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In the figure above, PQ is a diameter of circle O, PR = SQ,
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12 Jun 2020, 10:51
12 Jun 2020, 10:51
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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. \(\sqrt{3}\)
Re: In the figure above, PQ is a diameter of circle O, PR = SQ,
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12 Jun 2020, 10:51
12 Jun 2020, 10:51
Expert Reply
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Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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Re: In the figure above, PQ is a diameter of circle O, PR = SQ,
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12 Jun 2020, 12:32
12 Jun 2020, 12:32
2
Carcass wrote:
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:
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°
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
