Single Choice

A wire $$34$$ cm long is to 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?

A$$68\ \text{cm}^{2}$$
B$$70\ \text{cm}^{2}$$
C$$71.25\ \text{cm}^{2}$$
D$$72.25\ \text{cm}^{2}$$
Correct Answer

Solution

Let one side of quadrilateral be $$x$$ and another side be $$y$$
So, $$2(x + y) = 34$$
$$\Rightarrow (x + y) = 17$$ .... (i)
We know from the basic principle that for a given perimeter square has the maximum area.
So, $$x = y$$ and putting this value is equation (i), we get
$$x = y = \dfrac {17}{2}$$
$$\text{Area} = x . y = \dfrac {17}{2}\times \dfrac {17}{2} = \dfrac {289}{4} = 72.25$$

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