Ex. 6.6 Q6 Triangles Solution - NCERT Maths Class 10

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Question

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

 

Text Solution

 

Reasoning:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Steps:

In parallelogram \(ABCD \)

\(AB = CD \\ AD = BC\)

 Draw \(A E \perp C D\) , \(D F \perp A B\) 
\(​​​​ EA = DF\) (Perpendiculars drawn between same parallel lines)

In \(\Delta AEC\)

\begin{align}A{C^2} &= A{E^2} + E{C^2}\\ &= A{E^2} + {\left[ {ED + DC} \right]^2}\\ &= A{E^2} + D{E^2} + D{C^2} + 2DE.DC\\A{C^2} &= A{D^2} + D{C^2} + 2DE \cdot DC \ldots  \ldots   \left( i \right)\;\;\left[ {{\rm{Since, }}A{D^2} = A{E^2} + D{E^2}} \right]\end{align}

In \(\Delta \,DFB\)
\[\begin{align} B{D^2} &= D{F^2} + B{F^2}\\ &= D{F^2} + {\left[ {AB - AF} \right]^2}\\ &= D{F^2} + A{B^2} + A{F^2} - 2AB.AF\\ &= A{D^2} + A{B^2} - 2AB.AF\\ B{D^2} &= A{D^2} + A{B^2} - 2AB.AF \ldots  \left( \rm{ii} \right) \end{align}\] (Since \(A{D^2} = D{F^2} + A{F^2}\))

Adding (i) and (ii)

\[\begin{align} A{C^2} + B{D^2} &= A{D^2} + D{C^2} + 2DE \cdot DC + A{D^2} + A{B^2} - 2AB.AF\\ A{C^2} + B{D^2} &= B{C^2} + D{C^2} + A{D^2} + A{B^2} + 2AB.AF - 2AB.F \end{align}\]
 (Since \(AD = BC\) and \(DE = AF, CD = AB\))

 \[ \Rightarrow A{C^2} + B{D^2} = A{B^2} + B{C^2} + C{D^2} + A{D^2}\]

 

  
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