Question

USING Pythagoras theorem find the formula of equilateral triangle.​

Answer 1

Answer:

Answer with explanation is below~

Step-by-step explanation:

We could easily obtain the formulae/formula of an equilateral triangle not only through Pythagoras Theorem, but also through Heron's formula.

INTRODUCTION:

What's equilateral triangle?

• It's an triangle, whose all sides're of equal measurement.

What's Heron Formula?

• It's just a formula actually,take an example:Let,a,b and c denotes the lengths of 3 sides of any triangle,then the area will be given as:

$\boxed{\rm \: Area= \sqrt{s(s - a)(s - b)(s - c)} \: units ^{2}}$

Where,

• s = (a+b+c)/2 {Half of the perimeter, basically}

What's Pythagoras' Theorem?

• It's actually like a formula but a theorem introduced by Pythagoras.

SOLVING:

Let,ABC an equilateral triangle of sides a.

Now:Draw a perpendicular straight line AM to the side BC(Name each part of triangle)

So it's clear that ∆AMB is a right angled triangle at M, BM = (1/2)BC = a/2.

Please note AM here represents the height of ∆ ABC.

• Let's use Pythagoras' theorem now.

$\boxed{\rm \: AM = \sqrt{AB^2-BM^2}}$

• AB = a
• BM = a/2

$\rm \: AM = \sqrt{a {}^{2} - { \bigg( \cfrac{a}{2} \bigg) }^{2} }$

$\rm \: AM = 3 \: \cfrac{a {}^{2} }{4}$

$\rm \: AM = \cfrac{ \sqrt{3} }{2} \: a$

Now find the area of ∆ABC:

$\rm \triangle \: ABC = \cfrac{1}{2} \times \: BC \times A$

$\rm \: \triangle \: ABC = \cfrac{1}{2} \times a \times \cfrac{ \sqrt{3}}{4}a$

$\boxed{\rm\triangle \: ABC = \cfrac{ \sqrt{3 } }{4} a {}^{2} \: units {}^{2}}$

Hence,the formulae of equilateral triangle using Pythagoras' theorem is {√(3)/4} a^2

$\rule{225pt}{2pt}$

Extras:

Now let's find the area using Heron's Formula.

Solving:

Let each side of an equilateral triangle be a.

SO, then:

s = (3a/2)

We know that, (Heron's formula)

$\rm \: A = \sqrt{s(s - a)(s - b)(s - c)}$

Now the area A :

$\rm AR(A)= \sqrt{ \cfrac{ 3a } 2 \bigg(\cfrac{ 3a }{2} - a \bigg)\bigg(\cfrac{ 3a }{2} - a \bigg)\bigg(\cfrac{ 3a }{2} - a \bigg)}$

$\rm AR(A)= \sqrt{ \cfrac{3a}{2} \bigg( \cfrac{a}{2} \bigg)\bigg( \cfrac{a}{2} \bigg)\bigg( \cfrac{a}{2} \bigg)}$

$\rm \boxed{ \rm AR(A)= \cfrac{ \sqrt{ 3a}}{4} \: a {}^{2} \: \: units {}^{2}}$

And voila! we're done!

I hope this helps! :)

[Figure of Equilateral triangle is attached, it denotes the triangle we need to draw while finding the formulae through Pythagoras' theorem. ]