In an equilateral triangle ABC, If AD ⊥ BC, then
(a) $2 \mathrm{AB}^{2}=3 \mathrm{AD}^{2}$
(b) $4 \mathrm{AB}^{2}=3 \mathrm{AD}^{2}$
(c) $3 \mathrm{AB}^{2}=4 \mathrm{AD}^{2}$
(d) $3 \mathrm{AB}^{2}=2 \mathrm{AD}^{2}$
Given: In an equilateral $\triangle \mathrm{ABC}, \mathrm{AD} \perp \mathrm{BC}$
Applying Pythagoras theorem,
In ΔABD,
$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$
$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\left(\frac{1}{2} \mathrm{BC}\right)^{2}\left(\right.$ SinceBD $\left.=\frac{1}{2} \mathrm{BC}\right)$
$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\left(\frac{1}{2} \mathrm{AB}\right)^{2}($ Since $\mathrm{AB}=\mathrm{BC})$
$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\frac{1}{4} \mathrm{AB}^{2}$
$3 \mathrm{AB}^{2}=4 \mathrm{AD}^{2}$
We got the result as $(c)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.