:\mathrm{is}\:\mathrm{decreased}\:\mathrm{by}\:\mathrm{1}\:\mathrm{m},\:\mathrm{it}\: \\ $$$$\mathrm{becomes}\:\mathrm{a}\:\mathrm{square}.\:\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{square}\:\mathrm{is}\:\_\_\_\_\:\mathrm{m}. \\ $$
Answered by ajfour last updated on 20/May/17
x(x+1)=255 x^2 +x−255=0 x=((−1+(√(1+4(255))))/2) perimeter of square = 4x =(2(√(1021))−2)m ≈ 61.906m .
$${x}\left({x}+\mathrm{1}\right)=\mathrm{255} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{255}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}\left(\mathrm{255}\right)}}{\mathrm{2}} \\ $$$${perimeter}\:{of}\:{square}\:=\:\mathrm{4}{x} \\ $$$$\:\:\:\:\:\:\:=\left(\mathrm{2}\sqrt{\mathrm{1021}}−\mathrm{2}\right){m}\:\approx\:\mathrm{61}.\mathrm{906}{m}\:. \\ $$
Commented by aamaguestt last updated on 20/May/17
I slolve like this but the answer in the application is 64 m. y
$$\mathrm{I}\:{slolve}\:{like}\:{this}\:{but}\:{the}\:{answer}\:{in}\:{the}\:{application} \\ $$$${is}\:\mathrm{64}\:{m}. \\ $$$${y} \\ $$$$ \\ $$
Commented by prakash jain last updated on 20/May/17
I think question meant that length decreased by 1 with fixed perimeter so width increases by 1. (x−1)(x+1)=255 x^2 −1=255 x^2 =256⇒x=16 4x=64 rectangle length=17, width=15 rectangle area=255
$$\mathrm{I}\:\mathrm{think}\:\mathrm{question}\:\mathrm{meant}\:\mathrm{that}\:\mathrm{length} \\ $$$$\mathrm{decreased}\:\mathrm{by}\:\mathrm{1}\:\mathrm{with}\:\mathrm{fixed}\:\mathrm{perimeter} \\ $$$$\mathrm{so}\:\mathrm{width}\:\mathrm{increases}\:\mathrm{by}\:\mathrm{1}. \\ $$$$\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)=\mathrm{255} \\ $$$${x}^{\mathrm{2}} −\mathrm{1}=\mathrm{255} \\ $$$${x}^{\mathrm{2}} =\mathrm{256}\Rightarrow{x}=\mathrm{16} \\ $$$$\mathrm{4}{x}=\mathrm{64} \\ $$$$\mathrm{rectangle}\:\mathrm{length}=\mathrm{17},\:\mathrm{width}=\mathrm{15} \\ $$$$\mathrm{rectangle}\:\mathrm{area}=\mathrm{255} \\ $$
Commented by ajfour last updated on 20/May/17
good thought.i dint either like the answer.
$$\mathrm{good}\:\mathrm{thought}.\mathrm{i}\:\mathrm{dint}\:\mathrm{either}\:\mathrm{like} \\ $$$$\mathrm{the}\:\mathrm{answer}. \\ $$

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