Option 1 : 10 m

**GIVEN:**

Sum of area of the square formed on the diagonals of rhombus = 400 m^{2}

**Concept used:**

If the square is made on diagonals of rhombus,

then the area of squares = (d_{1})^{2} + (d_{2})^{2}

Where d_{1} and d_{2} = diagonals of Rhombus.

**Formula used:**

Side of rhombus = (d1/2)^{2} + (d2/2)^{2}

**CALCULATION:**

**According to the question,**

Side of Rhombus = √[(d1/2)2 + (d2/2)2]

⇒ Side of Rhombus = √[(d1)2 + (d2)2]/4

⇒ Side of Rhombus = √(400/4)

⇒ Side of Rhombus = 10 m

∴ S**ide of the rhombus = 10 m**

__Additional Information__

** **

The diagonals of a rhombus bisect each other at 90°, so triangle AOB, AOD, BOC and COD are right angled triangle.

In △AOB,

Pythagoras theorem

AO^{2} + OB^{2 }= AB^{2}

⇒ (AC/2)^{2} + (BD/2)^{2} = AB^{2}

⇒ AC^{2}/4 + BD^{2}/4 = AB2

⇒ AC^{2 }+ BD^{2} = 4AB^{2}

Similarly,

AC2 + BD2 = 4BC^{2}

AC2 + BD2 = 4CD^{2}

AC2 + BD2 = 4AD^{2}

Adding all these we get,

4(AB2 + CD2 + BC2 + AD2) = 4(AC2 + BD2)

**∴ The sum of the square of the sides of a rhombus is equal to the sum of the squares of its diagonals .**