Question

In a rhombus of side 2cm the sum of squares of its diagonals is​

Answer 1

sum of squares of its diagonals is 16

Step-by-step explanation:

ley say rhombus diagonals are

2a and 2b

we need to find sum of square of diagonals

${(2a)}^{2} + {(2b)}^{2} \\ = 4 {a}^{2} + 4 {b}^{2} \\ = 4( {a}^{2} + {b}^{2} )$

diagonals of rhombus bisect perpendicularly each other

2a/2 = a

2b/2 = b

${a}^{2} + {b}^{2} = {2}^{2} \\ {a}^{2} + {b}^{2} = 4$

sum of squares of its diagonals is

4(4)= 16

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Answer 2

$\bold {Sum\ of\ the\ square\ of\ the\ diagonal\ is\  16\ cm^2}$

Step-by-step explanation:

In a rhombus ABCD,

AB=CD=DA=BC=2 cm

We have to find,

Sum of the square of diagonals.

We know that ,diagonals AC and BD of a rhombus are always equal and intersect each other at 90° at point O.

Let AO =BO=CO=DO = x

Now,In ΔAOB,

By applying Pyhthagoras theorem,

$(AO)^2+(BO)^2=(AB)^2$

$x^2+x^2 = (2)^2$

$2x^2=4$

$x^2=2$

$x = \sqrt2$

So,$AO=BO=CO=DO = \sqrt 2$

Now,

AC = AO+CO

       =$\sqrt 2+\sqrt 2$

      = $2\sqrt2$

⇒$\bold {AC = BD = 2\sqrt2}$

Here,$(AC)^2+(BD)^2 = (2\sqrt2)^2 + (2\sqrt2)^2$

                                = 8 + 8

      $\bold { (AC)^2+(BD)^2 = 16}$

Hence,$\bold {Sum\ of\ the\ square\ of\ the\ diagonal\ is\  16\ cm^2}$