Question

show that the area of equilateral triangle is √3/4 x^2 where x is side of the triangle.​

Answer 1

GIVEN :-

ABC is an equilateral Triangle with each side measuring x

TO PROVE :- Its area is √3/4 x^2

SOLUTION :-

In order to solve this Question we have to keep to points in mind.

1. Properties Of Equilateral Triangle

The altitude of a equilateral triangle is its median

2. Area of a Triangle

Area of a triangle = 1/2 × base × height

3. Phythagoras Theorem

In a right angle Triangle,

${hypotenuse }^{2} = {base}^{2} + {height}^{2}$

Now, we are in a position to answer your Question.

Let AD be the median/altitude of the Triangle.

Now , let us consider Triangle ADC ,

angle D is 90° so,

AC^2 = AD^2 + DC^2

=> AD^2 = AC^2 - DC^2

=>AD^2 = x^2 - ( x/2 )^2

=> AD^2 = x^2 - x^2/4

=> AD^2 = 3x^2 / 4

=> AD = Sq. root of 3x^2/4

$= \sqrt{ \frac{3 {x}^{2}}{4} } \\ \\ = \frac{ \sqrt{3}x }{2}$

Now, Area of a Triangle

= 1/2 × Base × Height

= 1/2 × BC × AD

=

$\frac{1}{2} \times x \times \frac{ \sqrt{3} x}{2} \\ \\ = \boxed{ \frac{ \sqrt{3} {x}^{2} }{4} }$

Hence Proved.

Answer 2

Step-by-step explanation:

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