Question

The height of an equilateral triangle is 4√3. What is the perimeter of the equilateral triangle?a) 12 unitsb) 12√3 unitsc) 24 unitsd) 24√3 units

Answer 1

Given :

• Height of an equilateral triangle = 4√3

To Find :

• Perimeter of the triangle

                $\setlength{\unitlength}{2cm}\begin{picture}(6,2)\put(9.19,2.7){\large{A}}\put(8,1){\large{B}}\put(10.4,1){\large{C}}\put(9.23,0.8){\large{D}}\put(9.25,1){\line(0,1){1.55}}\put(8.2,1){\line(1,0){2.1}}\put(8.2,1){\line(2,3){1.05}}\put(10.3,1){\line(-2,3){1.05}}\put(9.25,1.1){\line(3,0){0.1}}\put(9.35,1){\line(0,1){0.1}}\put(9.3,1.6){\large{\bf{4} \sqrt{\bf{3}} }}\put(9.89,1.7){\tt{x}}\put(9.7,1.15){ \small{\dfrac{\tt{x}}{\tt{2}}} }\put(8.5,1.7){\tt{x}}\put(9.2,0.6){\tt{x}}\end{picture}$

Let sides of the triangle be x

We know that :

• All sides of an equilateral triangle are equal

$\qquad \longrightarrow \, \sf{AB=BC=AC=x}$

Now,

Taking Δ ADC to find the value of x

• Hypotenus (AC) = x
• Perpendicular (AD) = 4√3
• Base (DC) = $\sf{\dfrac{x}{2} }$

$\bf{H^{2}=P^{2}+B^{2} } \\\\ \rightarrow \sf{x^{2} =(4\sqrt{3} )^{2} +\bigg(\dfrac{x}{2}\bigg)^{2} } \\\\ \rightarrow \sf{x^{2} =48+\dfrac{x^{2}}{4} } \\\\ \rightarrow \sf{x^{2} -\dfrac{x^{2}}{4}=48} \\\\ \rightarrow \sf{\dfrac{4x^{2}-x^{2}}{4}=48 }\\\\ \rightarrow \sf{4x^{2}-x^{2}=192}\\\\ \rightarrow \sf{3x^{2}=192}\\\\ \rightarrow \sf{x^{2}=\dfrac{192}{3} }\\\\ \rightarrow \sf{x^{2}=64}\\\\ \rightarrow \sf{x=\sqrt{64} }\\\\\rightarrow\fbox{ \sf{x=8 }}$

Side of the triangle = 8 unit

We know that :

Perimeter of an equilateral triangle = 3 × side

→ 3 × 8 = 24

(c) Perimeter of the equilateral triangle is 24 units

Answer 2

$\huge\boxed{\fcolorbox{violet}{violet}{Answer}}$