Question

find the altitude of an equilateral triangle when each of side is 'a' cm​

Answer 1

Step-by-step explanation:

Altitude of equilateral triangle=

$\frac{ \sqrt{3} }{2} a$

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Answer 2

$\large{\underline{\underline{\mathfrak{\red{\sf{Answer-}}}}}}$

The altitude of an equilateral triangle when each of side is 'a' cm is $\green{\sf{\dfrac{\sqrt3a}{2}}}$

$\large{\underline{\underline{\mathfrak{\red{\sf{Explanation-}}}}}}$

Given that, AB = BC = AC = a cm

We have to find the length of altitude of ∆ABC,

Now, we know that, in an equilateral ∆, both median and altitude are the same things.

Therefore, D is the mid point of BC.

BD = $\sf{\dfrac{a}{2}}$ and DC = $\sf{\dfrac{a}{2}}$

Also, AD is perpendicular to BC.

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In rt ∆ABD,

By using Pythagoras theorem,

AB² = BD² + AD²

$\implies$ AB² - BD² = AD²

$\implies$ a² - $\sf{\dfrac{a}{2}^2}$ = AD²

$\implies$ $\sf{\dfrac{4a^2-a^2}{4}}$ = AD²

$\implies$ $\sf{\dfrac{3a^2}{4}}$ = AD²

$\implies$ $\sf\sqrt{\dfrac{3a^2}{4}}$ = AD

$\implies$ AD = $\sf{\dfrac{\sqrt3\:a}{2}}$

The length of altitude of equilateral ∆ABC is : $\sf{\dfrac{\sqrt3\:a}{2}}$

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