Math, asked by zodepriti1981, 9 months ago

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m ? If so, find
its length and breadth.

Answers

Answered by Anonymous
2

\bold\red{\underline{\underline{Answer:}}}

\bold{Yes, \ rectangular \ park \ can \ be \ made}

\bold{of \ length \ 20 \ m \ and \ breadth \ 20 \ m.}

\bold\orange{Given:}

\bold{=>Perimeter=80 \ m}

\bold{=>Area=400 \ m^{2}}

\bold\pink{To \ find:}

\bold{Length \ and \ breadth \ of \ the \ rectangle.}

\bold\green{\underline{\underline{Solution}}}

\bold{Let, \ length \ be \ x \ metre \ and \ breadth}

\bold{be \ y \ metre.}

\bold{According \ to \ first \ condition}

\bold{Perimeter=2(length+breadth)}

\bold{80=2(x+y)}

\bold{x+y=\frac{80}{2}}

\bold{x+y=40...(1)}

\bold{According \ to \ second \ condition}

\bold{Area=length×breadth}

\bold{400=xy}

\bold{xy=400...(2)}

\bold{By \ identity}

\bold{(a-b)^{2}=(a+b)^{2}-4ab}

\bold{\therefore{(x-y)^{2}=(x+y)^{2}-4(xy)}}

\bold{\therefore{(x-y)^{2}=40^{2}-4(400)}}

\bold{\therefore{(x-y)^{2}=1600-1600}}

\bold{\therefore{(x-y)^{2}=0}}

\bold{On \ taking \ square \ root \ of \ both \ sides}

\bold{x-y=0...(3)}

\bold{Add \ equation(1) \ and \ equation (3)}

\bold{x+y=40}

\bold{+}

\bold{x-y=0}

__________________

\bold{2x=40}

\bold{x=\frac{40}{2}}

\bold{x=20}

\bold{Substitute \ x=20 \ in \ equation (2)}

\bold{20×y=400}

\bold{y=\frac{400}{20}}

\bold{y=20}

\bold{But \ length=breadth}

\bold{\therefore{It \ is \ a \ square}}

\bold{But, \ all \ squares \ are \ the \ rectangle.}

\bold\purple{\tt{\therefore{Rectangular \ park \ can \ be}}}

\bold\purple{\tt{made \ of \ length \ 20 \ m \ and \ breadth \ 20 \ m}}

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