A park is shaped like a rhombus and has area $96$ sq m. If $40$ m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires along its two diagonals, at the rate of $125$ per m, is

We have a rhombus-shaped park with:

Area = 96 sq m

Perimeter = 40 m (fencing needed)

Cost of wire = ₹125 per meter

We need to find the cost of laying wires along both diagonals.

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Since a rhombus has 4 equal sides:

Side length = Perimeter ÷ 4 = 40 ÷ 4 = 10 m

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The area of a rhombus is given by:

Area = $\dfrac{1}{2} \times d_1 \times d_2$

Where $d_1$ and $d_2$ are the lengths of the two diagonals.

96 = $\dfrac{1}{2} \times d_1 \times d_2$

96 = $\dfrac{d_1 \times d_2}{2}$

$d_1 \times d_2 = 192$

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The diagonals bisect each other at right angles.

This means if we draw both diagonals, they form four right triangles inside the rhombus. Each right triangle has:

Hypotenuse = side of rhombus = 10 m

Two legs = $\dfrac{d_1}{2}$ and $\dfrac{d_2}{2}$ (half of each diagonal)

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Using the Pythagorean theorem for one of these right triangles:

$\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 10^2$

$\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 100$

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Now we have two equations:

1. $d_1 \times d_2 = 192$
2. $\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 100$

From equation 2:

$\dfrac{d_1^2}{4} + \dfrac{d_2^2}{4} = 100$

$d_1^2 + d_2^2 = 400$

We need to find two numbers whose:

Product = 192

Sum of squares = 400

Let's try some factor pairs of 192:

12 × 16 = 192

$12^2 + 16^2 = 144 + 256 = 400$

So $d_1 = 12$ m and $d_2 = 16$ m.

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Total length of wire needed = $d_1 + d_2 = 12 + 16 = 28$ m

Cost = 28 × ₹125 = ₹3500

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When dealing with rhombus problems, remember:

Diagonals bisect each other at right angles

This creates four congruent right triangles

Use both the area formula and Pythagorean theorem to find diagonal lengths

Answer: ₹3500