What is the perimeter of the triangle above?

tan \({30^o}\) = \(\frac{1}{\sqrt{3}}\) = \(\frac{1}{base}\)

Hence, base = \(\sqrt{3}\)

Applying pythagoras theorem,
hypoteneuse = \(\sqrt{1\ +\ \left(\sqrt{3}\right)^2}\) = 2

Hence, perimeter is 3 + \(\sqrt{3}\)

That is answer D

Vincen wrote: ↑

Thu Jul 30, 2020 8:33 am

2019-04-30_1128.png

What is the perimeter of the triangle above?

A. \(\dfrac{3+\sqrt3}2\)

B. \(3\)

C. \(2+\sqrt2\)

D. \(3+\sqrt3\)

E. \(3+\sqrt5\)

[spoiler]OA=D[/spoiler]

Source: Princeton Review

First, we can find the hypotenuse of the triangle as follows,
\(\sin{30^{\circ}} = \dfrac{1}{h}\)
\(\dfrac{1}{2}=\dfrac{1}{h} \Longrightarrow h=2\)

Next, we can find the base as follows,
\(\cos{30^{\circ}}=\dfrac{b}{2}\)
\(\dfrac{\sqrt{3}}{2}=\dfrac{b}{2} \Longrightarrow b=\sqrt{3}\)

Finally, the perimeter will be
\(P=2+1+\sqrt{3}=3+\sqrt{3}\)

Therefore, D