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## What is the perimeter of the triangle above?

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### What is the perimeter of the triangle above?

by Vincen » Thu Jul 30, 2020 8:33 am

00:00

A

B

C

D

E

## Global Stats 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

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### Re: What is the perimeter of the triangle above?

by terminator12 » Thu Jul 30, 2020 7:12 pm
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 + $$\sqrt{3}$$

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### Re: What is the perimeter of the triangle above?

by swerve » Fri Jul 31, 2020 4:37 am
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

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