Math, asked by MsQueen, 11 months ago

Derivation. of area of equilateral triangle ?

Answers

Answered by BrainlyQueen01
17
Derivation of Area of an equilateral triangle ;

Let ABC be an equilateral triangle with sides 'a'. Now, draw AD perpendicular to BC.

Here, we have ΔABD = ΔADC.

We will find area of ΔABD using pythagorean theorem, according to which, the square of hypotenuse is equal to the sum of the squares of the other two sides.

Here, we have ;

 \sf a {}^{2}  = h {}^{2}  + ( \frac{a}{2} ) {}^{2}  \\  \\ \sf h {}^{2}  = a {}^{2}  - ( \frac{a}{2} ) {}^{2}  \\  \\ \sf h {}^{2}  =  \frac{3a {}^{2} }{4}  \\  \\  \sf h =  \sqrt{ \frac{3a {}^{2} }{4} }  \\  \\  \sf h =  \frac{ \sqrt{3} }{2} a



Now, we get the height ;

 \sf area \: of \Delta =  \frac{1}{2}  \times base \times height \\  \\  \sf area \: of \Delta =  \frac{1}{2}  \times a \times  \frac{ \sqrt{3} }{2} a \\  \\   \sf area \: of \Delta =   \frac{ \sqrt{3} }{4} a {}^{2}



Hence, area of equilateral triangle is

 =  \sf  \frac{ \sqrt{3} }{2} a {}^{2}

rakeshmohata: great answer musi !!❣❣
BrainlyQueen01: Theku ^_^
Answered by AJThe123456
4
Heyy mate ❤✌✌✔

Here's your Answer...

Proof;

In triangle ABD,

BY PYTHAGORAS THEOREM

AB^2 = AD^2 + BD^2

=> a^2 = h^2 + (a/2)^2

=> h^2 = a^2 - a^2/4

=> h^2 = 3a^2/4

=> h = root 3a^2/ root 4

=> h = root 3a^2 /2-------(1)

Now, we know that

Area = 1/2 × b × h

putting eq 1 here, we get

=> 1/2 × a × root 3a^2/2

=> 1/2 × a × root 3 a/2

=> Area = root 3a^2/4
✔✔✔
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