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A wire $34 \mathrm{~cm}$ long is to be bent in the form of a quadrilateral of which each angle is $90^{\circ}$. What is the maximum area which can be enclosed inside the quadrilateral?
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$\begin{array}{lll}\text { } & 72.25 \mathrm{~cm}^{2} & {}\end{array}$
Let one side of quadrilateral be $x$ and another side be $y$ so, $2(x+y)=34$
or, $(\mathrm{x}+\mathrm{y})=17$
We know from the basic principle that for a given perimeter square has the maximum area, so, $x=y$ and putting this value in equation (i)
$\mathrm{x}=\mathrm{y}=\frac{17}{2}$
Area $=\mathrm{x} \cdot \mathrm{y}=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25$
or, $(\mathrm{x}+\mathrm{y})=17$
We know from the basic principle that for a given perimeter square has the maximum area, so, $x=y$ and putting this value in equation (i)
$\mathrm{x}=\mathrm{y}=\frac{17}{2}$
Area $=\mathrm{x} \cdot \mathrm{y}=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25$
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