In an equilateral triangle ABC, If AD ⊥ BC, then

Question:

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) $3 \mathrm{AB}^{2}=2 \mathrm{AD}^{2}$

Solution:

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 $(c)$

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