Question

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

Answer 1

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Answer 2

Answer:

Each altitude of the equilateral triangle is equal to $a\sqrt3$

Thus,

$\begin{aligned} A D & = a \sqrt { 3 } \\\\ B E & = a \sqrt { 3 } \\\\ C F & = a \sqrt { 3 } \end{aligned}$

To find:

Each of its altitudes = ?

Solution:  

Given : ABC is an equilateral triangle of side 2a.

In equilateral triangle, △ADB = △ADC

& ∠ADB = ∠ADC (Both 90 degree as AD ⊥ BC)

Thus,  △ ADB ≅ △ ADC (By R. H. S. Congruency)

As per CPCT i.e. they are corresponding parts of congruent triangle,

BD = DC  

$\begin{array} { c } { B D = D C = \frac { 1 } { 2 } B C } \\\\ { B D = D C = \frac { 2 a } { 2 } } \\\\ { B D = D C = a } \end{array}$

Using Pythagoras theorem,

$\begin{aligned} \text { (Hypotenuse } ) ^ { 2 } & = ( \text {Height} ) ^ { 2 } + ( \text {Base} ) ^ { 2 } \\\\ ( A B ) ^ { 2 } & = ( A D ) ^ { 2 } + ( B D ) ^ { 2 } \\\\ ( 2 a ) ^ { 2 } & = ( A D ) ^ { 2 } + ( a ) ^ { 2 } \\\\ 4 a ^ { 2 } & = ( A D ) ^ { 2 } + ( a ) ^ { 2 } \\\\ 4 a ^ { 2 } & - a ^ { 2 } = A D ^ { 2 } \\\\ & 3 a ^ { 2 } = A D ^ { 2 } \end{aligned}$

Hence,

$\begin{aligned} A D & = a \sqrt { 3 } \\\\ B E & = a \sqrt { 3 } \\\\ C F & = a \sqrt { 3 } \end{aligned}$

Thus, the altitudes of the given equilateral triangle is equal to $a\sqrt3$