Ex. 6.6 Q6 Triangles Solution - NCERT Maths Class 10
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 \(DE = AF, CD = AB\))
\[ \Rightarrow A{C^2} + B{D^2} = A{B^2} + B{C^2} + C{D^2} + A{D^2}\]