Question

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

Answer 1

The are of the rhombus is given by,
Area = \(\frac{1}{2}\) × d₁ × d₂
Where d₁ & d₂ are the diagonals of the rhombus.

Given:
Area of a rhombus = \(\frac{1}{2} \times d_1 \times d_2 = 96\)
So, \(2 \times 96 = d_1 \times d_2\) → \(d_1 \cdot d_2 = 192\)

Also, the rhombus is inscribed in a circle with radius 10 m. Since the diagonals of a rhombus bisect each other at right angles, the diagonals form a right triangle with hypotenuse = radius of circle = 10 m:
\[ \left( \frac{d_1}{2} \right)^2 + \left( \frac{d_2}{2} \right)^2 = 10^2 = 100 \] \[ \frac{d_1^2 + d_2^2}{4} = 100 \Rightarrow d_1^2 + d_2^2 = 400 \]

Now use identity: \[ (d_1 + d_2)^2 = d_1^2 + d_2^2 + 2d_1 d_2 \] \[ = 400 + 2(192) = 400 + 384 = 784 \Rightarrow d_1 + d_2 = \sqrt{784} = 28 \]

Now, cost of laying electric wire along diagonals at ₹125 per metre:
Total length = \(d_1 + d_2 = 28\) m
Cost = \(28 \times 125 = ₹3500\)

Final Answer: ₹3500

Answer 2

Perimeter of park = \(40\ \text{m}\) 

Area of rhombus = \(96\ \text{m}^2\)
Using formula for area of rhombus: \(\frac{1}{2} d_1 \cdot d_2 = 96\)
Therefore, \(d_1 \cdot d_2 = 192\)    (1)

Since the rhombus is inscribed in a circle with radius 10 m, diagonals bisect each other at 90° and form right-angled triangles.
Applying Pythagoras theorem to the triangle formed by half diagonals:
\(\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 10^2\)
\(\frac{d_1^2 + d_2^2}{4} = 100 \Rightarrow d_1^2 + d_2^2 = 400\)    (2)

Now, solving equations (1) and (2) simultaneously:

• \(d_1 \cdot d_2 = 192\)
• \(d_1^2 + d_2^2 = 400\)

Let’s solve using quadratic identities:
Assume \(d_1 = 12\) and \(d_2 = 16\)
Then, \(12 \cdot 16 = 192\) and \(12^2 + 16^2 = 144 + 256 = 400\)
✅ Both conditions satisfied.

Total length of electric wire = \(d_1 + d_2 = 12 + 16 = 28\) m

Cost per metre = ₹125
Total cost = \(28 \times 125 = ₹3500\)

Final Answer: \(₹3500\)