Mathematics: Area of a rectangle is 540 sq m. if the length was reduced by 5m and the width was increased by 5m, the are of the rectangle will be increased 35 sq m. calculate the length and the width of the first rectangle

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Area of a rectangle is 540 sq m. if the length

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nikitakhrabkov123

10.03.2022

$l-length\\w-width\\\\\text{The formula of an area of a rectangle:}\ V=lw\\\\\text{We have the area}\ 540\ m^2\to lw=540\\\\\text{New rectangle}\\\\l-5-length\\w+5-width\\\\\text{We have the area}\ 540\ m^2+35\ m^2=575\ m^2\to (l-5)(w+5)=575\\\\\text{Therefore we have the system of equation:}\\\\\left\{\begin{array}{ccc}lw=540&\to w=\dfrac{540}{l}&(*)\\(l-5)(w+5)=575&&(**)\end{array}\right\\\\\text{Substitute}\ (*)\ \text{to}\ (**)$

$(l-5)\left(\dfrac{540}{l}+5\right)=575\qquad\text{use distributive property}\\\\(l)\left(\dfrac{540}{l}\right)+(5)(l)-(5)\left(\dfrac{540}{l}\right)-(5)(5)=575\\\\540+5l-\dfrac{2700}{l}-25=575\\\\515+5l-\dfrac{2700}{l}=575\qquad\text{subtract 575 from both sides}\\\\-60+5l-\dfrac{2700}{l}=0\qquad\text{multiply both sides by }\ l\neq0\\\\-60l+5l^2-2700=0\\\\5l^2-60l-2700=0\qquad\text{divide both sides by 5}\\\\l^2-12l-540=0\\\\l^2+30l-18l-540=0\\\\l(l+30)-18(l+30)=0\\\\(l+30)(l-18)=0\iff l+30=0\ \vee\ l-18=0\\\\l=-30$

$\text{Substitute the value of}\ l\ \text{to}\ (*):\\\\w=\dfrac{540}{18}\\\\\boxed{w=30\ cm}$

$\ \boxed{length=18\ cm,\ width=30\ cm}$

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Area of a rectangle is 540 sq m. if the length was reduced by 5m and the width was increased by 5m, the are of the rectangle will be increased 35 sq m. calculate the length and the width of the first rectangle

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