Answer
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Hint: To solve this question, we should have the knowledge regarding the properties of a rhombus. Like, the diagonals of rhombus bisect each other at $90{}^\circ $. We should also know that for any right angled triangle, Pythagoras theorem states that, ${{\left( Perpendicular \right)}^{2}}+{{\left( Base \right)}^{2}}={{\left( Hypotenuse \right)}^{2}}$. By using these concepts, we can find the answer to this question.
Complete step-by-step answer:
In this question, we have been asked to find the value of the sum of the squares of the sides of the rhombus. To solve this question, let us first draw the figure of a rhombus ABCD.
Here, we have represented ABCD as the rhombus with O as the point of intersection of the diagonals AC and BD. Now, we know that the diagonals of rhombus bisects each other at an angle of $90{}^\circ $. So, we can say that, $\angle AOD=\angle BOA=\angle COB=\angle DOC=90{}^\circ $. And we can say that $\Delta AOD,\Delta BOA,\Delta COB,\Delta DOC$ are right-angled triangles, right angled at O. Now we know that for any right-angled triangle, Pythagoras theorem states that,${{\left( Perpendicular \right)}^{2}}+{{\left( Base \right)}^{2}}={{\left( Hypotenuse \right)}^{2}}$. So, we can say,
In $\Delta AOD$,
${{\left( AO \right)}^{2}}+{{\left( OD \right)}^{2}}={{\left( AD \right)}^{2}}\ldots \ldots \ldots \left( i \right)$
Similarly, we can say that in $\Delta BOA$,
${{\left( OA \right)}^{2}}+{{\left( OB \right)}^{2}}={{\left( BA \right)}^{2}}\ldots \ldots \ldots \left( ii \right)$
And in $\Delta COB$, we can say that,
${{\left( OB \right)}^{2}}+{{\left( OC \right)}^{2}}={{\left( CB \right)}^{2}}\ldots \ldots \ldots \left( iii \right)$
And in $\Delta DOC$, we can say,
${{\left( OD \right)}^{2}}+{{\left( OC \right)}^{2}}={{\left( DC \right)}^{2}}\ldots \ldots \ldots \left( iv \right)$
Now, we will add equations (i), (ii), (iii) and (iv). So, we get,
$\begin{align}
& {{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}={{\left( OA \right)}^{2}}+{{\left( OA \right)}^{2}}+{{\left( OB \right)}^{2}}+{{\left( OB \right)}^{2}}+{{\left( OC \right)}^{2}}+{{\left( OC \right)}^{2}}+{{\left( OD \right)}^{2}}+{{\left( OD \right)}^{2}} \\
& \Rightarrow {{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}=2{{\left( OA \right)}^{2}}+2{{\left( OB \right)}^{2}}+2{{\left( OC \right)}^{2}}+2{{\left( OD \right)}^{2}}\ldots \ldots \ldots \left( v \right) \\
\end{align}$
Now, we know that the diagonals of rhombus bisects each other. So, we can say,
$OA=OC$ and $OB=OD$
From the above 2 equations, we can write equation (v) as,
$\begin{align}
& {{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}=2{{\left( OA \right)}^{2}}+2{{\left( OB \right)}^{2}}+2{{\left( OA \right)}^{2}}+2{{\left( OB \right)}^{2}} \\
& \Rightarrow {{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}=4{{\left( OA \right)}^{2}}+4{{\left( OB \right)}^{2}} \\
\end{align}$
And we know that $4={{2}^{2}}$, so we can write $4{{\left( OA \right)}^{2}}={{\left( 2OA \right)}^{2}}$ and $4{{\left( OB \right)}^{2}}={{\left( 2OB \right)}^{2}}$. Hence, we can write the above equation as,
${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}={{\left( 2OA \right)}^{2}}+{{\left( 2OB \right)}^{2}}$
Now, we can say that if the diagonals bisect each other, then $2\left( OA \right)=AC$ and $2\left( OB \right)=BD$. Therefore, we can write the above equation as,
${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}={{\left( AC \right)}^{2}}+{{\left( BD \right)}^{2}}$
Hence, we can say that the sum of the squares of the sides of the rhombus is equal to the sum of the square of the diagonals. Therefore, we can say that option A is the correct answer.
Note: We should remember this as the identity of rhombus that the sum of the squares of the sides of a rhombus are equal to the sum of the diagonal of the rhombus. We should also remember that to prove this identity, we will consider each triangle that includes one side and 2 half diagonals and apply the Pythagoras theorem to get the answer.
Complete step-by-step answer:
In this question, we have been asked to find the value of the sum of the squares of the sides of the rhombus. To solve this question, let us first draw the figure of a rhombus ABCD.
Here, we have represented ABCD as the rhombus with O as the point of intersection of the diagonals AC and BD. Now, we know that the diagonals of rhombus bisects each other at an angle of $90{}^\circ $. So, we can say that, $\angle AOD=\angle BOA=\angle COB=\angle DOC=90{}^\circ $. And we can say that $\Delta AOD,\Delta BOA,\Delta COB,\Delta DOC$ are right-angled triangles, right angled at O. Now we know that for any right-angled triangle, Pythagoras theorem states that,${{\left( Perpendicular \right)}^{2}}+{{\left( Base \right)}^{2}}={{\left( Hypotenuse \right)}^{2}}$. So, we can say,
In $\Delta AOD$,
${{\left( AO \right)}^{2}}+{{\left( OD \right)}^{2}}={{\left( AD \right)}^{2}}\ldots \ldots \ldots \left( i \right)$
Similarly, we can say that in $\Delta BOA$,
${{\left( OA \right)}^{2}}+{{\left( OB \right)}^{2}}={{\left( BA \right)}^{2}}\ldots \ldots \ldots \left( ii \right)$
And in $\Delta COB$, we can say that,
${{\left( OB \right)}^{2}}+{{\left( OC \right)}^{2}}={{\left( CB \right)}^{2}}\ldots \ldots \ldots \left( iii \right)$
And in $\Delta DOC$, we can say,
${{\left( OD \right)}^{2}}+{{\left( OC \right)}^{2}}={{\left( DC \right)}^{2}}\ldots \ldots \ldots \left( iv \right)$
Now, we will add equations (i), (ii), (iii) and (iv). So, we get,
$\begin{align}
& {{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}={{\left( OA \right)}^{2}}+{{\left( OA \right)}^{2}}+{{\left( OB \right)}^{2}}+{{\left( OB \right)}^{2}}+{{\left( OC \right)}^{2}}+{{\left( OC \right)}^{2}}+{{\left( OD \right)}^{2}}+{{\left( OD \right)}^{2}} \\
& \Rightarrow {{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}=2{{\left( OA \right)}^{2}}+2{{\left( OB \right)}^{2}}+2{{\left( OC \right)}^{2}}+2{{\left( OD \right)}^{2}}\ldots \ldots \ldots \left( v \right) \\
\end{align}$
Now, we know that the diagonals of rhombus bisects each other. So, we can say,
$OA=OC$ and $OB=OD$
From the above 2 equations, we can write equation (v) as,
$\begin{align}
& {{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}=2{{\left( OA \right)}^{2}}+2{{\left( OB \right)}^{2}}+2{{\left( OA \right)}^{2}}+2{{\left( OB \right)}^{2}} \\
& \Rightarrow {{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}=4{{\left( OA \right)}^{2}}+4{{\left( OB \right)}^{2}} \\
\end{align}$
And we know that $4={{2}^{2}}$, so we can write $4{{\left( OA \right)}^{2}}={{\left( 2OA \right)}^{2}}$ and $4{{\left( OB \right)}^{2}}={{\left( 2OB \right)}^{2}}$. Hence, we can write the above equation as,
${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}={{\left( 2OA \right)}^{2}}+{{\left( 2OB \right)}^{2}}$
Now, we can say that if the diagonals bisect each other, then $2\left( OA \right)=AC$ and $2\left( OB \right)=BD$. Therefore, we can write the above equation as,
${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}+{{\left( CD \right)}^{2}}+{{\left( DA \right)}^{2}}={{\left( AC \right)}^{2}}+{{\left( BD \right)}^{2}}$
Hence, we can say that the sum of the squares of the sides of the rhombus is equal to the sum of the square of the diagonals. Therefore, we can say that option A is the correct answer.
Note: We should remember this as the identity of rhombus that the sum of the squares of the sides of a rhombus are equal to the sum of the diagonal of the rhombus. We should also remember that to prove this identity, we will consider each triangle that includes one side and 2 half diagonals and apply the Pythagoras theorem to get the answer.
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