MCQ
- AAC2 + BD2 = AB2
- BAC2 + BD2 = 2AB2
- CAC2 + BD2 = 4AB2
- D2(AC2 + BD2) = 3AB2
✓
Answer
- AC2 + BD2 = 4AB2
Solution:
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)
$\Rightarrow\text{AB}^2=\frac{1}{4}\text{AC}^2+\frac{1}{4}\text{BD}^2$
$\Rightarrow\text{AC}^2+\text{BD}^2=4\text{AB}^2$
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