Math, asked by aliza039, 9 months ago

answer this question pls its urgent ​

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Answered by Anonymous
3

\sf\blue{Question}

\sf{The \ length \ of \ the \ rectangle \ exceeds}

\sf{it's \ breadth \ by \ 3 \ cm. \ If \ the \ length}

\sf{and \ breadth \ are \ each \ increased \ by \ 2 \ cm, }

\sf{then \ the \ area \ of \ the \ new \ rectangle}

\sf{will \ be \ 70 \ sq.cm \ more \ than \ that \ of}

\sf{the \ given \ rectangle. \ Find \ the \ length}

\sf{and \ breadth \ of \ the \ given \ rectangle.}

___________________________________

\sf\red{\underline{\underline{Answer:}}}

\sf{Length \ and \ breadth \ of \ rectangle}

\sf{are \ 18 \ cm \ and \ 15 \ cm \ respectively.}

\sf\orange{Given:}

\sf{\implies{The \ length \ of \ the \ rectangle \ exceeds}}

\sf{it's \ breadth \ by \ 3 \ cm.}

\sf{\implies{If \ the \ length \ and \ breadth \ are}}

\sf{each \ increased \ by \ 2 \ cm, \ then \ the}

\sf{area \ of \ new \ rectangle \ will \ be \ 70 \ cm^{2}}

\sf{more \ than \ the \ given \ triangle.}

\sf\pink{\underline{\underline{To \ find:}}}

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

\sf\green{\underline{\underline{Solution:}}}

\sf{Let \ length \ of \ the \ rectangle \ be \ x}

\sf{and \ it's \ breadth \ be \ y.}

\sf{According \ to \ first \ condition.}

\sf{x=y+3}

\sf{\therefore{x-y=3...(1)}}

\sf{According \ to \ second \ condition.}

\sf{(x+2)(y+2)=xy+70}

\sf{xy+2x+2y+4=xy+70}

\sf{2x+2y=70-4}

\sf{2(x+y)=66}

\sf{\therefore{x+y=\frac{66}{2}}}

\sf{\therefore{x+y=33...(2)}}

\sf{Add \ equations \ (1) \ and \ (2)}

\sf{x-y=3}

\sf{+}

\sf{x+y=33}

____________________

\sf{2x=36}

\sf{\therefore{x=\frac{36}{2}}}

\boxed{\sf{\therefore{x=18}}}

\sf{Substitute \ x=18 \ in \ equation (2)}

\sf{18+y=33}

\sf{\therefore{y=33-18}}

\boxed{\sf{\therefore{y=15}}}

\sf{\therefore{Length=18 \ cm}}

\sf{\therefore{Breadth=15 \ cm}}

\sf\purple{\tt{\therefore{Length \ and \ breadth \ of \ rectangle}}}

\sf\purple{\tt{are \ 18 \ cm \ and \ 15 \ cm \ respectively.}}

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