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In a Rectangle, if the difference between the sum of the adjacent sides and the diagonal is 2/5 of the length of the longer side, what is the ratio of the shorter to the longer side?

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If $a$ is the longer side and $b$ the shorter one, the constraint states that:

$$a+b-\sqrt{a^2+b^2}=\frac25a$$

$$\sqrt{a^2+b^2}=\frac{3a+5b}5$$

$$25(a^2+b^2)=9a^2+25b^2+30ab$$

$$16a^2=30ab$$

$$\frac ba=\frac{16}{30}=\frac{8}{15}$$

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Let $L$ and $B$ be the length and breadth of the rectangle. We have, Sum of adjacent sides = $L + B$ with the length of diagonal $=\sqrt{L^2+B^2}$

We now have, $$(L + B) +\sqrt{L^2+B^2} - \sqrt{L^2+B^2} = \frac{2L}{5} \implies 3L + 5B = 0$$

Note that the only non-negative solution will be $L=B=0$ as we get $L=-\frac{5B}{3}<0$.

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  • $\begingroup$ why have you add $\sqrt{L^2+B^2}$ as the question does not say anything about adding the diagonals? $\endgroup$ – Ashutosh Kumar Dec 9 '17 at 13:49
  • $\begingroup$ even before changing the question is straightforward, but sorry for that $\endgroup$ – Ashutosh Kumar Dec 9 '17 at 14:03
  • $\begingroup$ You had written “difference between the sum of the two sides and the diagonal and the diagonal”!! $\endgroup$ – Rohan Dec 9 '17 at 14:04
  • $\begingroup$ sorry for that, mistyped by me, I don't change it willingly $\endgroup$ – Ashutosh Kumar Dec 9 '17 at 14:07
  • $\begingroup$ @AshutoshKumar What can I do for that? I will delete my answer, then. $\endgroup$ – Rohan Dec 9 '17 at 14:08

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