Question

prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagnols ?​

Answer 1

Answer:

Given:-

• ABCD is a rhombus
• AB = BC = DC = DA
• AC perpendicular to BD
• AO = OC = BO = OD

RTP:-

• AB²+BC²+CD²+DA²=AC²+BD²

Proof:-

In triangle AOB, angle AOB = 90°

So, OA²+OB²=AB²(By Pythagoras theorm)

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

${ \to{ \sf{ \frac{ {AC}^{2} + {BD}^{2} }{4} = {AB}^{2} }}}$

AC²+BD²=4AB²

AC²+BD²=AB²+AB²+AB²+AB²

AB²+BC²+CD²+DA²=AC²+BD²

${ \therefore{ \sf{AB = BC = CD = DA}}}$

Answer 2

$\implies$In rhombus ABCD, AB = BC = CD = DA

Diagonals of a rhombus bisect each other perpendicularly.

That is AC ⊥ BD,

∠AOB=∠BOC=∠COD=∠AOD=90° and

Consider right angled triangle AOB AB2 = OA2 + OB2

[By Pythagoras theorem]

$\implies$4AB2 = AC2+ BD2

$\implies$ AB2 + AB2 + AB2 + AB2 = AC2+ BD2

∴ AB2 + BC2 + CD2 + DA2 = AC2+ BD2

$\implies$Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.