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# Find the length of the altitude of an equilateral triangle with side 6cm.

Last updated date: 20th Sep 2024
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Hint: First of all, draw the diagram with the given data to get a clear idea of what we have to find. Then use Pythagoras theorem to find the altitude of the given triangle. An alternate method is also provided in the note. We can use any one of the methods to get the required answer.

Let the given equilateral triangle is $\Delta ABC$
The side length of equilateral $\Delta ABC$ is 6 cm.
Here we have to find the length of altitude of $\Delta ABC$.
Let the foot perpendicular to the altitude from point $A$ to the line $BC$ is $D$. So, the altitude is $AD$.
Since, $\Delta ABC$ is equilateral the altitude $AD$ divides the triangle into two equal right-angle triangles as shown in the below figure:

From Pythagoras theorem we have ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Adjacent side}}} \right)^2} + {\left( {{\text{Opposite side}}} \right)^2}$
So, in equilateral triangle $\Delta ABC$ we have
$\Rightarrow {\left( {AD} \right)^2} + {\left( {CD} \right)^2} = {\left( {AC} \right)^2} \\ \Rightarrow {\left( {AD} \right)^2} + {\left( 3 \right)^2} = {\left( 6 \right)^2} \\ \Rightarrow {\left( {AD} \right)^2} + 9 = 36 \\ \Rightarrow {\left( {AD} \right)^2} = 36 - 9 = 27 \\$
Rooting on both sides we get
$\Rightarrow \sqrt {{{\left( {AD} \right)}^2}} = \sqrt {27} = \sqrt {3 \times 3 \times 3} \\ \therefore AD = 3\sqrt 3 {\text{ cm}} \\$
Thus, the required length of altitude is $3\sqrt 3 {\text{ cm}}$.

Note: Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides” i.e., ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Adjacent side}}} \right)^2} + {\left( {{\text{Opposite side}}} \right)^2}$.
Alternate method:
We know that the length of the altitude of the equilateral triangle of side length $a{\text{ cm}}$ is given by $\dfrac{{\sqrt 3 }}{2}a{\text{ cm}}$.
So, the altitude of the equilateral triangle of length 6 cm is $\dfrac{{\sqrt 3 }}{2}\left( 6 \right) = \dfrac{{6\sqrt 3 }}{2} = 3\sqrt 3 {\text{ cm}}$