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C
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AB and CD are chords of the circle, and E and F are the midpoints of the chords, respectively. The line EF passes through the center O of the circle. If EF = 17, then what is radius of the circle?
(A) 10
(B) 12
(C) 13
(D) 15
(E) 25
Attachment:
#GREpracticequestion AB and CD are chords of the circle.jpg [ 17.99 KiB | Viewed 7386 times ]
03 Apr 2019, 11:46
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From the figure
AE = BE = 5
CF = FD = 12
EF = 17
OE = x, Then OF = 17-x
OAE is a right angle triangle, Apply Pythagorean theorem,
OA = \(\sqrt{25+x^2}\) = radius
Similarly, OCF is a right angle triangle, Apply Pythagorean theorem,
OC = \(\sqrt{(17-x)^2+144}\) = radius
Equating both equations as both are the radii, we get
\(25+x^2 = 289+x^2-34x+144\)
34x = 408
x = 12 = OE, then OF = 5
OA = OC = \(\sqrt{25+144}\) = 13
C is the answer.
AE = BE = 5
CF = FD = 12
EF = 17
OE = x, Then OF = 17-x
OAE is a right angle triangle, Apply Pythagorean theorem,
OA = \(\sqrt{25+x^2}\) = radius
Similarly, OCF is a right angle triangle, Apply Pythagorean theorem,
OC = \(\sqrt{(17-x)^2+144}\) = radius
Equating both equations as both are the radii, we get
\(25+x^2 = 289+x^2-34x+144\)
34x = 408
x = 12 = OE, then OF = 5
OA = OC = \(\sqrt{25+144}\) = 13
C is the answer.
Bunuel
AB and CD are chords of the circle, and E and F are the midpoints of the chords, respectively. The line EF passes through the center O of the circle. If EF = 17, then what is radius of the circle?
(A) 10
(B) 12
(C) 13
(D) 15
(E) 25
Attachment:
#GREpracticequestion AB and CD are chords of the circle.jpg
The hypotenuses of AEO and OCF must be the same (the radius) \(AO = OC\). \(EF = 17\); one can quickly see that \(OE=12\) and \(OF = 5\).
Pythagoras Theorem: \(25 + 144 = 169\) --> Answer: C.
Realizing that \(AO = OC\) and that the specific values for OE and OF can be determined quickly without any math saves a lot of time on this question.
