MCQ
In the given figure, $ABCD$ is a Rhombus. Then,
- A$A C^2+B D^2=2 A B^2$
- B$\left(A C^2+B D^2\right)=3 A B^2$
- C$\mathrm{AC}^2+\mathrm{BD}^2=\mathrm{AB}^2$
- ✓$\mathrm{AC}^2+\mathrm{BD}^2=4 \mathrm{AB}^2$
✓
Answer
Correct option: D.
$\mathrm{AC}^2+\mathrm{BD}^2=4 \mathrm{AB}^2$
$A B C D$ is a rhombus.
$A B=B C=C D=D A$
In Rhombus, diagonals bisect each other at right angles.
$\text { So, } AO=CO \text { and } BO=DO$
In triangle $\mathrm{AOB}, \mathrm{AO}^2+\mathrm{BO}^2=\mathrm{AB}^2$ (Pythagoras theorem)
$\Big(\frac{1}{2} \text{AC}\Big)^2 + \Big(\frac{1}{2} \text{BD}\Big)^2 = \text{AB}^2$
$\frac{\text{AC}^2}{4} + \frac{\text{BD}^2}{4} = \text{AB}^2$
$\text{AC}^2 + \text{BD}^2 = 4\text{AB}^2$
$A B=B C=C D=D A$
In Rhombus, diagonals bisect each other at right angles.
$\text { So, } AO=CO \text { and } BO=DO$
In triangle $\mathrm{AOB}, \mathrm{AO}^2+\mathrm{BO}^2=\mathrm{AB}^2$ (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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