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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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The correct answer is:
$72.25 \mathrm{~cm}^{2}$
Let one side of quadrilateral be $\mathrm{x}$ and another side be $y$ so, $2(x+y)=34$
or $(x+y)=17$
We know from the basic principle that for a given perimeter square has the maximum area, $\mathrm{so}, \mathrm{x}=\mathrm{y}$ and putting this value in equation (i)
$\mathrm{x}=\mathrm{y}=\frac{17}{2}$
Area $=x \cdot y=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25$
or $(x+y)=17$
We know from the basic principle that for a given perimeter square has the maximum area, $\mathrm{so}, \mathrm{x}=\mathrm{y}$ and putting this value in equation (i)
$\mathrm{x}=\mathrm{y}=\frac{17}{2}$
Area $=x \cdot y=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25$
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