| A wire 25 cm long is bent into the form of a rectangle of area \[25\text{ }c{{m}^{2}}.\]The length of the longer side is [WBSSC (CGL) 2014] |
A) 25 cm
B) 20 cm
C) 10 cm
D) 5 cm
Correct Answer: C
Solution :
| Perimeter of rectangle \[=2\,(l+b)\] |
| \[\Rightarrow \] \[25=2\,(l+b)\]\[\Rightarrow \]\[(l+b)=\frac{25}{2}\] (i) |
| Now, area of rectangle \[=\,\,l\times b\] |
| \[\Rightarrow \] \[25=l\times b\]\[\Rightarrow \]\[lb=25\] (ii) |
| Now, \[{{(l-b)}^{2}}=(l+{{b}^{2}})-4lb\] |
| \[={{\left( \frac{25}{2} \right)}^{2}}-4\times 25=\frac{625}{4}-100\] |
| \[=\,\,\frac{625-400}{4}\,\,=\,\,\frac{225}{4}\] |
| \[\Rightarrow \] \[l-b=\frac{15}{2}\] (iii) |
| On solving Eqs. (i) and (iii), we get |
| \[l=10\,cm\] and \[b=\frac{5}{2}\,cm\] |
| Hence, length of longest side is 10 cm. |
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