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.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
ABCD is a rhombus in which AB = BC = CD = DA = a (say). Its diagonals AC and BD are right bisectors of each other at O.
In $\Delta \mathrm{OAB}, \angle \mathrm{AOB}=90^{\circ}$
$\mathrm{OA}=\frac{1}{2} \mathrm{AC}$ and $\mathrm{OB}=\frac{1}{2} \mathrm{BD}$
By pythagoras theorem, we have
$\mathrm{OA}^{2}+\mathrm{OB}^{2}=\mathrm{AB}^{2}$
$\Rightarrow\left(\frac{1}{2} A C\right)^{2}+\left(\frac{1}{2} B D\right)^{2}=A B^{2}$
or $4 \mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BD}^{2}$
$\Rightarrow \mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}+\mathrm{DA}^{2}$
$=\mathrm{AC}^{2}+\mathrm{BD}^{2}$
Hence proved.
ABCD is a rhombus in which AB = BC = CD = DA = a (say). Its diagonals AC and BD are right bisectors of each other at O.
In $\Delta \mathrm{OAB}, \angle \mathrm{AOB}=90^{\circ}$
$\mathrm{OA}=\frac{1}{2} \mathrm{AC}$ and $\mathrm{OB}=\frac{1}{2} \mathrm{BD}$
By pythagoras theorem, we have
$\mathrm{OA}^{2}+\mathrm{OB}^{2}=\mathrm{AB}^{2}$
$\Rightarrow\left(\frac{1}{2} A C\right)^{2}+\left(\frac{1}{2} B D\right)^{2}=A B^{2}$
or $4 \mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BD}^{2}$
$\Rightarrow \mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}+\mathrm{DA}^{2}$
$=\mathrm{AC}^{2}+\mathrm{BD}^{2}$
Hence proved.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.