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# The length of an altitude of an equilateral triangle of side 2a is:-A. $\sqrt{2}a$ cmB. $\sqrt{3}a$cmC. 2 cmD. 1 cm

Last updated date: 29th Feb 2024
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Hint: For the solution of this question, we will be using the Pythagoras Theorem.
Pythagoras Theorem states that,
$\Rightarrow {{(Hypotenuse)}^{2}}={{(Perpendicular)}^{2}}+{{(Base)}^{2}}$
Now, since the triangle is an equilateral triangle, its altitude will bisect its base, and also, it will be perpendicular to its base.
So, as we get us Hypotenuse, Base and Perpendicular, we will put the values of Hypotenuse, Base and Perpendicular in the theorem to find the length of the altitude of the equilateral triangle.
The altitude will act like the perpendicular of the triangle which is equilateral.

Let the equilateral triangle be ABC.

We will now draw the altitude AD,

Now, in right angled ∆ABD, we will use the Pythagoras theorem as follows,
\begin{align} & \Rightarrow {{(H)}^{2}}={{(P)}^{2}}+{{(B)}^{2}} \\ & \Rightarrow {{(AB)}^{2}}={{(AD)}^{2}}+{{(BD)}^{2}} \\ & \Rightarrow {{(2a)}^{2}}={{(AD)}^{2}}+{{(a)}^{2}} \\ & \Rightarrow {{(2a)}^{2}}-{{(a)}^{2}}={{(AD)}^{2}} \\ & \Rightarrow 4{{a}^{2}}-{{a}^{2}}={{(AD)}^{2}} \\ & \Rightarrow 3{{a}^{2}}={{(AD)}^{2}} \\ & \Rightarrow \sqrt{(3{{a}^{2}})}=(AD) \\ & \Rightarrow \sqrt{3}a=AD \\ \end{align}.
So, the measure of AD is $\sqrt{3}a$ .

So, the correct answer is “Option B”.

Note: Let us now understand about Altitude and Median.
ALTITUDE: Altitude of a triangle is the perpendicular drawn from the vertex of the triangle to the opposite side. It is also known as the height of the triangle.
MEDIAN: A median of triangle is a line segment that joins a vertex to the mid-point of the side that is opposite to that vertex.
In the figure given below, in ∆ABC, AD is the altitude of the triangle ABC and AE is the median that divides BC into two equal halves, that is, BE = EC.