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  • Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.      

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

 

 
 
 
 
 

Answers (1)

Given \rightarrow Rhombus ABCD
point of intersection of diagonal = 0
To Prove \rightarrow AB^{2}+BC^{2}+CD^{2}+DA^{2}= AC^{2}+BD^{2}

Properties of Rhombus \rightarrow(i) All sides of Rhombus are equal
                                          (ii) diagonal bisect each other at 900
Proof \rightarrow \angle AOB= \angle BOC= \angle COD= \angle DOA= 90^{\circ} (properties of rhombus)
Also AO= CO= \frac{1}{2}AC---(i)\; and\; BO= DO= \frac{1}{2}BD---(ii) (property of rhombus)
Now in \triangle AOB
AB^{2}= AD^{2}+\left ( BO \right )^{2} (pythagorus theorem)
AB^{2}= \left ( \frac{1}{2} AC\right )^{2}+\left ( \frac{1}{2} BD\right )^{2} (using (i) and (ii))
AB^{2}= \frac{AC^{2}+BD^{2}}{4}
\Rightarrow 4AB^{2}= AC^{2}+BD^{2}
\Rightarrow AB^{2}+AB^{2}+AB^{2}+AB^{2}= AC^{2}+BD^{2}
\Rightarrow AB^{2}+BC^{2}+CD^{2}+DA^{2}= AC^{2}+BD^{2} (sides of rhombus are equal)
Hence proved

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Safeer PP

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