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) \[x=y=\frac{17}{2}\] Area \[=x.y=\frac{17}{2}\times \frac{17}{2}=\frac{289}{4}=72.25\]

"> 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) \[x=y=\frac{17}{2}\] Area \[=x.y=\frac{17}{2}\times \frac{17}{2}=\frac{289}{4}=72.25\]

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A wire 34 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?

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 11 months ago

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[d] 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) \[x=y=\frac{17}{2}\] Area \[=x.y=\frac{17}{2}\times \frac{17}{2}=\frac{289}{4}=72.25\]

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