MCQ
In the given figure, $\ce{ABCD}$ is a rhombus. Then:
- A$\ce{AC^2 + BD^2 = AB^2}$
- B$\ce{AC^2 + BD^2 = 2AB^2}$
- ✓$\ce{AC^2 + BD^2 = 4AB^2}$
- D$\ce{2\left(AC^2 + BD^2\right)=3AB^2}$
✓
Answer
Correct option: C.
$\ce{AC^2 + BD^2 = 4AB^2}$
The diagonals of a rhombus bisect each other at right angles.
$\Rightarrow\text{OA}=\frac{1}{2}\text{AC}$ and $\text{OB}=\frac{1}{2}\text{BD}$
Also, $\angle\text{AOB}=90^{\circ}$
In right $\triangle\text{AOB},$
$\therefore\text{AB}^2=\text{OA}^2+\text{OB}^2 ......($By Pythagoras theorem$)$
$\Rightarrow\text{AB}^2=\frac{1}{4}\text{AC}^2+\frac{1}{4}\text{BD}^2$
$\Rightarrow\text{AC}^2+\text{BD}^2=4\text{AB}^2$
$\Rightarrow\text{OA}=\frac{1}{2}\text{AC}$ and $\text{OB}=\frac{1}{2}\text{BD}$
Also, $\angle\text{AOB}=90^{\circ}$
In right $\triangle\text{AOB},$
$\therefore\text{AB}^2=\text{OA}^2+\text{OB}^2 ......($By Pythagoras theorem$)$
$\Rightarrow\text{AB}^2=\frac{1}{4}\text{AC}^2+\frac{1}{4}\text{BD}^2$
$\Rightarrow\text{AC}^2+\text{BD}^2=4\text{AB}^2$
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