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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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2425 Upvotes
Verified Answer
The correct answer is:
$72.25 \mathrm{~cm}^{2}$
Let one side of quadrilateral be $x$ and another side be $y$ so, $2(x+y)=34$
or, $(x+y)=17 ......(i)$
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)
$$
\begin{array}{l}
\mathrm{x}=\mathrm{y}=\frac{17}{2} \\
\text { Area }=\mathrm{x} \cdot \mathrm{y}=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25
\end{array}
$$
or, $(x+y)=17 ......(i)$
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)
$$
\begin{array}{l}
\mathrm{x}=\mathrm{y}=\frac{17}{2} \\
\text { Area }=\mathrm{x} \cdot \mathrm{y}=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25
\end{array}
$$
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