Ex.6.5 Q7 Triangles Solution - NCERT Maths Class 10

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Question

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

Diagram

Text Solution

Reasoning:

As we know, In a rhombus , diagonals bisect each other perpendicularly.

Steps:

In rhombus \({ABCD}\)

\(AC\bot BD\) and \( OA = OC ;OB = OD\)

In \(\Delta AOB\)

\(\begin{align}\angle A O B=90^{\circ} \end{align}\)

\(\Rightarrow A B^{2}=O A^{2}+O B^{2}\) .........(1)

Similarly, we can prove

\[\begin{align} B{{C}^{2}}=O{{B}^{2}}+O{{C}^{2}}\ldots \ldots \ldots \left( 2 \right) \\ C{{D}^{2}}=O{{C}^{2}}+O{{D}^{2}}\ldots \ldots \ldots \left( 3 \right) \\ A{{D}^{2}}=O{{D}^{2}}+O{{A}^{2}}\ldots \ldots \ldots \left( 4 \right) \\ \end{align}\]

Adding \((1), (2), (3)\) and \((4)\)

\(A{B^2} + B{C^2} + C{D^2} + A{D^2} = O{A^2} + O{B^2} + O{B^2} + O{C^2} + O{C^2} + O{D^2} + O{D^2} + O{A^2}\)

\(\begin{align}
A{B^2} + B{C^2} + C{D^2} + A{D^2} &= 2\,O{A^2} + 2\,O{B^2} + 2\,O{C^2} + 2\,O{D^2}\\
A{B^2} + B{C^2} + C{D^2} + A{D^2} &= 2[O{A^2} + O{B^2} + O{C^2} + O{D^2}]\\
A{B^2} + B{C^2} + C{D^2} + A{D^2} &= 2\left[ {{{\left( {\frac{{AC}}{2}} \right)}^2} + {{\left( {\frac{{BD}}{2}} \right)}^2} + {{\left( {\frac{{AC}}{2}} \right)}^2} + {{\left( {\frac{{BD}}{2}} \right)}^2}} \right]\\
&\left[ {OA = OC = \frac{{AC}}{2}\,\,and\,\,OB = OD = \frac{{BD}}{2}} \right]\\
A{B^2} + B{C^2} + C{D^2} + A{D^2} &= 2\left[ {\frac{{A{C^2} + B{D^2} + A{C^2} + B{D^2}}}{4}} \right]\\
A{B^2} + B{C^2} + C{D^2} + A{D^2} &= 2\left[ {\frac{{2A{C^2} + 2B{D^2}}}{4}} \right]\\
A{B^2} + B{C^2} + C{D^2} + A{D^2} &= 4\left[ {\frac{{A{C^2} + B{D^2}}}{4}} \right]\\
A{B^2} + B{C^2} + C{D^2} + A{D^2} &= A{C^2} + B{D^2}
\end{align}\)