Question

The area of an equilateral triangle having altitude of 3 cm is: $x^{2} cm^{2} \\3\sqrt{3} cm^{2} \\\sqrt{3 } } cm^{2} \\\\7\sqrt{3} cm^{2}$

Answer 1

◘ Given ◘

An equilateral triangle ABC with altitude 3 cm.

◘ To Find ◘

The area of the equilateral triangle ABC.

◘ Solution ◘

$\bigstar \underline{\underline{\sf{DIAGRAM:-}}}$

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We know,

♣ Altitude of an equilateral triangle,

$\sf{h=\dfrac{1}{2}\times \sqrt{3} \times a }$

$\sf{\longrightarrow 3=\dfrac{1}{2} \times \sqrt{3} \times a}\\\\\sf{\longrightarrow 6=\sqrt{3}a}\\\\\sf{\longrightarrow a=\dfrac{6}{\sqrt{3} } }\\\\\sf{\longrightarrow a=\dfrac{2 \times 3}{\sqrt{3}} }\\\\\sf{\longrightarrow a=2\sqrt{3}}$

$\rule{100}2$

Now we know for an equilateral triangle :-

$\sf{Area,\ A = \dfrac{\sqrt{3}}{4}a^2 }\\\\\sf{\longrightarrow A=\dfrac{\sqrt{3} }{4} (2\sqrt{3} )^2}\\\\\sf{\longrightarrow A=\dfrac{\sqrt{3}}{4}\times12 }\\\\\boxed{\sf{\longrightarrow A=3\sqrt{3}\ cm^2 }}$

Therefore, the area of the equilateral triangle with altitude of 3 cm is 3√3 cm².

Answer 2

Answer

• Option 2 is correct
• 3√3 cm²

Given

• Altitude of an equilateral triangle is 3 cm

To Find

• Area of an equilateral triangle

Concept Used

$\boxed{\begin{minipage}{7cm} \bf \bigstar \;\; Pythagoras\ Theorem \\\\\bf \bigstar \;\; (Opp.side)^2+(Adj.side)^2=(Hyp.)^2\\\\\bf \bigstar \;\; Area\ of\ triangle\\\\\bf \bigstar \;\; \dfrac{1}{2}\times base \times height \end{minipage}}$

Solution

Altitude of an equilateral triangle , h = 3 cm

Actually , Area of an equilateral triangle is ,

$\bigstar \;\; \bf A=\dfrac{\sqrt{3}}{4}x^2$

where , x is side of an equilateral triangle .

But here side of equilateral triangle is not given .

So , convert the area of an equilateral triangle interms of altitude of the triangle .

Find attachment for diagram .

Apply Pythgoras theorem for right angled triangle .

$\rm \implies h^2+\bigg(\dfrac{x}{2}\bigg)^2=x^2\\\\\implies \rm h^2=x^2-\dfrac{x^2}{4}\\\\\implies \rm h^2=\dfrac{3x^2}{4}\\\\\implies \rm h=\dfrac{\sqrt{3}x}{2}\\\\\implies \rm x=\dfrac{2h}{\sqrt{3}}$

Now , apply area of normal triangle .

$\implies \rm Area=\dfrac{1}{2}\times Base \times Height\\\\\implies \rm Area=\dfrac{1}{2}\times x\times h\\\\\implies \rm Area=\dfrac{1}{2}\times \dfrac{2h}{\sqrt{3}}\times h\;\;[Since,we\ find\ above]\\\\\implies \bf Area=\dfrac{h^2}{\sqrt{3}}\ \;\;\bigstar$

Now sub. value of altitude , h in the triangle . [ h = 3 cm ]$\implies \rm Area=\dfrac{(3)^2}{\sqrt{3}}\\\\\implies \rm Area=\dfrac{9}{\sqrt{3}}\\\\\implies \rm Area=\dfrac{3\times 3}{\sqrt{3}}\\\\\implies \rm Area=\dfrac{3 \times \sqrt{3}\times \cancel{\sqrt{3}}}{\cancel{\sqrt{3}}}\\\\\implies \bf Area=3\sqrt{3}\ cm^2\ \; \bigstar$

( Or ) [ Easy method ]

Find attachment

Apply trigonometric function sine .

$\implies \rm sin\ 60=\dfrac{h}{x}\\\\\implies \rm \dfrac{\sqrt{3}}{2}=\dfrac{3}{x}\ [Given\ h=3]\\\\\implies \rm x=2\sqrt{3}$

Now , Apply formula for equilateral triangle .

$\implies \rm Area=\dfrac{\sqrt{3}}{4}x^2\\\\\implies \rm Area=\dfrac{\sqrt{3}}{4}(2\sqrt{3})^2\\\\\implies \rm Area=\dfrac{\sqrt{3}}{4}\times 12\\\\\implies \bf Area=3\sqrt{3} \;\; \bigstar$