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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What is the perimeter of the triangle above?
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Vincen
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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
A. \(\dfrac{3+\sqrt3}2\)
B. \(3\)
C. \(2+\sqrt2\)
D. \(3+\sqrt3\)
E. \(3+\sqrt5\)
[spoiler]OA=D[/spoiler]
Source: Princeton Review
-
terminator12
- Junior | Next Rank: 30 Posts
- Posts: 13
- Joined: Tue Jul 28, 2020 9:01 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 + \(\sqrt{3}\)
That is answer D
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
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
