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03 Apr 2020, 08:36
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
79% (02:58) correct
21%
(01:49)
wrong
based on 29
sessions
History
Date
Time
Result
Not Attempted Yet
In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is
A. √11
B. √12
C. √13
D. √14
E. √15
Are You Up For the Challenge: 700 Level Questions
A. √11
B. √12
C. √13
D. √14
E. √15
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 09:11
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Bunuel
Let OE = \(x\), so OF = \(x + 1\)
Triangles OBE & ODF are right angled
--> \(OB^2 = OE^2 + EB^2\)
--> \(r^2 = 3^2 + x^2\) ....... (1)
--> \(OD^2 = OD^2 + DF^2\)
--> \(r^2 = 2^2 + (x + 1)^2\) ....... (2)
From (1) & (2),
\(9 + x^2 = 4 + x^2 + 2x + 1\)
--> \(2x = 4\)
--> \(x = 2\)
Substitute in (1),
--> \(r^2 = 3^2 + x^2 = 9 + 2^2 = 13\)
--> \(r = \sqrt{13}\)
Option C
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03 Apr 2020, 15:34
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Let the 6 cm long chord be x cm away from the centre of the circle. Let the radius of the circle be r cm.
The perpendiculars from the centre of the circle to the chords bisect the chords.
r^2 = x^2+3^2 = (x+1)^2+2^2
solving, x=2 , r= √13
correct answer is C
The perpendiculars from the centre of the circle to the chords bisect the chords.
r^2 = x^2+3^2 = (x+1)^2+2^2
solving, x=2 , r= √13
correct answer is C
