Question

in a rhombus, if sum of squares of it's sides is 240 then sum of the squares of diagonal is ?
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Answer 1

Pls mark this answer as the brainliest

Answer 2

The sum of the squares of diagonal is 240 sq units.

Solution:

Consider a rhombus ABCD.

AB=BC=CD=AD

They bisect each other at right angles.

$\therefore \angle AOB = \angle BOC = \angle COD = \angle AOD = 90^{\circ}$

$OA=OC=\frac{AC}{2}; OB=OD=\frac{BD}{2}$

Consider the right angled triangle AOB,

$\Rightarrow AB^2=OA^2+OB^2$

$\Rightarrow AB^2=(\frac{AC}{2})^2+(\frac{BD}{2})^2$

$\Rightarrow AB^2=\frac{(AC)^2}{4}+\frac{(BD)^2}{4}=\frac{(AC)^2+(BD)^2}{4}$

$\Rightarrow 4AB^2=AC^2+BD^2$

$\Rightarrow AB^2 + BC^2 + CD^2 + AD^2 = AC^2 + BD^2$

So, we can conclude that the sum of the squares of the sides in a rhombus = sum of the squares of diagonal.

Therefore, the sum of the squares of diagonal is 240 sq units.

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