Question

↓↓↓ Maths Question ↓↓↓

$prove \: that \: the \: area \: of \: an \\ equilateral \: triangle \: is \: equal \\ to \: \frac{ \sqrt{3} }{4} a ^{2} ..where \: a \: is \: the \: side \: of \\ the \: triangle$
× No Spam ×​

Answer 1

$\huge\mathfrak\purple{Bonjour!!}$

$\huge\mathcal{Solution!}$

♨ Given:-

An equilateral triangle ABC such that AB=BC=CA= a.

♨ To prove:-

ar (∆ABC) = √3/4 a²

♨ Construction:-

Draw AD perpendicular to BC.

♨ Proof:-

In ∆s ABD and ACD, we have,

AB=AC [Since ∆ABC is an equilateral]

angle ADB = angle ADC [Each equal to 90°]

and,

AD=AD [Common sides]

Therefore,

By RHS criterion of congruence, we have

∆ABD is congruent to ACD.

=> BD=DC

But,

BD + DC = a

=> BD= DC = a/2

Now, in right triangle ABD, we have

AB² = AD² + BD² [Using Pythagoras Theorem]

=> a² = AD² + [a/2]²

=>AD² = a² - a²/4 => AD² = 3a²/4

=>AD= √3a/2

Therefore,

ar (∆ABC) = 1/2 (BC × AD)

= 1/2 [a × √3/2 a]

= √3/4 a²

[Hence, Proved! ☃]

✔✔✔✔✔✔✔✔✔

Hope it helps...❣❣❣

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Be Brainly...✌✌✌

♣ WALKER ♠

Answer 2

Question :----- we have to prove area of equaliteral ∆ is ¼(√3a²) .....

Formula required :-----

• Area of ∆ = 1/2 * Base * Height
• pythagoras theoram = P²+B² = H²

Solution :::----( Refer to diagram also)

Area of Triangle = ½ × base × height

Here, base = a, and height = h

Now, apply Pythagoras Theorem in the triangle.

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

Now, put the value of “h” in the area of the triangle equation.

So,

Area of Triangle = ½ × base × height

⇒ A = ½ × a × ½(√3a)

Or,

Area of Equilateral Triangle = ¼(√3a²)

(Proved)