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Question Number 24459 by mustafasediqi last updated on 18/Nov/17

$$\mathrm{If}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{are}\mathrm{increased}$$$$\mathrm{each}\mathrm{by}10\mathrm{\%},\mathrm{find}\mathrm{the}\mathrm{\%}\mathrm{increase}\mathrm{in}\mathrm{its}$$$$\mathrm{diagonal}.$$

Answered by ajfour last updated on 18/Nov/17

$$d^2=l^2+b^2$$$$2d△d=2l△l+2b△b$$$$\Rightarrow \frac{△d}{d}=\frac{l△l+b△b}{d^2}$$$$=\frac{\frac{△l}{l}(l^2+\frac{l}{△l}\times \frac{△b}{b}\times b^2)}{l^2+b^2}$$$$=\frac{\frac{△l}{l}(l^2+b^2)}{l^2+b^2}=\frac{△l}{l}$$$$so\frac{△d}{d}\times 100=\frac{△l}{l}\times 100=10.$$

Answered by $@ty@m last updated on 18/Nov/17

$$Newdiagonal=\sqrt{{\left(\frac{110}{100}l\right)}^2+{\left(\frac{110}{100}b\right)}^2}$$$$=\frac{110}{100}\sqrt{l^2+b^2}$$$$increase=(\frac{110}{100}-1)\sqrt{l^2+b^2}$$$$=\frac{10}{100}\sqrt{l^2+b^2}$$$$\mathrm{\%}increase=\frac{\frac{10}{100}\sqrt{l^2+b^2}}{\sqrt{l^2+b^2}}\times 100$$$$=10$$$$\Rightarrow 10\mathrm{\%}increase$$

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