Question

A copper wire is bent into the shape of a rhombus whose diagonals measure 26cm and 13cm. The same wire is then bent into the shape of a square. Find the area of the square.​

Answer 1

Concept & Formula used :-

• when a wire is bent to form another shape, perimeter of both shapes remains same.
• Diagonals of a rhombus bisect at 90° .
• Pythagoras theorem .
• Perimeter of Rhombus = 4 * side .
• Perimeter of square = 4 * side .
• Area of Square = (side)² .

Solution :-

Given that, length of diagonals of rhombus are 26cm ans 13cm. Since , diagonals bisect each other at 90° .

By pythagoras theorem we can conclude that,

→ (26/2)² + (13/2)² = (side of rhombus)²

→ (13)² + (6.5)² = (side)²

→ (side)² = 169 + 42.25

→ (side)² = 211.25

→ side = √(211.25)

→ side of rhombus ≈ 14.53 cm.

So,

→ Perimeter of rhombus = (14*53 * 4) cm.

Now,

→ Perimeter of rhombus = Perimeter of square.

→ (14.53 * 4) = 4 * (side of Square)

→ side of Square = 14.53 cm.

Therefore,

→ Area of Square = (side)²

→ Area of Square = (14.53)² = 211.25 cm². (Ans.)