Rajasthan BoardEnglish MediumSTD 9MATHSQuadrilaterals1 Mark
Question
In the given figure$, \text{ABCD}$ is a Rhombus. Then,
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Answer
$\text{ABCD}$ is a rhombus.
$AB = BC = CD = DA$
In Rhombus, diagonals bisect each other at right angles.
So$, AO = CO$ and $BO = DO$
In triangle $AOB, AO^2 + BO^2 = AB^2 ($Pythagoras theorem$)$
$\Big(\frac{1}{2} \text{AC}\Big)^2 + \Big(\frac{1}{2} \text{BD}\Big)^2 = \text{AB}^2$
$\frac{\text{AC}^2}{4} + \frac{\text{BD}^2}{4} = \text{AB}^2$
$\text{AC}^2 + \text{BD}^2 = 4\text{AB}^2$
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