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Question-75627




Question Number 75627 by liki last updated on 14/Dec/19
Answered by $@ty@m123 last updated on 14/Dec/19
Initially,  Length=l  Breadth=b  ∴ Diagonal   d=(√(l^2 +b^2 ))  ......(1)  After inceasing sides by 10%,  Length=l(1+((10)/(100)))=((110)/(100))l=((11)/(10))l  Breadth=b(1+((10)/(100)))=((110)/(100))b=((11)/(10))b  ∴ Diagonal    D=(√((((11)/(10))l)^2 +(((11)/(10))b)^2 ))   ⇒  D=((11)/(10))(√(l^2 +b^2 )) .......(2)  ∴ Percentage increase in diagonal   =((D−d)/d)×100  =(((((11)/(10))(√(l^2 +b^2 )) −(√(l^2 +b^2 ))))/( (√(l^2 +b^2 ))))×100  =(((((11)/(10)) −1)(√(l^2 +b^2 ))))/( (√(l^2 +b^2 ))))×100  =((11−10)/(10))×100  =(1/(10))×100  =10%
$$\boldsymbol{{Initially}}, \\ $$$${Length}={l} \\ $$$${Breadth}={b} \\ $$$$\therefore\:{Diagonal}\:\:\:\boldsymbol{{d}}=\sqrt{{l}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:……\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{After}}\:\boldsymbol{{inceasing}}\:\boldsymbol{{sides}}\:\boldsymbol{{by}}\:\mathrm{10\%}, \\ $$$${Length}={l}\left(\mathrm{1}+\frac{\mathrm{10}}{\mathrm{100}}\right)=\frac{\mathrm{110}}{\mathrm{100}}{l}=\frac{\mathrm{11}}{\mathrm{10}}{l} \\ $$$${Breadth}={b}\left(\mathrm{1}+\frac{\mathrm{10}}{\mathrm{100}}\right)=\frac{\mathrm{110}}{\mathrm{100}}{b}=\frac{\mathrm{11}}{\mathrm{10}}{b} \\ $$$$\therefore\:{Diagonal}\:\:\:\:\boldsymbol{{D}}=\sqrt{\left(\frac{\mathrm{11}}{\mathrm{10}}{l}\right)^{\mathrm{2}} +\left(\frac{\mathrm{11}}{\mathrm{10}}{b}\right)^{\mathrm{2}} } \\ $$$$\:\Rightarrow\:\:\boldsymbol{{D}}=\frac{\mathrm{11}}{\mathrm{10}}\sqrt{{l}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:…….\left(\mathrm{2}\right) \\ $$$$\therefore\:\boldsymbol{{Percentage}}\:\boldsymbol{{increase}}\:\boldsymbol{{in}}\:\boldsymbol{{diagonal}}\: \\ $$$$=\frac{{D}−{d}}{{d}}×\mathrm{100} \\ $$$$=\frac{\left(\frac{\mathrm{11}}{\mathrm{10}}\sqrt{{l}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:−\sqrt{{l}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}{\:\sqrt{{l}^{\mathrm{2}} +{b}^{\mathrm{2}} }}×\mathrm{100} \\ $$$$=\frac{\left.\left(\frac{\mathrm{11}}{\mathrm{10}}\:−\mathrm{1}\right)\sqrt{{l}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}{\:\sqrt{{l}^{\mathrm{2}} +{b}^{\mathrm{2}} }}×\mathrm{100} \\ $$$$=\frac{\mathrm{11}−\mathrm{10}}{\mathrm{10}}×\mathrm{100} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}×\mathrm{100} \\ $$$$=\mathrm{10\%} \\ $$
Commented by liki last updated on 14/Dec/19
....Thank you sir be blessed
$$….\boldsymbol{{Thank}}\:\boldsymbol{{you}}\:\boldsymbol{{sir}}\:\boldsymbol{{be}}\:\boldsymbol{{blessed}} \\ $$

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