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.

Area = \(\frac{1}{2}\) × d₁ × d₂
96 = \(\frac{1}{2}\) × d₁ × d₂
96 × 4 = 2 × d₁ × d₂

Also,
\((\frac{d_1}{2})^2+(\frac{d_2}{2})^2=10^2\)

d₁² + d₂² = 400
(d₁ + d₂ )² = d₁² + d₂² + 2 × d₁ × d₂
(d₁ + d₂ )² = 400 + 4(96)
(d₁ + d₂ )² = 4(100 + 96)
(d₁ + d₂ )² = 4(196)
(d₁ + d₂) = 2(14)
d₁ + d₂ = 28

The cost of laying electric wires along the diagonals at the rate of per meter ₹125.
= 28 × 125
= ₹3500.

Answer 2

Perimeter of park \(= 40 \ m\)
Area of rhombus \(= 96\ m^2\)
\(\frac 12 d_1.d_2 = 96\)
\(d_1.d_2 = 192\)    \(…… (1)\)
We know that diagonals of rhombus are perpendicular to each other,
Then, \(\frac {d_1^2}{4}+\frac {d_2^2}{4}= 100\)
\(d_1^2+d_2^2 = 400\)   \(……. (2)\)
On solving both equations,
\(d_1 = 12\)
\(d_2 = 16\)
Total length of electric wires = 16+12 = 28\(= 16+12 = 28\)
Total cost \(= 28 \times 125 = ₹\ 3500\)

So, the answer is \(₹\ 3500\).