Some Derivation

Derivation of Area of Equilateral Triangle

June 19, 2021
edusingh

Derivation of Area of Equilateral Triangle :-

Derivation of Area of Equilateral Triangle

            There are three methods to Derivation of Area of Equilateral Triangle :-

 

 

 

 

 

 

  • Using Heron Formula :- Mainly the Heron formula is used when all the three sides of a triangle are known.

We have a equilateral  ΔABC 

Derivation of Area of Equilateral Triangle

 

We know that equilateral have equal sides 

∴ AB=BC=AC= a      (we let that)

Now,

Area of Triangle = \fn_cm \sqrt{s(s-a)(s-b)(s-c)}

  and   s  =\fn_cm \frac{a+b+c}{2}

 

 

Since, all sides are equal so  a=b=c

Area of Triangle = \fn_cm \sqrt{s(s-a)(s-a)(s-a)}

 

    and s =\fn_cm \frac{a+a+a}{2} = \fn_cm \frac{3a}{2}

Now we put the value of ‘S’

Area of Triangle = \fn_cm \sqrt{\frac{3a}{2}\left ( \frac{3a}{2}-a \right )\left ( \frac{3a}{2}-a \right )\left ( \frac{3a}{2}-a \right )}

Area of Triangle  = \fn_cm \sqrt{\frac{3a}{2}\left ( \frac{3a-2a}{2} \right )\left ( \frac{3a-2}{2} \right )\left ( \frac{3a-2a}{2} \right )}

 

Area of Triangle  = \fn_cm \sqrt{\frac{3a}{2}\left ( \frac{a}{2} \right )\left ( \frac{a}{2} \right )\left ( \frac{a}{2} \right )}       (we can write it  \fn_cm \sqrt{\frac{3a}{2}}= \sqrt{3\times \frac{a}{2}}  like this )

Area of Triangle =  \fn_cm \sqrt{3\times \frac{a}{2}\times \frac{a}{2}\times \frac{a}{2}\times \frac{a}{2}}       (Now make pairs of them and take them out of the root)

Area of Triangle =  \fn_cm \sqrt{3}\times \frac{a}{2}\times \frac{a}{2}

 we proved that Area of equilateral Triangle = \fn_cm \frac{\sqrt{3}}{4}a^{2}

Derivation of Area of Equilateral Triangle

(First, Derivation of Area of Equilateral Triangle is complete)

Second Way to Find Area of equilateral Triangle

Using Area of right angle triangle :- 

We know that

Area of right angle triangle  =   \fn_cm \frac{1}{2} × base × height =  \fn_cm \frac{1}{2} × b × h     …….      (i)

Since, The perpendicular drawn from vertex of the equilateral triangle to the opposite side bisects it into equal halves

∴  AD = CD = \fn_cm \frac{a}{2}

In ΔADB , ∠D = 90   (so, We can use Pythagoras Theorem)

“In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“

AB2 = AD2 + BD2

a2= BD2 + \fn_cm \left ( \frac{a}{2} \right )^{2}         

BD2=  \fn_cm a^{2}-\frac{a^{2}}{4}

BD2= \fn_cm \frac{4a^{2}-a^{2}}{4}  = \fn_cm \frac{3a^{2}}{4}

BD =\fn_cm \sqrt{\frac{3a^{2}}{4}}   = \fn_cm \frac{\sqrt{3} a}{2}

and now

Area of right angle triangle = \fn_cm \frac{1}{2} × base × height

(base = AC = a  and Height = BD = \fn_cm \frac{\sqrt{3} a}{2}

Area of right angle triangle =  \fn_cm \frac{1}{2} × a × \fn_cm \frac{\sqrt{3} a}{2}  

Now, we proved that Second time  Area of equilateral Triangle = \fn_cm \frac{\sqrt{3}}{4}a^{2}

(Second, Derivation of Area of Equilateral Triangle are complete)

Now, Third Way to Find Area of equilateral Triangle

Area of right angle triangle  =   \fn_cm \frac{1}{2} × base × height =  \fn_cm \frac{1}{2} × b ×h     …….      (i)

In ΔADB , ∠D = 90 

We know 

Sin A  = \fn_cm \frac{BD}{AB} = \fn_cm \frac{P}{H}          and         Cos A = \fn_cm \frac{AD}{AB} = \fn_cm \frac{B}{H}

Since, each angle of equilateral triangle are 600

So,  ∠ BAD = 600

Sin 600 =\fn_cm \frac{P}{H}  =  \fn_cm \frac{P}{a}                       As Sin 60 = \fn_cm \frac{\sqrt{3}}{2}

Than \fn_cm \frac{\sqrt{3}}{2}  =  \fn_cm \frac{P}{a}             

After cross multiply we get             P   =  \fn_cm \frac{\sqrt{3}a}{2}

Hence,    

Area of right angle triangle = \fn_cm \frac{1}{2} × base × height  = \fn_cm \frac{1}{2} × B × P

Area of right angle triangle = \fn_cm \frac{1}{2} × a × \fn_cm \frac{\sqrt{3}a}{2}   

Thus we proved that third time Area of equilateral Triangle = \fn_cm \frac{\sqrt{3}}{4}a^{2}

(Third, Derivation of Area of Equilateral Triangle are complete)

“हम सीखेंगे, समझेंगे,और करेंगे खेल खेल मे”

 

 

 

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