What is the perimeter of an equilateral triangle whose height is 2(radical 3)?

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Answer 1 · 2016-04-28T19:21:05

Socratic Formatting for radical is : hashsymbol sqrt(3) hashsymbol giving: $\sqrt{3}$ $sqrt(3)$ . Look at https://socratic.org/help/symbols.

Perimeter = 4

Let each triangle side be of length $x$ $x$

Let height be $h$ $h$

Then, by using Pythagoras

$h^{2} + {(\frac{x}{2})}^{2} = x^{2}$ $h^2+(x/2)^2=x^2$

subtract ${(\frac{x}{2})}^{2}$ $(x/2)^2$ from both sides

$h^{2} = x^{2} - {(\frac{x}{2})}^{2}$ $h^2=x^2-(x/2)^2$

$h^{2} = \frac{4 x^{2}}{4} - \frac{x^{2}}{4}$ $h^2=(4x^2)/4-x^2/4$

$h^{2} = \frac{3}{4} x^{2}$ $h^2=3/4x^2$

Multiply both sides by $\frac{4}{3}$ $4/3$

$\frac{4}{3} h^{2} = x^{2}$ $4/3 h^2=x^2$

Square root both sides

$x = \frac{2 h}{\sqrt{3}}$ $x=(2h)/sqrt(3)$

Mathematicians do not like the denominator to be a radical

Multiply the right by 1 but in the form of $1 = \frac{\sqrt{3}}{\sqrt{3}}$ $1=sqrt(3)/(sqrt(3)$

$x = \frac{2 h \sqrt{3}}{3}$ $x=(2hsqrt(3))/3$

But $h = 2 \sqrt{3}$ $h=2sqrt(3)$ so by substitution for $h$ $h$

$x = \frac{2 (2 \sqrt{3}) \sqrt{3}}{3}$ $x=(2(2sqrt(3))sqrt(3))/3$

$x = \frac{12}{3} = 4$ $x=12/3=4$

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Triangle has 3 sides and each side is 4

Perimeter is $3 \times 4 = 12$ $3xx4=12$