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A park is shaped like a rhombus and has area $96 \; \text{sq m}.$ If $40 \; \text{m}$ of fencing is needed to enclose the park, the cost, in $\text{INR},$ of laying electric wires along its two diagonals, at the rate of $ ₹ \; 125 \; \text{per m},$ is
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Let the length of the diagonal of a rhombus be $2a$ meter, $2b$ meter.



Area of the rhombus $= \dfrac{\text{Product of diagonal}}{2}$

$\Rightarrow 96 = \frac{2a \times 2b}{2}$

$\Rightarrow \boxed{ab = 48}$

In $\triangle \text{AOB},$ we can apply the Pythagorean theorem.

$\text{Hypotenuse}^{2} = \text{Perpendicular}^{2} + \text{Base}^{2}$

$\Rightarrow \text{AB}^{2} = \text{OA}^{2} + \text{OB}^{2}$

$\Rightarrow 10^{2} = a^{2} + b^{2}$

$\Rightarrow \boxed{a^{2} + b^{2} = 100}$

Now, $(a+b)^{2} – 2ab =100$

$\Rightarrow (a+b)^{2} – 96 =100 \quad [\because ab=48]$

$\Rightarrow (a+b)^{2} =196$

$\Rightarrow \boxed{a+b =14} \; \longrightarrow (1)$

And $(a-b)^{2} + 2ab =100$

$\Rightarrow (a-b)^{2}+96 =100$

$\Rightarrow (a-b)^{2} =4$

$\Rightarrow \boxed{a-b =2} \; \longrightarrow (2)$

Adding the equation $(1)$, and equation $(2).$

$ \qquad \require{cancel} \begin{array} {} a+\cancel{b} = 14 \\ a-\cancel{b} = 2 \\\hline 2a \quad = 16 \end{array}$

$\Rightarrow \boxed{a=8}$

From equation $(1),$

$\Rightarrow a+b=14$

$\Rightarrow 8+b=14$

$\Rightarrow \boxed{b=6}$

So,

  • The length of the one diagonal $=2a=16$ meters.
  • The length of the other diagonal $=2b=12$ meters.

$\therefore$ The cost in INR, of laying electric wires along its two diagonal at the rate of ₹$125$ per meter $=28 \times 125= ₹3500.$

Correct Answer $:3500$

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