MCQ
In an equilateral triangle $\text{ABC}$ if $\text{AD}\perp\text{BC},$ then :
- A$ 5 A B^2=4 A D^2$
- ✓$ 3 A B^2=4 A D^2 $
- C$ 4 A B^2=3 A D^2 $
- D$ 2 A B^2=3 A D^2 $
✓
Answer
Correct option: B.
$ 3 A B^2=4 A D^2 $
$\triangle\text{ABC}$ is an equilateral triangle and $\text{AD}\perp\text{BC}.$
In $\triangle\text{ABD},$ applying Pythagoras theorem, we get,
$\text{AB}^2=\text{AD}^2+\text{BD}^2$
$\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\Big(\because\text{BD}=\frac{1}{2}\text{BC}\Big)$
$\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{AB}\Big)^2\Big(\because\text{AB}=\text{BC}\Big)$
$\text{AB}^2=\text{AD}^2+\frac{1}{4}\text{AB}^2$
$3\text{AB}^2=4\text{AD}^2$
We got the result as $B.$
In $\triangle\text{ABD},$ applying Pythagoras theorem, we get,
$\text{AB}^2=\text{AD}^2+\text{BD}^2$
$\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\Big(\because\text{BD}=\frac{1}{2}\text{BC}\Big)$
$\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{AB}\Big)^2\Big(\because\text{AB}=\text{BC}\Big)$
$\text{AB}^2=\text{AD}^2+\frac{1}{4}\text{AB}^2$
$3\text{AB}^2=4\text{AD}^2$
We got the result as $B.$
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