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27 Oct 2017, 00:20
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D
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Difficulty:
15%
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Question Stats:
80% (01:37) correct
20%
(01:21)
wrong
based on 109
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In the figure, OR and PR are radii of circles. The length of OP is 4. If OR = 2, what is PR ? PR is tangent to the circle with center O.
(A) 2
(B) 5/2
(C) 3
(D) \(2 \sqrt{3}\)
(E) \(3 \sqrt{2}\)
Attachment:
2017-10-27_1118.png [ 9.82 KiB | Viewed 3672 times ]
27 Oct 2017, 02:33
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Tangent to the circle creates a right angle to the radius of the circle.
Hence angle ORP is 90.
By using Pythagoras theorem,
We have OR^2 +PR^2 = OP^2
This gives option D
Sent from my iPhone using GMAT Club Forum
Hence angle ORP is 90.
By using Pythagoras theorem,
We have OR^2 +PR^2 = OP^2
This gives option D
Sent from my iPhone using GMAT Club Forum
27 Oct 2017, 08:47
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Bunuel
In the figure, OR and PR are radii of circles. The length of OP is 4. If OR = 2, what is PR ? PR is tangent to the circle with center O.
(A) 2
(B) 5/2
(C) 3
(D) \(2 \sqrt{3}\)
(E) \(3 \sqrt{2}\)
Attachment:
2017-10-27_1118.png
We can infer: this is a 30-60-90 right triangle with side lengths in ratio
\(x: x\sqrt{3}: 2x\)
Rule: If one leg and a hypotenuse of a right triangle are in the ratio of a 30-60-90 triangle, it is a 30-60-90 triangle.*
OR = 2: corresponds with x
OP = 4: corresponds with 2x
PR therefore corresponds with \(x\sqrt{3}\), which = \(2\sqrt{3}\)
Answer D
**
By the Pythagorean theorem, this rule could not be otherwise, so if you recognize the relationship, you do not have to do this math:
Let OR = a
Let PR = b
Let OP = c
\(a^2 + b^2 = c^2\)
\(2^2 + b^2 = 4^2\)
\(b^2 = 16 - 4\)
\(b^2 = 12 = 2\sqrt{3} =\\
PR\)
