A thin metal iron-sheet is a rhombus in shape, with each side 10 m. If one of its diagonals is 16 m, find the cost of painting its both sides at the rate of ₹ 6 per m².

Also, find the distance between the opposite sides of this rhombus.

Given:

Side of iron sheet (AB) = 10 m

Diagonal of the sheet (AC) = 16 m

Join BD.

Then, OA = OC = $\dfrac{AC}{2}$ = $\dfrac{16}{2}$ = 8 m

Since the diagonals of a rhombus bisect each other at 90°, applying the pythagoras theorem in Δ AOB, we get:

AB² = OA² + OB²

⇒ (10)² = (8)² + OB²

⇒ 100 = 64 + OB²

⇒ OB² = 100 - 64

⇒ OB² = 36

⇒ OB = $\sqrt{36}$

⇒ OB = 6 m

Thus, BD = 2 x OB = 2 x 6 m = 12 m

The area of rhombus = $\dfrac{1}{2}$ x product of diagonals

= $\dfrac{1}{2}$ x 16 x 12 m²

= $\dfrac{1}{2}$ x 192 m²

= 96 m²

The rate of painting is ₹ 6 per m².

Therefore, the total cost of painting is:

Total cost = 2 x area of sheet x rate of painting

= 2 x 96 x 6

= 192 x 6

= ₹ 1,152

We also know that the area of the rhombus = base x height

⇒ 96 = 10 x height

⇒ height = $\dfrac{96}{10}$

⇒ height = 9.6 m

Hence, the total cost of painting is ₹ 1,152 and the distance between the opposite sides of the rhombus is 9.6 m.