Question

ello!!❤❤


In an equilateral triangle with side a, prove that

$1) \: altitude \: = \frac{a \sqrt{3} }{2}$
$2) \: area \: = \frac{ \sqrt{3} }{4} {a}^{2}$

Answer 1

hiii dear friend_______

An equilateral triangle is symmetrical so if you draw a perpendicular from point A to BC it bisects the side BC.

so CD=a/2

AD is the altitude(height) of the triangle ABC

In triangle ADC by pythagoras theorem :

AD^2 = a^2 - a^2/4 = 3a^2/4

AD =a√3/2

We know that area of a triangle = 1/2 base×height

so  area of this triangle =

1/2 x (a√3/2) x a

= √3 a^2 / 4

Answer 2

Heya !!


Here is your solution :


Consider a equilateral ∆ABC in that AD is altitude on line BC.


Hence, in ∆ABC we shall get two other triangles named ∆ABD and ∆ACD.


Considering ∆ABD,


⇒AB = a ( Given ) [ Hypotenuse ( Side opposite to 90° ) ]


⇒BD = ( a/2 ) [ Altitude of an equilateral ∆ bisects its base ]


⇒<ADB = 90°


⇒AD = ?


Using Pythagoras Theorem,


⇒Hypotenuse² = Base² + Altitude²


⇒ a² = ( a/2 )² + AD²


⇒a² = ( a²/4 ) + AD²


⇒a² - ( a²/4 ) = AD²


⇒( 4a² - a² ) /4 = AD²


⇒ 3a²/4 = AD²


⇒ AD = √( 3a²/4 )


•°• AD = ( a√3 / 2 )


Proved !!


Now , For ∆ABC , we have,


⇒AC = Base = a


⇒AD = Corresponding Altitude = ( a√3 / 2 )


We know that,


⇒Area of an equilateral ∆ = ( 1/2 ) × Base × Corresponding Altitude


⇒Ar. of an equilateral ∆ = ( 1/2 ) × a × ( a√3/2 )


•°• Ar. of an equilateral ∆ = ( √3/4 )a²


Proved !!