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AB and CD are chords of the circle, and E and F are the mid
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23 Mar 2019, 08:56
23 Mar 2019, 08:56
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44% (02:04) correct
55% (02:21) wrong based on 36 sessions
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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
Re: AB and CD are chords of the circle, and E and F are the mid
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11 Apr 2020, 11:01
11 Apr 2020, 11:01
Expert Reply
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.
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.
Re: AB and CD are chords of the circle, and E and F are the mid
[#permalink]
31 Oct 2020, 04:34
31 Oct 2020, 04:34
How do you know OAE and OCF is a right triangle? there is no clue.
