Answer: Choice D
HG= 11 square root 3/3 and HI = 22 square root 3/3
In other words, [tex]\text{HG} = \frac{11\sqrt{3}}{3} \ \text{ and } \ \text{HI} = \frac{22\sqrt{3}}{3}\\\\[/tex]
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Explanation:
Let's say that x is the short leg and y is the long leg
For any 30-60-90 triangle, we have this connection: [tex]y = x\sqrt{3}[/tex]
The long leg y is exactly sqrt(3) times longer compared to the short leg x.
Let's solve for x and then plug in y = 11
[tex]y = x\sqrt{3}\\\\x = \frac{y}{\sqrt{3}}\\\\x = \frac{y*\sqrt{3}}{\sqrt{3}*\sqrt{3}}\\\\x = \frac{y\sqrt{3}}{3}\\\\x = \frac{11\sqrt{3}}{3}\\\\[/tex]
Side HG, the shorter leg, has an exact length of [tex]\text{HG} = \frac{11\sqrt{3}}{3}\\\\[/tex]
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Once we know the short leg, we double that expression to get the length of the hypotenuse. Like before, this only applies to 30-60-90 triangles.
[tex]\text{hypotenuse} = 2*(\text{short leg})\\\\\text{HI} = 2*\text{HG}\\\\\text{HI} = 2*\frac{11\sqrt{3}}{3}\\\\\text{HI} = \frac{22\sqrt{3}}{3}\\\\[/tex]
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Since [tex]\text{HG} = \frac{11\sqrt{3}}{3}\\\\[/tex] and [tex]\text{HI} = \frac{22\sqrt{3}}{3}\\\\[/tex], this shows that choice D is the final answer.
