Question

The sides of an equilateral triangle ABC are 12 cm each. D is the foot of the perpendicular from A to BC and E is the mid point of AD BE is

• A. $4\sqrt{3}cm$
• B. $6\sqrt{2}cm$
• C. $\sqrt{63}cm$
• D. None of these

Answer 1

Answer: (C)

$\sqrt{63}cm$

In equilateral triangle ABC,
$AD\perp BC$
Since, it is an equilateral triangle, hence, D is the mid point of BC.
$BD=6$ cm
Now, In $△ABD$
$A{B}^2=B{D}^2+A{D}^2$
${12}^2=6^2+A{D}^2$
$AD=\sqrt{108}$
Now, E is the mid point of AD. Hence, $DE=\frac{\sqrt{108}}{2}$
In triangle BED,
$B{E}^2=D{E}^2+B{D}^2$
$B{E}^2=(\frac{\sqrt{108}}{2}{)}^2+6^2$
$B{E}^2=\frac{108}{4}+36$
$B{E}^2=63$
$BE=\sqrt{63}$