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VITEEE 2019 Paper

Section: Mathematics

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Question : 104 of 125

Marks: +1, -0

A wire 34 cm long is to be bent in the form of a quadrilateral of which each angle is 90°. What is the maximum area which can be enclosed inside the quadrilateral?

$68 \text{ cm}^2$
$70 \text{ cm}^2$
$71.25 \text{ cm}^2$
$72.25 \text{ cm}^2$

Solution:

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}=72.25

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