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 midpoint of AD. Then BE is
A.$4\sqrt{3}$cm
B.$6\sqrt{2}$cm
C.$\sqrt{63}$cm
D.None of these

Answer 1

Hint: Since we are given an equilateral triangle ABC and AD is perpendicular to BC. By using the property the height of an equilateral triangle bisects its base we get BD=6cm and then by using Pythagoras theorem we can find AD and then considering the small triangle BDE and using Pythagoras theorem we can find BE

Complete step-by-step answer:
Given ABC is an equilateral triangle .
Hence all its sides are equal.

And it's also given that D is the foot of the perpendicular from A to BC
It means that AD is perpendicular to BC

Step 2 :
Since ABC is an equilateral triangle , we know that the height or altitude of an equilateral triangle bisects its base.
Here AD is the height of the triangle .
Therefore it bisects the base and now we have BD=BC=6cm
Step 3:
Now let's find AD by using Pythagoras theorem
Consider the triangle ABD . It is a right triangle .

By using Pythagoras theorem
$$\begin{gathered} \Rightarrow (\text{Hypotenuse }{)}^2=\text{ Sum of squares of other two sides} \\ \Rightarrow {\left(AB\right)}^2={\left(AD\right)}^2+{\left(BD\right)}^2 \\ \Rightarrow {12}^2={\left(AD\right)}^2+6^2 \\ \Rightarrow 144={\left(AD\right)}^2+36 \\ \Rightarrow 144-36={\left(AD\right)}^2 \\ \Rightarrow 108={\left(AD\right)}^2 \\ \Rightarrow AD=\sqrt{108} \end{gathered}$$
Now we have that $AD=\sqrt{108}cm$
Step 4 :
We are given that E is the midpoint of AD.

Since E is the midpoint of AD , we have AE=ED=$\frac{\sqrt{108}}{2}$cm
We need to find BE , so let's consider the right triangle BDE
Lets use Pythagoras theorem to find BE
 $$\begin{gathered} \Rightarrow (\text{Hypotenuse }{)}^2=\text{ Sum of squares of other two sides} \\ \Rightarrow {\left(BE\right)}^2={\left(ED\right)}^2+{\left(BD\right)}^2 \\ \Rightarrow {\left(BE\right)}^2={\left(\frac{\sqrt{108}}{2}\right)}^2+6^2 \\ \Rightarrow {\left(BE\right)}^2=\frac{108}{4}+36 \end{gathered}$$
Now by taking lcm we get
$$\begin{gathered} \Rightarrow {\left(BE\right)}^2=\frac{108+144}{4} \\ \Rightarrow {\left(BE\right)}^2=\frac{252}{4} \\ \Rightarrow {\left(BE\right)}^2=63 \\ \Rightarrow BE=\sqrt{63}cm \end{gathered}$$
Therefore $BE=\sqrt{63}cm$
The correct option is C

Note: The sides of an equilateral triangle are congruent.
An equilateral triangle is a special case of a triangle where all 3 sides have equal length and all 3 angles are equal to 60 degrees.