Question

A wire 34 cm long is 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. $68c{{m}^{2}}$
B. $70c{{m}^{2}}$
C. $71.25c{{m}^{2}}$
D. $72.25c{{m}^{2}}$

Answer 1

Hint: Quadrilateral formed is a rectangle when each angle is ${{90}^{\circ }}$. We know perimeter of the rectangle can be given as 2 (length + breadth) and area of it can be given by relation $\text{length}\times \text{breadth}$.By substituting the value of variable from perimeter of rectangle in the area of rectangle we get quadratic equation and express the quadratic equation in perfect square forms and simplify it to get the required maximum area of quadrilateral.

Complete step-by-step answer:
Here, it is given that quadrilateral formed using the wire of 34cm has each angles of ${{90}^{\circ }}$ , and we know that if each angles of a quadrilateral is ${{90}^{\circ }}$ then it will be a rectangle because rectangle is a quadrilateral with each angles as ${{90}^{\circ }}$ . Hence, we can draw diagram as

Where x and y are the length and breadth of the rectangle. As opposite sides are equal in rectangle, so we get
AB = CD = x
BC = AD = y
Now, it is given that the wire is to bend in the form of a rectangle, hence the perimeter of the rectangle should be equal to the length of the wire. Now, perimeter of the rectangle is given as:
Perimeter of rectangle = 2(length +breadth).
We have length and breadth of rectangle ABCD as,
Length = x cm and breadth = y cm
Hence, perimeter of rectangle formed = 2(x + y). So, perimeter off rectangle should be equal to 34cm. Hence, we get
2(x + y) =34 or x + y = 17………………….(i)
Now, we need to get the maximum area of the rectangle with condition between sides as x + y = 17 from equation (i). So, we know the area of the rectangle is $\text{length}\times \text{breadth}$ .
Hence, area of rectangle formed by wire can be given as
Area of rectangle formed $=x\times y=xyc{{m}^{2}}$ .
Let the area of the rectangle be ‘M’. So, we get
M = xy…………………..(ii)
Now, put the value of ‘y’ from equation (i) in the equation (ii) to get the equation in a single variable. Hence, value of ‘y’ from equation (i) can be calculated as
x + y = 17
y = 17 – x ……………………..(iii)
Hence, we get from equation (ii) and (iii) as
$\begin{aligned} & M=x\left( 17-x \right) \\\ & M=17x-{{x}^{2}} \\\ & M=-\left( {{x}^{2}}-17x \right) \\\ \end{aligned}$
Now, add and subtract ${{\left( \dfrac{17}{2} \right)}^{2}}$ inside the bracket to form a perfect square. Hence, we get
$\begin{aligned} & M=-\left( {{x}^{2}}-17x+{{\left( \dfrac{17}{2} \right)}^{2}}-{{\left( \dfrac{17}{2} \right)}^{2}} \right) \\\ & M=-\left( {{x}^{2}}-2\times \dfrac{17}{2}\times x+{{\left( \dfrac{17}{2} \right)}^{2}}-{{\left( \dfrac{17}{2} \right)}^{2}} \right) \\\ \end{aligned}$
Now, we can observe that $\left( {{x}^{2}}-\dfrac{17}{2}\times 2\times x+{{\left( \dfrac{17}{2} \right)}^{2}} \right)$ is a perfect square of ${{\left( x-\dfrac{17}{2} \right)}^{2}}$ by comparing the identity of ${{\left( a-b \right)}^{2}}$ which can be given as
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Hence, we get the value of M as
$\begin{aligned} & M=-\left[ {{\left( x-\dfrac{17}{2} \right)}^{2}}-{{\left( \dfrac{17}{2} \right)}^{2}} \right] \\\ & M={{\left( \dfrac{17}{2} \right)}^{2}}-{{\left( x-\dfrac{17}{2} \right)}^{2}}..................(iv) \\\ \end{aligned}$
Now, we can observe that value of M will decrease if value of ${{\left( x-\dfrac{17}{2} \right)}^{2}}$ will increase and we know that square of any number will never be a negative value. Hence, we can get the maximum value of M by minimizing ${{\left( x-\dfrac{17}{2} \right)}^{2}}$ . So, we get a minimum value of ${{\left( x-\dfrac{17}{2} \right)}^{2}}$ as ‘0’ because the minimum value of any square can be 0. Hence, at $x=\dfrac{17}{2}$ , term ${{\left( x-\dfrac{17}{2} \right)}^{2}}$ of equation (iv) will become 0. Hence, maximum value of M can be given as
${{M}_{\max }}={{\left( \dfrac{17}{2} \right)}^{2}}=\dfrac{289}{4}=72.25$
As M is representing the area of the rectangle, ${{M}_{\max }}$ is the maximum possible area. Hence, maximum possible area of the quadrilateral mentioned in the question is $72.25c{{m}^{2}}.$
So, option (D) is correct.

Note: One may take the quadrilateral formed by the wire as square, which is wrong. We don’t know about the sides of the quadrilateral; we have each angle of the quadrilateral as ${{90}^{\circ }}$ . So, we can suppose the quadrilateral as a rectangle only. We need to know the property that if sum of two positive numbers are given the products of them will be maximum if numbers are equal. Hence, another direct approach would be, We have x + y =17.
So, ${{M}_{\max }}=\max \left( xy \right)$
Since, x = y, then $x=y=\dfrac{17}{2}$
Hence, ${{M}_{\max }}=\dfrac{289}{4}$
One more approach for finding maximum value of $M=17x-{{x}^{2}}$ would be that we can differentiate M w.r.t x and put it ‘0’ to get a value of x at which M will be maximum. Hence, we get
$\begin{aligned} & \dfrac{dM}{dx}=17-2x=0 \\\ & x=\dfrac{17}{2} \\\ \end{aligned}$
Now, put $x=\dfrac{17}{2}$ in equation of M and get ${{M}_{\max }}.$