Q67 / 125 · BITSAT 2016 · Mathematics

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?

  • A
    $68 \mathrm{~cm}^{2}$
  • B
    $70 \mathrm{~cm}^{2}$
  • C
    $71.25 \mathrm{~cm}^{2}$
  • Correct answer
    D
    $72.25 \mathrm{~cm}^{2}$

Solution

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 $=x \cdot y=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25$

Hence, the correct answer is option: 'D'.

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