Question

ABC is an equilateral triangle with each side as 2a. Find each of its altitudes.

Answer 1

Hint: Draw an equilateral triangle with sides 2a. Draw an altitude perpendicular to the base. Prove the triangles are similar using RHS congruence. Find the altitude of the triangle by using Pythagoras theorem. Similarly, prove that altitudes drawn from all vertices are of the same length.

Complete step-by-step answer:
Consider the given figure which is an equilateral triangle ABC with each side 2a.

Now let us draw an altitude to side BC. Such that AD is perpendicular to BC. i.e.\[AD\bot BC\].
We need to find the value of AD.
Let us consider \[\Delta ADB\] and \[\Delta ADC\].
\[\therefore AB=AC\],as both are 2a and the sides of an equilateral triangle are equal.
AD = AD, which is common on both \[\Delta ADB\] and \[\Delta ADC\].
\[\angle ADB=\angle ADC\], both are \[{{90}^{\circ }}\] as, \[AD\bot BC\].
Hence we can say that\[\Delta ADB\cong \Delta ADC\], by R.H.S congruence. RHS congruence states that if two right triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.
\[\therefore \Delta ADB\cong \Delta ADC\]
Hence, we can say that BD = DC and thus D becomes the midpoint of side BC.
We know, BC = 2a, from the figure.

BC = BD + DC
BC = 2DC
Substitute, BC = 2a.
2DC = 2a
$ \therefore DC=\dfrac{2a}{2}=a $
$ \therefore BD=DC=a $
Hence, BD = a.
Now, let us apply Pythagoras theorem in \[\Delta ADB\].
$ {{\left( hypotenuse \right)}^{2}}={{\left( height \right)}^{2}}+{{\left( base \right)}^{2}} $
$  A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}} $
We know, AB = 2a and BD = a.
$ \therefore AD=\sqrt{A{{B}^{2}}-B{{D}^{2}}}=\sqrt{{{\left( 2a \right)}^{2}}-{{a}^{2}}} $
$  AD=\sqrt{4{{a}^{2}}-{{a}^{2}}}=\sqrt{3{{a}^{2}}}=\sqrt{3}a $
$  \therefore AD=\sqrt{3}a $
Similarly, if we are considering \[\Delta AEB\] and \[\Delta CEB\].

By RHS congruence,
\[\Delta AEB\cong \Delta CEB\] and AE = EC = a.
\[\therefore BE\bot AC\]
\[\therefore \]Altitude\[BE=a\sqrt{3}\], by applying Pythagoras theorem on \[\Delta AEB\].
Similarly,\[CF\bot AB\], altitude \[CF=a\sqrt{3}\].
Length of altitude AD = BE = CF =\[a\sqrt{3}\].

Note: As it is an equilateral triangle, we know all sides are the same and all angles are \[{{60}^{\circ }}\]. So if we draw altitude \[AD\bot BC\]. \[\Delta ADB\] and \[\Delta ADC\] will be similar. So, BD = DC = a.
You can directly apply Pythagoras theorem to one of the similar triangles to find the altitude AD.
In case of an equilateral triangle, all altitudes drawn are medians. And all medians are altitude.