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Re: In the figure above, PQ is a diameter of circle O, PR = SQ, [#permalink]
Hi Brent,just 1 question. If we know this is equilateral triangle means all sides and angles are equal, and we got one side also that is x=1 so cant we just compare 1=2x ? why do we have to compare 1=x root3 if sides are equal?

GreenlightTestPrep wrote:
Carcass wrote:
Image
In the figure above, PQ is a diameter of circle O, PR = SQ, and ΔRST is equilateral. If the length of PQ is 2, what is the length of RT ?

A. \(\frac{1}{2}\)

B. \(\frac{1}{\sqrt{3}}\)

C. \(\frac{\sqrt{3}}{2}\)

D. \(\frac{2}{\sqrt{3}}\)

E. \(\sqrt{3}\)

GIVEN: The length of PQ is 2
In other words, the DIAMETER = 2
From this, we can conclude that the RADIUS = 1
So, we can add this information to our diagram:
Image


Since ΔRST is equilateral, we know the altitude (TO) will be a perpendicular bisector of side RS
Also, since ΔRST is equilateral, we know that ∠ORT = 60°, which also means ∠RTO = 30°
Image

At this point, we can see that ΔTRO is a special 30-60-90 triangle.
If we compare ΔTRO with the BASE 30-60-90 triangle, we can create an equation by comparing corresponding sides

We can write: x/2 = 1/(√3)

Multiply both sides by 2 to get: x = 2/(√3)

Answer: D
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Re: In the figure above, PQ is a diameter of circle O, PR = SQ, [#permalink]
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